in the figure calculates the acceleration of the block friction not today

In The Figure Calculates The Acceleration Of The Block Friction Not Today

Answers

Answer 1

Answer:

A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.

The hot air displacing the cold air is an example of  transfer by

Explanation:

also the answer is hit my dm on ig


Related Questions

Which of these parameters is directly related to sound frequency?

Answers

Answer:Velocity

Explanation:

Velocity is directly proportional to the frequency of a wave.

Velocity=frequency x wavelength

Q1. What is the frequency of rotation of 1000 loop coil of area 20cm2 in a magnetic field of 5T to

generate an emf that has a maximum value of 15.7V?​

Answers

Answer:

Explanation:

Emf e generated in a coil with no of turn n and area A rotating in a magnetic field B  with angular speed of ω is given by the expression

e = e₀ sinωt

where e₀ = nωAB which is the maximum emf generated

Putting the given values

15.7 = 1000xω x 20 x 10⁻² x 5

ω = .0157

frequency of rotation

= ω / 2π

= .0157 / 2 x 3.14

= .0025 /s

9 rotation / hour .

Match these items.


1 . pls help


asteroids

between Mars and Jupiter

2 .

fission

ice, dust, frozen gases

3 .

energy

sun's atmosphere

4 .

fusion

ability to do work

5 .

corona

splitting atoms

6 .

comets

the combining of atomic nuclei to form one nucleus

Answers

Answer:

Here's your answer :

Asteroids - Between mars and JupiterFission - splitting atomsEnergy - Sun's atmosphereFusion - The combining of atomic nuclei to form one nucleusCorona - Ability to do workComets - Ice, dust, frozen gases

hope it helps!

A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming to rest. what was the deceleration experienced by the survivor? Use g = 9.8 m/s2 Calculate to one decimal.

Answers

Answer:

The deceleration is  [tex]a = - 76.27 m/s^2[/tex]

Explanation:

From the question we are told that

   The height above  firefighter safety net is [tex]H = 14 \ m[/tex]

   The length by which the net is stretched is [tex]s = 1.8 \ m[/tex]

   

From the law of energy conservation

    [tex]KE_T + PE_T = KE_B + PE_B[/tex]

 Where [tex]KE_T[/tex] is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

   and  [tex]PE_T[/tex] is the potential energy of the before jumping  which is mathematically represented at

          [tex]PE_T = mg H[/tex]

and  [tex]KE_B[/tex] is the kinetic energy of the person just before landing on the safety net  which is mathematically represented at

        [tex]KE_B = \frac{1}{2} m v^2[/tex]

and  [tex]PE_B[/tex] is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

   So the above equation becomes

          [tex]mgH = \frac{1}{2} m v^2[/tex]

=>           [tex]v = \sqrt{2 gH }[/tex]

    substituting values

                [tex]v = 16.57 m/s[/tex]

Applying the equation o motion

             [tex]v_f = v + 2 a s[/tex]

Now the final velocity is zero because the person comes to rest

      So

         [tex]0 = 16.57 + 2 * a * 1.8[/tex]

            [tex]a = - \frac{16.57^2 }{2 * 1.8}[/tex]

            [tex]a = - 76.27 m/s^2[/tex]

         

         

A disk with a rotational inertia of 2.5 kg-m2 and a radius 1.1 m rotates on a frictionless fixed axis perpendicular to the disk faces and through its center. A force of 7.7 N is applied tangentially to the rim. The angular acceleration of the disk is _____ rad/s2. Round your answer to the nearest tenth.

Answers

Answer:

3.4 rad/sec^2

Explanation:

rotational inertia = 2.5 kg-m^2   radius = 1.1 m   force = 7.7 N

t = rotational inertia * angular acceleration    equation 1

also t = force * radius

therefore to calculate angular acceleration equation 1 becomes

f * r = inertia * angular acceleration   hence

angular acceleration = f * r / inertia = [tex]\frac{7.7 * 1.1 }{2.5}[/tex]   8.47 / 2.5 = 3.388 ≈ 3.4 rad/sec^2

Materials that allow electricity to pass through them are called ?

Answers

Answer:

Materials that allow electricity to pass through them are called conductors.

Copper wire is a good conductor. Materials that do not allow electricity to pass through them are called insulators.

Explanation:

Can I have brainliest?

(20) A rocket is launched vertically. At time t = 0 seconds, the rocket’s engine shuts down. At the time, the rocket has reached an altitude of 500m and is rising at a velocity of 125 m/s. Gravity then takes over. The height of the rocket as a function of time is h(t)=-9.8/2 t^2+125t+500,t>0. Using your function file from HW2A: Generate a plot of height (vertical axis) vs. time (horizontal axis) from 0 to 30 seconds. Include proper axis labels. Find the maximum height and the time at which it occurs: Analytically, showing your steps and equations. This part should be done entirely in the write-up: no coding Using the data cursor on the plot. Using the MAX function on your data from part (a) Using FMINSEARCH on your m file Comment on the differences between the methods. How closely does each method match the "true" (analytical) value? Find the time when the rocket hits the ground: Analytically, showing your equations. This part should be done entirely in the write-up: no coding Using the data cursor on the plot. Using FZERO on your m file Comment on the differences between the methods in each of part (B) and (C). How closely does each method match the "true" (analytical) value? Use a quantitative comparison to make your argument.

Answers

Answer:

Explanation:

Given that,

h(t) = -9.8t² / 2 + 125t + 500

for t > 0.

At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.

We want to find the maximum height reached by rocket

Using mathematics maxima and minima

let find the turning point when dh/dt = 0

dh/dt = -9.8t + 125

dh / dt = 0 = -9.8t + 125

9.8t = 125

t = 125 / 9.8

t = 12.76s

Let find the turning point to know if this time t = 12.76 is maximum or minimum point

Let find d²h / dt²

d²h / dt² = -9.8

Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.

Then, the maximum height reached is

h =  -9.8t² / 2 + 125t + 500

h =  -9.8(12.76)² / 2 + 125(12.76) + 500

h = -797.80 + 1595 + 500

h = 1297.2 m

The maximum height reached is 1297.2 m

From the attachment, the maximum height is 1297.2m at t = 12.76sec.

Comment, the result are the same for both the analysis aspect and the graphical aspect.

Distributions of electric charges in a cell play a role in moving ions into and out of a cell. In this situation, the motion of the ion is affected by two forces: the electric force due to the non-uniform charge distribution in the cell membrane, and the resistive force (viscosity) due to colliding with the fluid molecules. In order to begin our analysis of this, let's consider a toy model in which the ion is moving in response to electric forces alone.

Charges in a cell membrane are distributed along the opposite sides of the membrane approximately uniformly. This leads to an (on the average) constant electric field inside the membrane. A simple model that gives this kind of field is two large parallel plates close together. The field between the plates is approximately constant pointing from the negative to the parallel plate. This results in a charge feeling a constant force anywhere between the plates (sort of like flat-earth gravity turned sideways). Outside of the plates the electric fields from the two plates cancel and there is no force.

2. The electric field between the plates (inside the membrane) is about 107 N/C and the thickness of the membrane is about 7 nm. Estimate:

2.1 The electric force on the ion when it is in the center of the channel.
F = N

Explain your reasoning.



2.2 The acceleration of the ion when it is in the center of the channel.
a = nm/s2
Explain your reasoning.



2.3 The magnitude of the change in the ion's potential energy as it crosses from one side of the plates to the other.
U = J
Explain your reasoning.



2.4 The kinetic energy the ion would gain as it crosses from one side of the plates to the other.
KE = J
Explain your reasoning.

Could you explain 2.3!

Answers

Answer:

An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element. Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers.Explanation:

An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element.

What is atom?

Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers.

Each atom is made up of a nucleus and one or more electrons that are linked to it. One or more protons and a significant number of neutrons make up the nucleus. Only the most prevalent type of hydrogen is neutron-free.

Atoms that are neutral or ionized make up every solid, liquid, gas, and form of plasma. Atoms are incredibly tiny, measuring typically 100 picometers across. The nucleus of an atom contains more than 99.94% of its mass.

Therefore, An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element.

To learn more about atom, refer to the link:

https://brainly.com/question/1566330

#SPJ2

The throwing back by a wall or barrier of a sound wave without absorbing
it. *
1 point

Answers

Answer:Reflection

Explanation:

The throwing back of a sound wave without absorbing it is called reflection

In acoustic reflection of sound is termed as echo i.e. sound arrived at the listener after a particular delay depending upon the position of barrier to the observer.

The reflection of sound is used in many devices like megaphone, trumpets, etc. It is also used in auditorium such that the ceiling of the auditorium is curved for multiple reflections of sound so that sound can be reached at every corner of the auditorium.

Pendulum clock. Your friend is trying to construct a clock for a craft show and asks you for some advice. She has decided to construct the clock with a pendulum. The pendulum will be a very thin, very light wooden bar with a thin, but heavy, brass ring fastened to one end. The length of the rod is 80 cm and the diameter of the ring is 10 cm. She is planning to drill a hole in the bar to place the axis of rotation 15 cm from one end. She wants you to tell her the period of this pendulum.

Answers

Answer:

The time period for this pendulum is 1.68 seconds

Explanation:

Solution

Given that:

The length of the pendulum is measured from the axis of rotation to the center of mass of the bob of the pendulum

Now,

In this case, the length becomes:

L= 80 - 15+5

L = 70 cm

The time period = T = 2π √L/g

T = 2* 3.14 *√0.7/9.8

= 1.68 seconds

Note: Kindly find an attached work to the part of the solution of the given question

The uniform slender bar of mass m and length l is released from rest in the vertical position and pivots on its square end about the corner at O. (a) If the bar is observed to slip when   30 , find the coefficient of static friction s between the bar and the corner. (b)If the end of the bar is notched so that it cannot slip, find the angle  at which contact between the bar and the corner ceases.

Answers

Answer:

A) 0.188

B) 53.1 ⁰

Explanation:

taking moment about 0

∑ Mo = Lo∝

mg 1/2 sin∅ = 1/3 m L^2∝

note ∝ = w[tex]\frac{dw}{d}[/tex]∅

forces acting along t-direction ( ASSUMED t direction)

∑ Ft = Ma(t) = mr∝

mg sin ∅ - F = m* 1/2 * 3g/2l sin∅

therefore F = mg/4 sin∅

forces acting along n - direction ( ASSUMED n direction)

∑ Fn = ma(n) = mr([tex]w^{2}[/tex])

= mg cos∅ - N = m*1/2*3g/1 ( 1 - cos∅ )

hence N = mg/2 ( 5cos∅ -3 )

A ) Angle given = 30⁰c find coefficient of static friction

∪ = F/N

  = [tex]\frac{\frac{mg}{4}sin30 }{\frac{mg}{2}(5cos30 -3) }[/tex]  = 0.188

B) when there is no slip

N = O

   = 5 cos ∅ -3 =0

   therefore cos ∅ = 3/5  hence ∅ = 53.1⁰

Dogs keep themselves cool by panting, rapidly breathing air in and out. Panting results in evaporation from moist tissues of the airway and lungs, which cools the animal. Measurements show that, on a 35∘C day with a relative humidity of 50%, a 12 kg dog loses 1.0 g of water per minute if it is panting vigorously. What rate of heat loss, in watts, does this achieve?

Answers

The rate of heat loss, in watts, does this achieve is 37.66 W

Evaporation:

It leads in cooling since water absorbs heat equivalent to mass times latent heat of evaporation to get converted into vapor .

So,

latent heat of evaporation of water = 2260 x 10³ J / kg or 2260 J / g

Now

in the evaporation of 1 g of water , heat lost = 2260 J

And,

heat lost per minute = 2260 J

So,

heat lost per second = 2260 / 60

= 37.66 J /s

= 37.66 W

Learn more about heat here: https://brainly.com/question/9636950

which one of the following statements is true? A.in an elastic collision,only momentum is conserved B. in any collision,both momentum & kinetic energy are conserved C.in an inelastic collision,both momentum & kinetic energy are conserved D.in an elastic collision,only kinetic energy is conserved ​

Answers

Answer:

option C is correct

................

Answer:

C- in an inelastic collision, both momentum & kinetic energy are conserved

Explanation:

Took the test

The current in the wires of a circuit is 60 milliamps. If the resistance of the circuit were doubled (with no change in voltage), then it’s new current would be _____ milliamps

Answers

Answer:30

Explanation:

Current=60 milliamps

Current=(voltage)/(resistance)

60=(voltage)/(resistance)

Doubling the resistance means multiplying both sides by 1/2

60x1/2=(voltage)/(resistance) x 1/2

30=(voltage)/2(resistance)

Therefore the resistance would be 30 milliamp if we double the resistance

19
Which gas is the most abundant greenhouse gas?
A.
ozone
B.
chlorofluorocarbon
C.
carbon dioxide
OD.
methane
E.
water vapor
Reset
Next

Answers

Carbon dioxide is the most abundant greenhouse gas in the atmosphere.

Answer:C

Explanation:

Carbon dioxide is the most abundance greenhouse gas in The atmosphere.

The figure shows a crane whose weight is 12.5 kN and center of gravity in G. (a) If the crane needs to suspend the 2.5kN drum, determine the reactions on the wheel in A and B when the boom is in the position shown.(b) Considering the same situation illustrated, what is the value of the maximum weight that the crane can suspend without tipping over?

Answers

Answer:

(a) Ra = 9.25 kN; Rb = 5.75 kN

(b) 26.7 kN

Explanation:

(a) Draw a free-body diagram of the crane.  There are four forces:

Reaction Ra pushing up at A,

Reaction Rb pushing up at B,

Weight force 12.5 kN pulling down at G,

and weight force 2.5 kN pulling down at F.

Sum of moments about B in the counterclockwise direction:

∑τ = Iα

-Ra (0.66 m + 0.42 m + 2.52 m) + 12.5 kN (2.52 m + 0.42 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0

-Ra (3.6 m) + 12.5 kN (2.94 m) − 2.5 kN (1.38 m) = 0

Ra = 9.25 kN

Sum of moments about A in the counterclockwise direction:

∑τ = Iα

Rb (0.66 m + 0.42 m + 2.52 m) − 12.5 kN (0.66 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° + 0.66 m + 0.42 m) = 0

Rb (3.6 m) − 12.5 kN (0.66 m) − 2.5 kN (4.98 m) = 0

Rb = 5.75 kN

Alternatively, you can use sum of the forces in the y direction as your second equation.

∑F = ma

Ra + Rb − 12.5 kN − 2.5 kN = 0

Ra + Rb = 15 kN

9.25 kN + Rb = 15 kN

Rb = 5.75 kN

However, you must be careful.  If you make a mistake in the first equation, it will carry over to this equation.

(b) At the maximum weight, Ra = 0.

Sum of the moments about B in the counterclockwise direction:

∑τ = Iα

12.5 kN (2.52 m + 0.42 m) − F ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0

12.5 kN (2.94 m) − F (1.38 m) = 0

F = 26.7 kN

Which of the following is not true about of the use of MRI in medicine?
1) It produces no negative side-effects on the human body
2) It produces high resolution images of soft tissues
3) It is very cheap
4) It requires very strong magnetic fields

Answers

Answer:

3) False. It is expensive since it requires sophisticated equipment and very low temperatures

Explanation:

Nuclear magnetic resonance imaging measurements consist of magnetic resonance imaging to analyze tissues by the transition of the unpaired electron at carbon 13, giving information on the structure and composition of tissues. This information is processed in computers and transformed into images.

So the physical measurement is the MRN

Now we can analyze the statements in the problem

1) True by itself a magnetic measurement is non-invasive

2) True. Measuring carbon transitions has information about the soft tissue of the body

3) False. It is expensive since it requires sophisticated equipment and very low temperatures

4) Right. The applied magnetic field is high to be able to induce carbon transaction

Which of the following statements are true? a. Kinematics is the science that studies forces and motion of particles and bodies. b. Speed is a vector quantity. c. The units of velocity are length divided by time. d. The term deceleration is commonly used to describe a negative acceleration.

Answers

Answer:

true. b, c and d

Explanation:

Let's review each statement separately  

a) False. The kinematics studies the position, speed and acceleration of the bodies, but not what causes these changes

b) True. Velocity is the displacement between time, displacement is a vector, and time is a scalar, so the division between them gives a vector

c) True. speed is the displacement that is a length between time, so its unit is length / time

d) true  desaceleration = - aceleration

A cobalt-60 source with activity 2.60×10-4 Ci is embedded in a tumor that has
mas 0.20 kg. The source emits gamma photons with average energy 1.25 MeV.
Half the photons are absorbed in the tumor, and half escape.
i. What energy is delivered to the tumor per second? [4 marks]
ii. What absorbed dose, in rad, is delivered per second? [2 marks]
iii. What equivalent dose, in rem, is delivered per second if the RBE for
these gamma rays is 0.70? [2 marks]
Page 6 of 7
iv. What exposure time is required for an equivalent dose of 200 rem? [2
marks]
B. A laser with power output of 2.0 mW at a wavelength of 400 nm is projected
onto a Calcium metal. The binding energy is 2.31 eV.
i. How many electrons per second are ejected? [6 marks]
ii. What power is carried away by the electrons? [4 marks]
C. A hypodermic needle of diameter 1.19 mm and length 50 mm is used to
withdraw blood from a patient? How long would it take for 500 ml of blood to be
taken? Assume a blood viscosity of 0.0027 Pa.s and a pressure in the vein of
1,900 Pa. [10 marks]
D. A person with lymphoma receives a dose of 35 gray in the form of gamma
radiation during a course of radiotherapy. Most of this dose is absorbed in 18
grams of cancerous lymphatic tissue.
i. How much energy is absorbed by the cancerous tissue? [2 marks]
ii. If this treatment consists of five 15-minute sessions per week over the
course of 5 weeks and just one percent of the gamma photons in the
gamma ray beam are absorbed, what is the power of the gamma ray
beam? [4 marks]
iii. If the gamma ray beam consists of just 0.5 percent of the photons
emitted by the gamma source, each of which has an energy of 0.03
MeV, what is the activity, in Curies, of the gamma ray source? [4 marks]
E. A water heater that is connected across the terminals of a 15.0 V power supply
is able to heat 250 ml of water from room temperature of 25°C to boiling point
in 45.0 secs. What is the resistance of the heater? The density of water is 1,000
kg/m2 and the specific heat capacity of water is 4,200 J/kg/°C. [10 marks]

Answers

Answer:

A i. E = 9.62 × 10⁻⁷ J/s

ii. The absorbed dose is 4.81 × 10⁻⁶ Gy

iii. The equivalent dose is  3.37 × 10⁻⁴ rem/s

iv.  t = 593471.81 seconds

B. i. 4.025 × 10¹⁵/s

ii. 0.512 mW

C. 7218092.2 seconds

D. i. 6.3 × 10⁻¹ J

ii. 1.4 × 10⁻² W

iii. 1.57 × 10³ Curie

E. 0.129 Ω

Explanation:

The given parameters are;

Mass of tumor = 0.20 kg

Activity of Cobalt-60 = 2.60 × 10⁻⁴ Ci

Photon energy = 1.25 MeV

(i) The energy, E, delivered to the tumor is given by the relation;

[tex]E = \frac{1}{2}\left (Number \, of \, decay / seconds \right )\times \left (Energy \, of \, photon \right )[/tex]

[tex]E = \frac{1}{2}\left (2.6\times 10^{-4}Ci )\times \left (\frac{3.70\times 10^{10}decays/s}{1 Ci} \right )\times 1.25\times 10^{6}eV\times \frac{1.6\times 10^{-19}J}{1eV}[/tex]

E = 9.62 × 10⁻⁷ J/s

(ii) The equation for absorbed dose is given as follows;

Absorbed dose, D, in Grays Gy = (Energy Absorbed Joules J)/Mass kg

Therefore, absorbed dose = (9.62 × 10⁻⁷ J/s)/( kg) = 4.81 × 10⁻⁶ Gy

1 Gray = 100 rad

4.81 × 10⁻⁷ Gy = 100 × 4.81 × 10⁻⁶ = 4.81 × 10⁻⁴ rad/s

(iii) Equivalent dose, H, is  given by the relation;

H = D × Radiation factor, [tex]w_R[/tex]

∴ H = 0.7 × 4.81 × 10⁻⁴ rad/s = 3.37 × 10⁻⁴ Sv = 3.37 × 10⁻⁴ rem/s

(iv) The exposure time required for an equivalent dose of 200 rem is given as follows;

[tex]\dot{H} = \dfrac{H}{t}[/tex]

Therefore;

[tex]t= \dfrac{200}{{3.37 \times 10^{-4}} } = 593471.81 \, s[/tex]

∴ t = 6.9 days

B. The number of electrons ejected is given by the relation;

[tex]N = \frac{P}{E} = \frac{P \times \lambda}{hc}[/tex]

[tex]N = \dfrac{2.0 \times 10^{-3} \times 400 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} = 4.025 \times 10^{15}/s[/tex]

(ii) The power carried by the electron

The energy carried away by the electrons is given by the relation;

[tex]KE_e = hv - \Phi[/tex]

[tex]KE_e = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} - 2.31 \times \frac{1.6 \times 10 ^{-19} }{1}[/tex]

[tex]KE_e = 4.9695 \times 10^{-19} - 3.696 \times 10 ^{-19} = 1.2735 \times 10^{-19} J[/tex]

Power, P[tex]_e[/tex], carried away by the electron = 4.025 × 10¹⁵ × 1.2735 × 10⁻¹⁹ = 0.512 mW

C. The given parameters are;

d = 1.19 mm, ∴ r = 1.19/2 = 0.595 × 10⁻³ m

l = 50 mm = 5 × 10⁻³ m

V = 500 ml = 5 × 10⁻⁴ m³

η = 0.0027 Pa

p = 1,900 Pa.

[tex]\dfrac{V}{t} = \dfrac{\pi }{8} \times \dfrac{P/l}{\eta } \times r^4[/tex]

[tex]t = \dfrac{8\times \eta\times V\times l }{\pi \times P \times r^4}[/tex]

[tex]t = \dfrac{8\times 0.0027 \times 5 \times 10^{-4} \times 5 \times 10^{-2} }{\pi \times 1900 \times (0.595 \times 10^{-4} )^4}[/tex]

t = 7218092.2 seconds

D) i. Energy absorbed is given by the relation;

E = m×D

Where:

D = 35 Gray = 35 J/kg

m = 18 g = 18 × 10⁻³ kg

∴ E = 35 × 18 × 10⁻³ = 6.3 × 10⁻¹ J

ii. Total time for treatment = 15 × 5 = 75 minutes

Energy absorbed = 6.3 × 10⁻¹ × 100 = 63 J

Power = Energy(in Joules)/Time (in seconds)

∴ Power = 63/(75×60) = 1.4 × 10⁻² W

iii. Whereby the power is provided by 0.5% of the photons emitted by the source, we have;

[tex]P_{source}= \frac{P_{beam}}{0.005} =\frac{0.0014}{0.005} =0.28 \, W[/tex]

1 MeV = 1.60218 × 10⁻¹³ J

0.03 MeV = 0.03 × 1.60218 × 10⁻¹³ J = 4.80654 × 10⁻¹⁵ J/photon

Therefore, the number of disintegration per second = 0.28 J/s ÷  4.80654 × 10⁻¹⁵ J/photon = 5.83 × 10¹³ disintegrations per second

1 Curie = 3.7 × 10¹⁰  disintegrations per second

Hence, 5.83 × 10¹³ disintegrations per second = (5.83 × 10¹³)/(3.7 × 10¹⁰) Curie

= 1.57 × 10³ Curie

E. The parameters given are;

Density of water = 1000 kg/m³

Volume of water = 250 ml = 0.00025 m³

Initial temperature, T₁, = 25°C

Final temperature, T₂, = 100°C

Change in temperature, ΔT = 100 - 25 = 75°

Specific heat capacity of the water = 4200 J/kg/°C

Mass of water = Density × Volume = 1000 × 0.00025 = 0.25 kg

∴ Heat supplied = 4200 × 0.25 × 75 = 78,750 J

Time to heat the water = 45.0 sec

Therefore, power = Energy/time = 78750/45 = 1750 W

The formula for electrical power = I²R =VI = V²/R

Therefore, where V = 15.0 V, we have;

15²/R = 1750

R = 15²/1750 = 0.129 Ω.

The resistance of the heater = 0.129 Ω.

I need help plz help me out 10 points!!!!!!!

Answers

Answer:

The answer is diffraction

Explanation:

Answer:

The answer is diffraction

Explanation:

I did the test! HOPE THIS HELPS!

(10 points) A spring with a 7-kg mass and a damping constant 12 can be held stretched 1 meters beyond its natural length by a force of 4 newtons. Suppose the spring is stretched 2 meters beyond its natural length and then released with zero velocity. In the notation of the text, what is the value c2−4mk? m2kg2/sec2 Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t of the form c1eαt+c2eβt where α= (the larger of the two) β=

Answers

Answer:

......................

Which term defines the distance from crest to crest

Answers

Answer:

The horizontal distance between two adjacent crests or troughs is known as the wavelength.

Answer: Wavelength

Explanation:

From crest to crest, it is one full wavelength

The voltage in a circuit is given by the equation V= IR.in this equation v is the voltage Iis correct and R is the resistance which answer shows this equation solved for current?

Answers

Answer:I=V/R

Explanation:

V=IR

Divide both sides by R

V/R=IR/R

V/R=I

I=V/R

The voltage in a circuit is given by the equation V= IR, in this equation v is the voltage Iis correct and R is the resistance the solution for the current is given as follows,

V= IR

I = V/R

What is resistance?

Resistance is the obstruction of electrons in an electrically conducting material. The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.

V=IR

As for the given problem  if we have to solve for the current from the equation V=IR

V=IR

I = V/R

Let us suppose a 60-volt battery connected in a closed circuit with a resistor of 15 ohms then we have o find out the amount of current flowing in the circuit,

Voltage = 60V

Resistance = 15 Ohm

Current =?

By using Ohm,s Law,

V=IR

I = V/R

By substituting the respective values,

I = 60/15

I = 4 Ampere

Hence, we solved for the current from the equation V=IR.

Learn more about resistance from here, refer to the link;

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Under electrostatic conditions, the electric field just outside the surface of any charged conductor

A. is always zero because the electric field is zero inside conductors
B. can have non zero components perpendicular to and parallel to the surface of the conductor
C. is always perpendicular to the surface of the conductor
D. is always parallel to the surface
E. is perpendicular to the surface of the conductor only if it is a sphere, a cylinder, or a flat sheet.

Answers

Answer:

C. is always perpendicular to the surface of the conductor

Explanation:

On a charged conductor , electric charge is uniformly distributed on its surface . The lines of forces are also uniformly  distributed on all directions . They repel each other so they emerge perpendicular to the surface so that they do nor cut each other and at the same time they remain at maximum distance from each other.

Two forces are applied on a body. One produces a force of 480-N directly forward while the other gives a 513-N force at 32.4-degrees above the forward direction .Find the magnitude and direction(relative to forward direction of the resultant force that these forces exert on the body)​

Answers

Answer:

F = (913.14 , 274.87 )

|F| = 953.61 direction 16.71°

Explanation:

To calculate the resultant force you take into account both x and y component of the implied forces:

[tex]\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N[/tex]

Thus, the net force over the body is:

[tex]F=(913.14N)\hat{i}+(274.87N)\hat{j}[/tex]

Next, you calculate the magnitude of the force:

[tex]F=\sqrt{(913.14N)+(274.87N)^2}=953.61N[/tex]

and the direction is:

[tex]\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°[/tex]

Electric motors convert electrical energy to mechanical energy. When the current-carrying coil is placed between the magnetic poles, a force acts on it that causes it to rotate. The image below shows a simple electric motor. The motor is used to lift metal boxes. How can the motor be changed to be able to lift a heavier box? A. rotate the coil in a counter-clockwise direction B. add more loops of wire between the magnets C. change the polarity of the magnet D. decrease the size of the magnets

Answers

Answer:

B. add more loops of wire between the magnets

Explanation:

this would increase the magnetic force acting on the rod therefore increasing

By adding more loops of wire between the magnets the motor is changed to be able to lift a heavier box.

What is an electric motor?

An electric motor is a mechanism that turns electricity into mechanical energy.

The interaction between the motor's magnetic field and electric current in a wire winding generates force in the form of torque imparted to the motor's shaft in most electric motors.

An electric generator is physically equivalent to an electric motor, but it converts mechanical energy into electrical energy using a reversed flow of power.

The load capacity in the motor can be increased by increasing the no of loops. So by adding more loops of wire between the magnets the motor is changed to be able to lift a heavier box.

Hence option B is correct.

To learn more about the electric motor refer to the link;

https://brainly.com/question/15409160

A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 14.9 cm. How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of h0? (d) If the block were released from height 3h0 above the spring, what would be the maximum compression of the spring?

Answers

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= [tex]\frac{1}{2}[/tex]kx² = [tex]\frac{1}{2}[/tex](530)(0.149)² =  5.88 J

b)  Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mg[tex]h_o[/tex] = [tex]\frac{1}{2}[/tex]kx² - mgx

And, [tex]h_o[/tex]= ( [tex]\frac{1}{2}[/tex]kx² - mgx )/(mg) = [tex][\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)[/tex]]/(0.645x9.8)    

   [tex]h_o[/tex]=   0.78m            

d) Now, if the initial initial height of block is 3[tex]h_o[/tex]

[tex]h_o[/tex] = 3 x 0.78 = 2.34m

then, [tex]\frac{1}{2}[/tex]kx² - mgx - mg[tex]h_o[/tex] =0

 

[tex]\frac{1}{2}[/tex](530)x²  - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0  

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum  compression of the spring)

An astronaut is being tested in a centrifuge. The centrifuge has a radius of 11.0 m and, in starting, rotates according to θ = 0.260t2, where t is in seconds and θ is in radians. When t = 2.40 s, what are the magnitudes of the astronaut's (a) angular velocity, (b) linear velocity, (c) tangential acceleration, and (d) radial acceleration?

Answers

Answer:

a) 1.248 rad/s

b) 13.728 m/s

c) 0.52 rad/s^2

d) 17.132m/s^2

Explanation:

You have that the angles described by a astronaut is given by:

[tex]\theta=0.260t^2[/tex]

(a) To find the angular velocity of the astronaut you use the derivative og the angle respect to time:

[tex]\omega=\frac{d\theta}{dt}=\frac{d}{dt}[0.260t^2]=0.52t[/tex]

Then, you evaluate for t=2.40 s:

[tex]\omega=0.52(2.40)=1.248\frac{rad}{s}[/tex]

(b) The linear velocity is calculated by using the following formula:

[tex]v=\omega r[/tex]

r: radius if the trajectory of the astronaut = 11.0m

You replace r and w and obtain:

[tex]v=(1.248\frac{rad}{s})(11.0m)=13.728\frac{m}{s}[/tex]

(c) The tangential acceleration is:

[tex]a_T=\alpha r\\\\\alpha=\frac{\omega^2}{2\theta}=\frac{(1.248rad/s)^2}{2(0.260(2.40s)^2)}=0.52\frac{rad}{s^2}[/tex]

(d) The radial acceleration is:

[tex]a_r=\frac{v^2}{r}=\frac{(13.728m/s)^2}{11.0m}=17.132\frac{m}{s^2}[/tex]

Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance ddd is


|F|=K|QQ′|d2|F|=K|QQ′|d2,


where K=14πϵ0K=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2)ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space.


Consider two point charges located on the x axis: one charge, q1q1q_1 = -15.0 nCnC , is located at x1x1x_1 = -1.660 mm ; the second charge, q2q2q_2 = 34.5 nCnC , is at the origin (x=0.0000)(x=0.0000).


What is the net force exerted by these two charges on a third charge q3q3q_3 = 47.0 nCnC placed between q1q1q_1 and q2q2q_2 at x3x3x_3 = -1.240 mm ?


Your answer may be positive or negative, depending on the direction of the force.

Answers

Answer:

Explanation:

Force between two charges of q₁ and q₂ at distance d is given by the expression

F = k q₁ q₂ / d₂

Here force between charge q₁ = - 15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = (1.66 - 1.24 ) = .42 mm

k = 1/ 4π x 8.85 x 10⁻¹²

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 35969.4 x 10⁻³ N .

force between charge q₂ =  34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x  34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 82729.6  x 10⁻³ N

Both these forces will act in the same direction towards the left (away from the origin towards - ve x axis)

Total force = 118699 x 10⁻³

= 118.7 N.

A 25kg box in released on a 27° incline and accelerates down the incline at 0.3 m/s2. Find the friction force impending its motion? What is the coefficient of kinetic friction?
A block is given an initial speed of 3m/s up a 25° incline. Coefficient of friction

Answers

Answer:

a)  μ = 0.475 , b)   μ = 0.433

Explanation:

a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it

X axis

     Wₓ - fr = m a

the friction force has the expression

     fr = μ N

y Axis

     N - [tex]W_{y}[/tex] = 0

let's use trigonometry for the components the weight

     sin 27 = Wₓ / W

     Wₓ = W sin 27

     cos 27 = W_{y} / W

     W_{y} = W cos 27

     N = W cos 27

     W sin 27 - μ W cos 27 = m a

     mg sin 27 - μ mg cos 27 = m a

      μ = (g sin 27 - a) / (g cos 27)

      very = tan 27 - a / g sec 27

      μ = 0.510 - 0.0344

      μ = 0.475

b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result

         μ = tan 25 - 0.3 / 9.8 sec 25

         μ = 0.466 -0.03378

         μ = 0.433

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