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Lesechka [4]
2 years ago
13

If the torque required to loosen a nut on a wheel has a magnitude of 40.0 N·m and the force exerted by a mechanic is 133 N, how

far from the nut must the mechanic apply the force?
Question 7 options:

1.) 0.15 m
2.) 0.6 m
3.) 0.3 m
4.) 1.2 m
Physics
1 answer:
MakcuM [25]2 years ago
3 0

Answer:

3.) 0.3 m

Explanation:

Given parameters:

Torque  = 40Nm

Force = 133N

Unknown:

Distance from the nut that the mechanic must apply the force  = ?

Solution:

Torque is the force that causes an object to rotate on its axis;

  Torque  = force x distance

   Distance  = \frac{40}{133}  = 0.3m

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A capacitor of capacitance C has charge Q and stored energy is U if the charge is increased to 2Q then energy will be A)4U B)2U
Lubov Fominskaja [6]

Answer:

2u

Explanation:

2u

W = Vq

q = CV

W = V.CV

W = CV²

W/C = V²

V = √(W/C)

√(W1/C1) = √(W2/C2)

√(u/c) = √(x/2c)

x = 2u

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3 years ago
You punt a ball straight up at 20 m/s. What is the balls hangtime
Natalija [7]

The hang time of the ball is 4.08 s

Explanation:

The ball is in free fall motion: this means that it is acted upon gravity only, so its acceleration is the acceleration of gravity,

a=g=-9.8 m/s^2

downward (the negative sign refers to the downward direction).

Since this is a uniformly accelerated motion, we can solve the problem by using the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

First we calculate the time it takes for the ball to reach the maximum height, where the velocity is zero:

v = 0

Substituting:

u = +20 m/s

a=-9.8 m/s^2

we find t

t=\frac{v-u}{a}=\frac{0-20}{-9.8}=2.04 s

The motion of the ball is symmetrical, so the total time of flight is just twice the time needed to reach the maximum height, therefore:

T=2t=2(2.04)=4.08 s

Learn more about free fall:

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4 0
2 years ago
To determine the location of the center of mass (or center of gravity) of a car, the car is drivenover a scale on a horizontal f
just olya [345]

Explanation:

It is given that,

When the front wheels are over the scale, the weight recorded by the scale is 5800 N, F₁ = 5800 N

When the rear wheels are over the scale, the scale reads 6500 N, F₂ = 6500

The distance between the front and rear wheels is measured to be 3.20 m, x₂ = 3.2 m

We need to find the location of center of mass behind the front wheels. Let the center of is located at a distance of x₁. Thus balancing the torques we get :

5800\times x_2=6500\times (3.2-x_2)

On solving the above equation we get, x₂ = 1.69 m

So, the center of mass is located at a distance of 1.69 meters behind the front wheels.

4 0
2 years ago
A 12cm candle is placed 6cm from a converging lens with a focal length of 15cm. What is the height of the image of the candle? S
amm1812

Answer:

The height of the image of the candle is 20 cm.

Explanation:

Given that,

Size of the candle, h = 12 cm

Object distance from the candle, u = -6 cm

Focal length of converging lens, f = 15 cm

To find,

The height of the image of the candle.

Solution,

Firstly, we will find the image distance of the candle. Let it is equal to v. Using lens formula to find the image distance.

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

v is image distance

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-6)}\\\\v=-10\ cm

If h' is the height of the image. Magnification is given by :

m=\dfrac{h'}{h}=\dfrac{v}{u}

h'=\dfrac{vh}{u}\\\\h'=\dfrac{-10\times 12}{-6}\\\\h'=20\ cm

So, the height of the image of the candle is 20 cm.

3 0
2 years ago
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