In an exciting game, a baseball player manages to safely slide into second base. The mass of the baseball player is 88.9 kg and the coefficient of kinetic friction between the ground and the player is 0.53. (a) Find the magnitude of the frictional force in newtons. N (b) It takes the player 1.7 s to come to rest. What was his initial velocity (in m/s)

Answers

Answer 1

Answer:

Look at explanation

Explanation:

a) Kinetic Friction= μmg

μmg=0.53*88.9*9.8=461.75N

b)  -461.75N=ma

a= -5.19m/s^2

v=v0+at

5.19*1.7=v0

v0=8.81m/s^2

Answer 2

(a) The magnitude of the frictional force will be 461.75N

(b)The initial velocity will be 8.81 m/s.

What is kinetic friction?

A force that acts among sliding parts is referred to as kinetic friction. A body moving on the surface is subjected to a force that opposes its progressive motion

The size of the force will be determined by the kinetic friction coefficient between the two materials.

The given data in the problem is;

μ is the coefficient of kinetic friction= 0.53.

m is the mass = 88.9 kg

g is the acceleration due to gravity= 9.81 m/s²

v is the speed =?

The formula for friction force is;

[tex]\rm F= \mu R \\\\ R=mg \\\\ F= \mu mg \\\\\ F=0.53 \times 88.9 \times 9.81 \\\\ F= 461.75 \ N[/tex]

Mechanical force is found as;

F=ma

-461.75=(88.9)a

(-ve shows the -ve work done)

a=-5.19 m/s

From the Newton's first equation of motion;

v=u+at

0=u+at

u=-at

u=(- (-5.19)(1.7)

u=8.81 m/s²

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Related Questions

A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the loop. The magnitude of the magnetic field is 3.35 T , and the radius of the wire loop is 0.240 m . Find the magnetic flux Φ through the loop.

Answers

Answer:

0.5849Weber

Explanation:

The formula for calculating the magnetic flus is expressed as:

[tex]\phi = BAcos \theta[/tex]

Given

The magnitude of the magnetic field B = 3.35T

Area of the loop = πr² = 3.14(0.24)² = 0.180864m²

angle of the wire loop θ = 15.1°

Substitute the given values into the formula:

[tex]\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb[/tex]

Hence the magnetic flux Φ through the loop is 0.5849Weber

You are working on a project to make a more efficient engine. Your team is investigating the possibility of making electrically controlled valves that open and close the input and exhaust openings for an internal combustion engine. Determine the stability of the valve by calculating the force on each of its sides and the net force on the valve.

The valve is made of a thin but strong rectangular piece of non-magnetic material that has a current-carrying wire along its edges. The rectangle is 0.35 cm x 1.83 cm. The valve is placed in a uniform magnetic field of 0.15 T such that the field lies in the plane of the valve and is parallel to the short sides of the rectangle. The region with the magnetic field is slightly larger than the valve. When a switch is closed, a 1.7 A current enters the short side of the rectangle on one side and leaves on the opposite short side of the rectangle. At the suggestion of a colleague, who is hoping to ensure different currents along the sides of the valve, resistors have been included along the wire on each of the short sides of the valve. The value of the resistor on one side is twice that on the other side.

Answers

Answer:

The answer is "0.00466 N".

Explanation:

[tex]F=(B \times i) L\\\\[/tex]

therefore the smaller side is parallel to magnetic field  

[tex]\therefore \\\\F= B i L\ \sin\ 'o'=0 \ N[/tex]

calculating the force on the layer side:

[tex]\to F=0.15 \times 1.7 \times 0.0183 \times \sin 90^{\circ}=0.00466\ N\\\\[/tex]

Therefore [tex]F_o[/tex] the net force on the  rectangular loop [tex]= 0.00466 \ N[/tex]

A force of 1000N is used to kick a football of mass 0.8kg find the velocity with which the ball moves if it takes 0.8 sec to be kicked.​

Answers

The velocity of the ball is 100m/s

The first step is to write out the parameters;

The force used to kick the ball is 1000N

The mass of the ball is 0.8 kg

Time is 0.8 seconds

Therefore the velocity can be calculated as follows

F= Mv-mu/t

1000= 0.8(v) - 0.8(0)/0.8

1000= 0.8v- 0.8/0.8

Cross multiply both sides

1000(0.8) = 0.8v

800= 0.8v

divide both sides by the coefficient of v which is 8

800/0.8= 0.8v/0.8

v= 1000m/s

Hence the velocity is 1000m/s

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A block weighing 400 kg rest on a horizontal surface and supports on top of it another block of weight 100 kg placed on the top of it as shown. The block W2 is attached to a vertical wall by a string 6 m long. Ifthe coefficient of friction between all surfaces is 0.25 and the system is in equilibrium find the magnitude of the horizontal force P applied to the lower block.

Answers

The horizontal force applied to the lower block is approximately 1,420.85 Newtons

The known parameters are;

The mass of the block, m₁ = 400 kg, weight, W₁ = 3,924 N

The mass of the block resting on the first block, m₂ = 100 kg, weight, W₂ = 981 N

The length of the string attached to the block, W₂, l = 6 m

The horizontal distance from the point of attachment of the second block to the block W₂, x = 5 m

The coefficient of friction between the surfaces, μ = 0.25

Let T represent the tension in the string

The upward force on W₂ due to the string = T × sin(θ)

The normal force of W₁ on W₂, N₂ = W₂ - T × sin(θ)

The tension in the string, T = N₂ × μ × cos(θ)

∴ T = (W₂ - T × sin(θ)) × μ × cos(θ)

sin(θ) = √(6² - 5²)/6

cos(θ) = 5/6

T = (981 - T × √(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T ≈ 183.27 N

The normal reaction on W₂, N₂ = T/(μ × cos(θ))

∴ N₂ = 183.27/(0.25 × 5/6) = 879.7

N₂ ≈ 879.7 N

The friction force, [tex]F_{f2}[/tex] = N₂ × μ

∴ [tex]F_{f2}[/tex] = 879.7 N × 0.25 = 219.925 N

The total normal reaction on the ground, [tex]\mathbf{N_T}[/tex] = W₁ + N₂

[tex]N_T[/tex] = 3,924 N + 879.7 N = 4,803.7 N

The friction force, on the ground [tex]\mathbf{F_T}[/tex] = [tex]\mathbf{N_T}[/tex] × μ

∴  [tex]F_T[/tex] = 4,803.7 N × 0.25 = 1,200.925 N

The horizontal force applied to the lower block, P = [tex]\mathbf{F_T}[/tex] + [tex]\mathbf{F_{f2}}[/tex]

Therefore;

P = 1,200.925 N + 219.925 N = 1,420.85 N

The horizontal force applied to the lower block, P ≈ 1,420.85 N

Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 55.3 kg, down a theta= 79.6º slope at constant acceleration a=-4.3 m/s2, as shown in Figure (here we assume the positive direction is going down the slope. So the given acceleration is a negative value, it means its direction is going up the slope, slowing down as it moving downward). So, the coefficient of friction between the sled and the snow is 0.100. How many Joules of work is done by the tension in the rope as the sled moves 2.1 m along the hill? Use g= 10 m/s2.

Answers

The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.

In this case, we need to construct the Free Body Diagram of the sled-victim System in order to determine what Forces are doing Work. Then, we construct the respective Energy equation by Newton's Laws of Motion, Work-Energy Theorem and definition of Work.

Given that system experiments an uniform Acceleration, we must solve the resulting model for the work done by the Tension in the rope.

From the Free Body Diagram (see image attached), we see that both Weight of the sled and Friction between sled and snow are doing work in favor of gravity, whereas Tension forces is against gravity. Normal force is not doing work as its direction is perpendicular to the direction of motion. The energy equation of this system is:

[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex] (1)

Where:

[tex]W_{T}[/tex] - Work done by tension, in joules.

[tex]m[/tex] - Mass of the sled-victim system, in kilograms.

[tex]\mu[/tex] - Coefficient of kinetic friction, no unit.

[tex]g[/tex] - Gravitational acceleration, in meters per square second.

[tex]s[/tex] - Travelled distance, in meters.

[tex]\theta[/tex] - Slope angle, in sexagesimal degrees.

[tex]a[/tex] - Net acceleration of the sled-victim system, in meters per square second.

If we know that [tex]\mu = 0.100[/tex], [tex]m = 55.3\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}}[/tex], [tex]s = 2.1\,m[/tex], [tex]\theta = 79.6^{\circ}[/tex] and [tex]a = -4.3\,\frac{m}{s^{2}}[/tex], then the work done by the tension in the rope is:

[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex]

[tex]W_{T} = \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta -m\cdot a\cdot s[/tex]

[tex]W_{T} = (0.100)\cdot \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \cos 79.6^{\circ} + \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \sin 79.6^{\circ} - (55.3\,kg)\cdot \left(-4.3\,\frac{m}{s^{2}} \right) \cdot (2.1\,m)[/tex]

[tex]W_{T} = 1662.544\,J[/tex]

The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.

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Find the intensity of the electromagnetic wave described in each case. (a) an electromagnetic wave with a wavelength of 655 nm and a peak electric field magnitude of 1.5 V/m. 0.002984 W/m2 (b) an electromagnetic wave with an angular frequency of 6.5 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 T. 1.19366E-6 W/m2

Answers

The intensity of the electromagnetic wave in terms of the electric field is 0.00298 W/m² and the intensity of the electromagnetic wave in terms of the magnetic field is 1.193x10⁻⁶  W/m².

The intensity of the electromagnetic wave is related to the electric field as well as to the magnetic field.    

a) Intensity of the electromagnetic wave for the electromagnetic field.

The intensity of the electromagnetic wave (I) in terms of the electromagnetic field is given by:

[tex] I = \frac{E^{2}*c*\epsilon_{0}}{2} [/tex]   (1)

Where:

c: is the speed of light = 3.00*10⁸ m/s  

E: is the magnitude of the electric field = 1.5 V/m

ε₀: is the permittivity of free space = 8.85*10⁻¹² C²/Nm²

Hence, the intensity of the electromagnetic wave (eq 1) is:

[tex] I = \frac{(1.5 V/m)^{2}*3.00 \cdot 10^{8} m/s*8.85 \cdot 10^{-12} C^{2}/(N*m^{2})}{2} = 0.00298 W/m^{2} [/tex]                                                                                          

b) Intensity of the electromagnetic wave for the magnetic field

We can calculate the intensity of the electromagnetic wave (I) in terms of the magnetic field with the following equation:

[tex] I = \frac{cB^{2}}{2\mu_{0}} [/tex]   (2)

Where:

B: is the magnitude of the magnetic field = 10⁻¹⁰ T

μ₀: is the vacuum permeability = 4π*10⁻⁷ m*T/A

Therefore, the intensity of the electromagnetic wave (eq 2) is:

[tex] I = \frac{3.00 \cdot 10^{8} m/s*(1\cdot 10^{-10} m*T/A)^{2}}{2*4\pi \cdot 10^{-7} T/A} = 1.193 \cdot 10^{-6} W/m^{2} [/tex]

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In a rolling race, two objects are released from the top of two identical ramps. They then roll without slipping to the bottom of the ramp. If the two objects are 2 hoops of the same radius but different masses, which reaches the bottom first?

a. The lighter one reaches the bottom first
b. The heavier one reaches the bottom first
c. We don’t have enough information
d. They reach the bottom at the same time

Answers

Answer:

b. The heavier one reaches the bottom first.

Answer:

B

Explanation:

The answer is B the heavier item has more g force pushing it making it roll faster reaching the bottom of the ramp first.

Solar System - Scaling. When you look at Neptune in a telescope, you are actually looking into the past as the light has to travel from Neptune to your eyes. If the speed of light is ~300,000 km/s, how far back into the past are you looking (or put another way, how long does it take light to travel from Neptune to your eyes on Earth)

Answers

Answer:

Distance from sun to Neptune = 4.495E9 km

Time for light to travel = 4.495E9 / 3E5 sec = 14,980 sec

That is from sun to Neptune time fof light = 250 min

Time for light to travel from sun to earth is about 8 min

So the time from Neptune would be 242 to 258 min depending on position of Neptune - Note that Neptune is about 30X as far from the sun as earth and

250 min / 8 min is roughly 30

The uniform motion of kinematics allows us to find the time it takes for light to arrive from Neptune to Earth, which varies between:

          t₁ = 1.45 10⁴ s and t₂₂= 1.55 10⁴ s

depending on the relative distance of the two planets

given parameters

The speed of light c = 300,000 km / s = 3 10⁸ m / s The distance from Neptune to Sum

to find

The time when light arrives from Neptune to Earth

They  velocit of an electromagnetic wave is constant, so we can use the uniform motion relationships

             v = d / t

             t = d / v

where v is the speed of light, d the distance and y time, in this case the speed of the wave is the speed of light (v = c)

We look in the tables for the distances and the rotation periods around the sun

                           distance ( m)         period (s)

Sun Neptunium     4.50 10¹²             5.2 10⁹

Sun - Earth             1.5    10¹¹              3.2 10⁷

With the data of the period it is observed that the rotation of Neptune is much greater than that of Eart rotation around the sun, for which we will assume that Neptunium is fixed in space and the Earth may be in its aphelion or perihelion, maximum approach o away distance from the sun, consequently we calculate the time for the two cases:

Maximum approach

positions relative distance from the dos Plantetas is

         Δd = [tex]x_{Neptuno - Sum} - x_{Earth - Sum}[/tex]d  

     

         Δd = 4.50 10¹²  - 1.5 10¹¹

         Δd = 43.5 10¹¹ m

the time it takes for Neptune's light to reach Earth is

        Δt = [tex]\frac{ 43.5 \ 10^{11} }{3 \ 10^8}[/tex]  

        Δt = 14.5 10³ s

        Δt = 1.45 10⁴ s

       

We reduce to hours

        Δt = 1.45 10⁴ s (1 h / 3600 s) = 4.03 h

Maximum away

         Δd = [tex]x_{Neptune - Sum} + x_{Neptune-Sum}[/tex]  

         Δd = 4.50 10¹² + 1.5 10¹¹

         Δd = 46.5 10¹¹

The time is

         Δt = [tex]\frac{46.5 \ 10^{11}}{ 3 \ 10^8}[/tex]  

         Δt = 15.5 10³

         Δt = 1.55 10⁴ s

We reduce to hours

         Δt = 1.55 10⁴ s (1 h / 3600 s) = 4.31 h

In conclusion, the time it takes for light to arrive from Neptune to Earth varies between:

          t₁ = 1.45 10⁴ s and t₂ = 1.55 10⁴ s

depending on the relative distance of the two plants

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No matter how far you stands from a mirror your image appear errect .the mirror is

Answers

Answer:

convex mirror

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

Answer:

convex mirror..........

Hi Friends!

please help me with these questions!

SUBJECT: Chemistry, Physics,Biology

Answers

Answer:

q.1 : Air near candle gets heated up and after this it rises by convection so the thermometer B will receive more heat than the thermometer A So, according to the given condition thermometer B will show a greater rise in temperature.

q.2 : x is the pure sample of compound . y is the pure sample of element . z is the mixture of different elements

q.3 : the saliva contains an enzyme salivary amylase (ptyalin) which converts starch in roti into maltose, isomaltose and small dextrins called a-dextrin.

cyclist always bends when moving the direction opposite to the wind. Give reasons​

Answers

When he bends he kinda off his feet and light but if it’s not i’m so sorry this is just my thinking.

​Determine usando ecuación de Bernoulli la Presión P1 necesaria para mantener la condición mostrada dentro del sistema mostrado en la figura, sabiendo que el aceite tiene un s.g =0.45 y el valor de d=90mm.

Answers

Answer:

PlROCA

Explanation:

find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4​

Answers

The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex].

The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:

Q(t) = Aeσ[tex]T^{4}[/tex]

where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.

To determine the rate of energy radiated by the man in the given question;

[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ

But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.

So that;

[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]

     = 3.8556 x [tex]10^{-8}[/tex]

     = 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex]

Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex]  J/s. [tex]m^{2}[/tex].

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Which of the following choices is not an example of climate?
0000
San Diego has mild, warm temperatures and sea breezes year-round.
Anchorage has short, cool summers and long, snowy winters.
It will be 78° on Friday in Clovis.
Florida is tropical, with a significant rainy season.

Answers

Answer:

Florida is tropical, with a significant rainy seson

A bicycle tire with a volume of 0.00210 m^3 is filled to its recommended absolute pressure of 495 kPa on a cold winter day when the tire's temperature is -14°C. The cyclist then brings his bicycle into a hot laundry room at 32°C.

a. If the tire warms up while its volume remains constant, will the pressure increase be greater than, less than, or equal to the manufacturer's stated 10% overpressure limit?
b. Find the absolute pressure in the tire when it warms to 32 degrees Celcius at constant volume.

Answers

(A) The pressure will be greater than 10% overpressure limit.

(B) The final pressure will be "582.915 kPa".

Given:

Volume,

[tex]V = 0.0021 \ m^3[/tex]

Initial pressure,

[tex]P_o= 495 \ kPa[/tex]

Initial temperature,

[tex]T_o = -14^{\circ} C[/tex]

            [tex]= 259 \ K[/tex]

Final temperature,

[tex]T = 32^{\circ} C[/tex]

(B)

Number of moles,

→ [tex]n = (\frac{P_o V}{RT_o} )[/tex]

then,

The final absolute pressure,

→ [tex]P = \frac{nRT}{V}[/tex]

      [tex]= (\frac{P_o V}{RT_o} )(\frac{RT}{V} )[/tex]

      [tex]=(\frac{T}{T_o} )P_o[/tex]

      [tex]= (\frac{305}{259} )\times 495[/tex]

      [tex]= 582.915 \ kPa[/tex]

Thus the above approach is correct.

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5. A bicyclist is finishing her repair of a flat tire when a friend rides by at a constant velocity of
3.5 m/s. Three seconds later, the bicyclist hops on her bike and accelerates at 3.6 m/s² until she
catches her friend.
a. How much time does it take until she catches her friend?
b. How far has she traveled in this time?
c. What is her speed when she catches up?

Answers

Answer:

a) t = 3.6 s

b) d = 23 m

c) v = 13 m/s

Explanation:

Let t be the time the accelerating rider rides

the distance she travels is

d = ½3.6t²

the distance for the other cyclist is

d =3.5(t + 3)

½3.6t² = 3.5(t + 3)

1.8t² - 3.5t - 10.5 = 0

quadratic formula, positive answer

t = (3.5 + √(3.5² - 4(1.8)(-10.5))) / (2(1.8))

t = 3.575786...

d = ½(3.6)(3.575786²) = 23.015...

v = 3.6(3.575786) = 12.8728...

define nortons theorem​

Answers

Answer:

In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.

Determine the point of contraflexure

Answers

Answer:

The point of contraflexure (PoC) occurs where bending is zero and at the point of change between positive and negative (or between compression and tension). In a beam that is flexing (or bending), the point where there is zero bending moment is called the point of contraflexure.

I only need help with e (bottom of the page).​

Answers

Explanation:

The box is accelerating along the y-axis at a rate of [tex]+2.5\:\text{m/s}^2[/tex] as well as along the x-axis at a rate of [tex]+5.1\:\text{m/s}^2.[/tex] So the magnitude of the box's total acceleration is given by

[tex]a_T = \sqrt{a_x^2 + a_y^2}[/tex]

[tex]\:\:\:\:= \sqrt{(5.1\:\text{m/s}^2)^2 + (2.5\:\text{m/s}^2)^2}[/tex]

[tex]\:\:\:\:=5.7\:\text{m/s}^2[/tex]

The direction of the acceleration [tex]\theta[/tex] with respect to the horizontal direction is given by

[tex]\theta = \tan^{-1}\!\left(\dfrac{a_y}{a_x}\right) = \tan^{-1}\!\left(\dfrac{2.5\:\text{m/s}^2}{5.1\:\text{m/s}^2}\right)[/tex]

[tex]\:\:\:\:= 26.1°[/tex]

The graph below shows a cycle of a heat engine. Add the following labels to the graph. Some labels are used more than once.
Labels: Isobaric process; W= 0J; Work done on the system; Work done by the system.

I will give brainliest!

P.S. AL2006 if you see this please help!​

Answers

I'm not very good at this material. I'll try it, but if I were you, I wouldn't bet money on these answers.

"Isobaric" means constant pressure. So those are the horizontal lines, where every point on the line is at the same pressure. Those are the processes 1>2 and 3>4 .

I'm going around and around in my mind with the other labels, and I can't decide. So I'm afraid I can't answer any more of them ... they might be wrong.

Answer:

1 -> 2 & 3 -> 4: Isobaric process

4 -> 1: Work done BY the system

2 -> 3: Work done ON the system

W(total): W = 0J

Explanation:

The two horizontal lines (1 -> 2 & 3 -> 4) are "Isobaric" since isobaric processes take place at constant pressure. I believe 4 -> 1 is "Work done BY the system" since pressure increases when there is an increase of thermal energy, in other words, the system is absorbing heat. This is why the volume increases from 1 -> 2 after the system has absorbed heat in 4 -> 1. Following the directions of the arrows, 2 -> 3 would be "Work done ON the system" since pressure is DECREASING, meaning temperature is also exiting the system. That's why the next step (3 -> 4) shows a decrease in volume. This model depicts a process that has a W(total) of 0 J because this is a cycle.

I hope this helps :))

The timing device in an automobile’s intermittent wiper system is based on an RC time constant and utilizes a 0.500 micro F capacitor and a variable resistor. Over what range must R be made to vary to achieve time constants from 2.00 to 15.0 s?

Answers

Answer:

check 2 photos for answer

check 2 photos for answer

PLEASE ANSWER IF YOU CAN AND NOT FOR THE SAKE OF GAINING POINTS!

Answers

A is the correct answer because the the plane mirror would move the same speed in the same direction

Una persona de 76 kg está siendo retirada de un edificio en llamas mientras se muestra en la figura. Calcule la tensión
en las dos cuerdas si la persona está momentáneamente inmovil.
Ayuda por favor.

Answers

Answer:

T1 = 736.6 N, T2 = 193.5 N

Explanation:

W = 76 N

The tension is T1 and T2.

By use of Lami's theorem

[tex]\frac{T_1}{Sin100}=\frac{T_2}{Sin165}=\frac{W}{Sin 95}\\\\So, \\\\T_1 = \frac{76\times 9.8\times Sin 100}{Sin 95} = 736.6 N \\And\\T_2 = \frac{76\times 9.8\times Sin 165}{Sin 95} = 193.5 N \\[/tex]

The source of sound moves away from the listener.The listener has the impression that the source is lower in pitch. Why?

Answers

When the source is moving away from the observer the velocity of the source is added to the speed of light. This increases the value of the denominator, decreasing the value of the observed frequency. Frequency corresponds to pitch or tone; a lower observed frequency will result in a lower observed pitch.

Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B​

Answers

Explanation:

Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]

The sum of the two vectors is

[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]

[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]

The difference between the two vectors can be written as

[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]

[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]

Mention the importance of occupation??​

Answers

Answer:

ln the contemporary time , farming can be considered as comparitively important occupation as it can feed the population , So agriculture is having a greater importance than any other occupation.

how interfacial angles are determined using contact goinometer​

Answers

Interfacial angles can be measured using a contact goniometer or more precisely using a reflection goniometer. See Klein & Hurlbut's figure 2.40. The interfacial angle between two faces of a mineral crystal is identical for all crystals of the same mineral that exhibit the corresponding two faces.

What is an internal resistance?

Answers

Explanation:

some thing inside a resistor

The instrument includes a light source, which is passed through a Choose... , which isolates a single wavelength to pass through an aperture to reach the Choose... . Then, the light travels to the Choose... , which measures the intensity of light reaching it.

Answers

Answer:

Following are the response to the given question:

Explanation:

It's being used to measure the amount of light absorbed after traveling through a test tube (the amount of solar radiation received). For several quantitative estimations, this technique is widely employed. Spectrometer and Spectrometer were two devices that are used together to light intensity and light intensity.

It creates and diffuses phosphorescent light into the selected frequency, while the Spectrometer measures the strength of attenuation by the sample solution.

Diffraction beams or prisms are being used to convert polychromatic illumination into monochrome light.

Afterward, the sunlight has a certain hue. Once it reaches the specimen cuvette, it begins absorption. It falls on a sensor that transforms its intensity into such an electronic current.

Here are some ways to fill in such gaps:

In order to reach the specimen cuvette, the light from the light source must be routed via an aperture in order to be isolated by either a diffraction pattern. Light travels to the detector, which detects its intensity.

the number of significant figures in the measurement 4.300×10^5 km are​

Answers

Answer:

6

Explanation

Any numbers in scientific notation are considered significant. For example, 4.300 x 10-4 has 4 significant figures.

Answer From Gauth Math

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