Answer:30N
Explanation:
Mass=5kg
Acceleration=6m/s^2
Force=mass x acceleration
Force=5 x 6
Force=30N
Answer:
30n
Explanation:
Mass=5kg
Acceleration=6m/s^2
Force=mass x acceleration
Force=5 x 6
Force=30N
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!
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.
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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
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
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
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
a) When we were examining the Electromagnetic Tab, we saw that a flow of electrons or a current as we say it, creates a magnetic field. What about the converse, can a magnetic field be involved in the creation of a flow of electrons/current? Therefore is it reasonable to suggest that we can create a magnetic field by having a flow of current and this can be used to make more current? Explain how this can occur
Answer:
Magnetic field can be used to produce current, infact a changing magnetic field can produce current.
A changing magnetic field in a loop causes the flux linked with the loop to change in turn generating a emf in the loop and therefore a current.
For a loop of area A and resistance R.
I =dPhi/dt/R
В. А
I = AcosФ/R .dB /dt
But it isn't reasonable to say that we can create a magnetic field by having a flow of current and this can be used to make more current because the current generated due to change in magnetic field created by increase/decrease in flow of current will be in a direction such that it will counter act the change in magnetic field caused by increase/decrease in current flow.(lenz's law).
We were unable to transcribe this image
Ф= В. А
I = Acos dB Rd
Mr. Dunn drives 64.8km from work at a speed of 48km/h. Mrs. Dunn drives 81.2km from work
at a speed of 58km/h. They both leave work at the same time. Show complete working to secure
full credits. [4]
i. Who arrives home first?
ii. How many minutes later is it before the second person gets home?
iii. A Coyote is chasing its meal (the Road Runner). Unfortunately, the Coyote has difficulty
adjusting to the Road Runner’s speed but we have a good idea of what it is.
plz help me i will mark you as brainliest
Answer:
i) Mr. Dunn arrives to home first.
ii) 3 min
Explanation:
i. To find who arrives first to home you calculate the time, by using the following formula:
[tex]t=\frac{x}{v}[/tex]
x: distance
v: velocity
Mr. Dunn:
[tex]t=\frac{64.8km}{48km/h}=1.35h[/tex]
Mrs. Dunn:
[tex]t=\frac{81.2km}{58km/h}=1.4h[/tex]
Hence, Mr. Dunn arrives to home first.
ii. To calculate the difference in minutes, you convert hours to minutes:
[tex]1.35h*\frac{60min}{1h}=81min\\\\1.40h*\frac{60min}{1h}=84min\\\\\Delta\ t=(84-81)min=3min[/tex]
the difference between the times is 3min
(i) Mr. Dunn takes less time so he arrives at home first.
(ii) The second person arrives 3 min late.
Time taken to arrive home:
(i) We have to calculate the time taken to reach home by Mr. Dunn and Mrs. Dunn.
t = x/v
where x is the distance
and v is the velocity
Time taken by Mr. Dunn:
distance x = 64.8 km
speed v = 48 km/h
t = 64.8 / 48
t = 1.35 h
Time taken by Mrs. Dunn:
distance x = 81.2 km
speed v = 58 km/h
t' = 81.2 / 58
t' = 1.4 h
Hence, Mr. Dunn arrives at home first.
(ii) To calculate the difference in minutes, you convert hours to minutes:
The time taken by Mr. Dunn in minutes is:
t = 1.35×60 = 81 minutes
The time taken by Mrs. Dunn in minutes is:
t' = 1.4×60 = 84 minutes
the difference between the times is 3min
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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]
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.
An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 30.8 N, the spring is stretched by 17.7 cm from its equilibrium length. Calculate the additional work required by F to stretch the spring by an additional 12.4 cm from that position.
Answer:
[tex]W=5.16 J[/tex]
Explanation:
Using the Hooke's law we can find the elasticity constant:
[tex]F=-k\Delta x[/tex]
[tex]30.8=-k*0.177[/tex]
[tex]k=|-\frac{30.8}{0.177}|[/tex]
[tex]k=174 N/m[/tex]
Now, we know that the work done is equal to the elastic energy, so we will have:
[tex]W=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})[/tex]
x2 is the final distance (x2 = 0.177+0.124 = 0.301 m)
x1 is the initial distance (x1 = 0.177 m)
[tex]W=\frac{1}{2}*174(0.301^{2}-0.177^{2})[/tex]
[tex]W=5.16 J[/tex]
I hope it helps you!
A car speeds up from 18.54 m/s to
29.52 m/s in 13.84 s.
The acceleration of the car is:
Answer:
.7934[tex]m/s^{2}[/tex]
Explanation:
Acceleration = change in velocity / change in time
A = 10.98[tex]m/s[/tex] / 13.84[tex]s[/tex]
A = .7934[tex]m/s^{2}[/tex]
Answer:0.8 m/s^2
Explanation:
initial velocity(u)=18.54m/s
Final velocity(v)=29.52m/s
Time(t)=13.84 sec
Acceleration =(v-u)/t
acceleration =(29.52-18.54)/13.84
Acceleration =10.98/13.34
Acceleration=0.8 m/s^2
When a high‑energy photon passes near a heavy nucleus, a process known as pair production can occur. As a result, an electron and a positron (the electron's antiparticle) are produced. In one such occurrence, a researcher notes that the electron and positron fly off in opposite directions after being produced, each traveling at speed 0.941c. The researcher records the time that it takes for the electron to travel from one position to another within the detector as 15.7 ns. How much time would it take for the electron to move between the same two positions as measured by an observer moving along with the positron?
Answer:
1.47*10^{-8}s
Explanation:
You first calculate the distance traveled by the electron:
[tex]x=vt\\\\x=(0.941(3*10^8m/s))(15.7*10^{-9}s)=4.43m[/tex]
Next, you calculate the relative speed as measure by an observer in the positron, of the electron:
[tex]u'=\frac{u+v}{1+\frac{uv}{c^2}}\\\\u'=\frac{0.941c+0.941c}{1+\frac{(0.941)^2c^2}{c^2}}\\\\u'=0.99c[/tex]
with this relative velocity you calculate the time:
[tex]t=\frac{x}{u'}\\\\t=\frac{4.43m}{0.99c}=1.47*10^{-8}s[/tex]
Which of these parameters is directly related to sound frequency?
Answer:Velocity
Explanation:
Velocity is directly proportional to the frequency of a wave.
Velocity=frequency x wavelength
You expend 1000 W of power in moving a piano 5 meters in 5 seconds. How much force did you exert?
Answer:B
Explanation:
Power=1000 watts
Time=5 seconds
Distance=5 meters
Force=(power x time) ➗ distance
Force=(1000 x 5) ➗ 5
Force=5000 ➗ 5
Force=1000
Force=1000N
Answer:1,000
Explanation:
ape.x
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)
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]
The friends now feel prepared for a homework problem. Consider a cylinder initially filled with 9.30 10-4 m3 of ideal gas at atmospheric pressure. An external force is applied to slowly compress the gas at constant temperature to 1/6 of its initial volume. Calculate the work that is done. Note that atmospheric pressure is 1.013 105 Pa
Answer:
Explanation:
Initial volume of gas V₁ = 9.30 x 10⁻⁴ m³
final volume V₂ = 1 / 6 x 9.30 x 10⁻⁴
= 1.55 x 10⁻⁴ m³
Atmospheric pressure P = 1.013 x 10⁵ Pa .
temperature T .
PV = n RT
nRT = 1.013 x 10⁵ x 9.3 x 10⁻⁴
= 94.21
work done in isothermal process
= 2.303 nRT log V₁ / V₂
= 2.303 x 94.21 log 6
= 168.83 J .
Modified Newtonian dynamics(MoND)proposes that, for small accelerations, Newton’s second law, F = ma, approaches the form F = ma2/a0, where a0 is a constant.
(a) (10 points) Show how such a modified version of Newton’s second law can lead to flat rotation curves, without the need for dark matter.
(b) (10 points) Alternatively, propose a new law of gravitation to replace F = GMm/r2 at distances greater than some characteristic scale r0 so that again, you can explain the observed flat rotation curved of galaxies without dark matter.
Answer:
Explanation:
The two pictures attached here shows the solution to the two questions from the problem. thank you and I hope it helps you
(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.
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.
Which element is malleable and ductile
Answer:
Gold, silver, platinum. Gold is the most malleable and ductile.
Explanation:
The elements which are malleable and ductile include the following:
CopperIronCobalt etc.What is Malleability and Ductility?Malleability is the ability of a substance to be hammered into thin sheets
while ductility involves the deformation of a substance without any
breakage occurring in it.
Transition metals are the group of elements which have both
characteristics and examples are listed above.
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Photons of light scatter off molecules, and the distance you can see through a gas is proportional to the mean free path of photons through the gas. Photons are not gas molecules, so the mean free path of a photon is not equal to that of a molecule, but its dependence on the number density of the gas and on molecular radius is the same. Suppose you are in a smoggy city and can barely see buildings 500 m away.
(a) How far would you be able to see if all the molecules around you suddenly doubled in volume?
(b) How far would you be able to see if the temperature suddenly rose from 20◦C to a blazing hot 1500◦C with the pressure unchanged?
Answer:
a) 315 m
b) 3025.6 m
Explanation:
The picture attached shows the full explanation for the problem.
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?
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]
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.
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
If A = (6i-8j) units, B = (-8i-3j) units, and C = (26i-19j) units, determine a and b
such that aA + bB + C = 0
Answer:
Explanation:
given equation
aA + bB + C = 0
Putting the given values
a(6i-8j) +b (-8i-3j) +(26i-19j) = 0
i ( 6a - 8b ) - j ( 8a + 3 b ) = - 26 i + 19 j
comparing the coefficients of i and j
6a - 8b = -26
8a + 3b = -19.
multiplying first equation by 4 and second equation by 3
24a - 32 b = - 104
24a + 9b = -57
9b + 32b = -57 + 104
41 b = 47
b = 1.41
6 a - 8 x 1.41 = -26
6a = -14.72
a = - 2.45
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
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 gaseshope it helps!
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.
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.
Official (Closed) - Non Sensitive
MEF Tutorial 2 Q3
A train with a maximum speed of 29.17 m/s has an
acceleration rate of 0.25 m/s2 and a deceleration
rate of 0.7 m/s2. Determine the minimum running
time, if it starts from rest at one station and stops
at the next station 7 km away.
Answer:
The minimum running time is 319.47 s.
Explanation:
First we find the distance covered and time taken by the train to reach its maximum speed:
We have:
Initial Speed = Vi = 0 m/s (Since, train is initially at rest)
Final Speed = Vf = 29.17 m/s
Acceleration = a = 0.25 m/s²
Distance Covered to reach maximum speed = s₁
Time taken to reach maximum speed = t₁
Using 1st equation of motion:
Vf = Vi + at₁
t₁ = (Vf - Vi)/a
t₁ = (29.17 m/s - 0 m/s)/(0.25 m/s²)
t₁ = 116.68 s
Using 2nd equation of motion:
s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²
s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²
s₁ = 1701.78 m = 1.7 km
Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.
We have:
Final Speed = Vf = 0 m/s (Since, train is finally stops)
Initial Speed = Vi = 29.17 m/s (The train must maintain max. speed for min time)
Deceleration = a = - 0.7 m/s²
Distance Covered to stop = s₂
Time taken to stop = t₂
Using 1st equation of motion:
Vf = Vi + at₂
t₂ = (Vf - Vi)/a
t₂ = (0 m/s - 29.17 m/s)/(- 0.7 m/s²)
t₂ = 41.67 s
Using 2nd equation of motion:
s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²
s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²
s₂ = 607.78 m = 0.6 km
Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.
The remaining distance is:
s₃ = 7 km - s₂ - s₁
s₃ = 7 km - 0.6 km - 1.7 km
s₃ = 4.7 km
Now, for uniform speed we use the relation:
s₃ = vt₃
t₃ = s₃/v
t₃ = (4700 m)/(29.17 m/s)
t₃ = 161.12 s
So, the minimum running time will be:
t = t₁ + t₂ + t₃
t = 116.68 s + 41.67 s + 161.12 s
t = 319.47 s
In a shipping company distribution center, an open cart of mass 50 kg is rolling to the left at a speed of 5 m/s. You can ignore friction between the cart and the floor. A 15 kg package slides down a chute that makes an angle of 27 degrees below the horizontal. The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds. The package comes to a stop in the cart after 4 seconds. What is:a) the speed of the package just before it lands in the cart
Answer:
Explanation:
The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds . During .75 second duration . package undergoes free fall due to which additional vertical velocity is added
velocity added = a x t
= 9.8 x .75
= 7.35 m /s
Total vertical velocity
= 3 sin27 + 7.35
= 8.71 m /s
Horizontal component = 3 cos 27
= 2.67 m /s
If v be the resultant velocity of these components
v² = 2.67² + 8.71²
v² = 7.13 + 75.86
v = 9.11 m /s .
What spectacles are required for reading purposes by a person whose near point is 2.0m
Answer:Convex lens spectacles is required for reading purpose..
Explanation:
I don't say you have to mark my ans as brainliest but if it has really helped you please don't forget to thank me...
Which term defines the distance from crest to crest
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
A firearms company is testing a new model of rifle by firing a 7.50-g lead bullet into a block of wood having a mass of 17.5 kg. The bullet embeds into the block and the collision generates heat. As a consequence, the temperature rises by 0.040°C, as measured with a high-precision thermometer. Assuming that all the kinetic energy of the bullet goes into heating the system, what is the bullet’s speed when it enters the block? The initial temperatures of bullet and wooden block can be considered identical and the specific heats of lead and wood are cPb = 130 J/(kg ⋅ C°) and c wood = 1700 J/(kg ⋅ C°), respectively.
Answer:
Explanation:
Let the bullets speed be V .
Kinetic energy = 1/2 mV² where m is mass of bullet
This energy is converted into heat Q which raises the temperature of target by Δ T .
Q = mc Δ T , m is mass , c is specific heat and Δ T is rise in temperature .
heat absobed by bullet
= .0075 x 130 x .040
= .039 J
heat absorbed by block of wood
= 17.5 x 1700 x .04
= 1190 J
Total heat absorbed
= 1190.039 J
So kinetic energy = heat absobed
= 1/2 x .0075 x V² = 1190.039
V² = 317343.73
V = 563.33 m /s
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
Answer:
option C is correct
................
Answer:
C- in an inelastic collision, both momentum & kinetic energy are conserved
Explanation:
Took the test
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.
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.
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.
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
