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3 years ago

A stuntman of mass 48 kg is to be launched horizontally out of a spring-

Physics

1 answer:

Damm [24]3 years ago

6 0

The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

The Kinetic energy of the stuntman is equal to the elastic potential energy of the spring.

<h3 /><h3>Velocity: </h3>

This is the ratio of displacement to time. The S.I unit of Velocity is m/s. The velocity of the stuntman can be calculated using the formula below.

⇒ Formula:

mv²/2 = ke²/2
mv² = ke².................. Equation 1

⇒ Where:

m = mass of the stuntman
v = velocity of the stuntman
k = force constant of the spring
e = compression of the spring

⇒ Make v the subject of the equation

v = √(ke²/m)................. Equation 2

From the question,

⇒ Given:

m = 48 kg
k = 75 N/m
e = 4 m

⇒ Substitute these values into equation 2

v = √[(75×4²)/48]
v = √25
v = 5 m/s.

Hence, The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

Learn more about velocity here: brainly.com/question/10962624

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A salt shaker sits 0.102 m from

Ivahew [28]

The maximum speed is 0.55 m/s

Explanation:

For an object in uniform circular motion, the force of friction between the object and the ground provides the centripetal force required to keep the body in motion. Therefore we can write:

$\mu_s mg = m\frac{v^2}{r}$

where the term on the left is the frictional force and the term on the right is the centripetal force, and where

$\mu_s$ is the coefficient of static friction

m is the mass of the body

g is the gravitational acceleration

v is the speed of the body

r is the radius of the circular path

In this problem, we have:

$\mu_s = 0.307$

r = 0.102 m

$g=9.8 m/s^2$

Substituting and re-arranging, we find the maximum speed v at which the salt shaker can rotate:

$v=\sqrt{\mu gr}=\sqrt{(0.307)(9.8)(0.102)}=0.55 m/s$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

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3 years ago

Suppose that a sled is accelerating at a rate of 2 m/s^2. if the net force is tripled and the mass is doubled, then what is the

Gwar [14]

So new acceleration is 3 m/s^2

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You are standing 10 meters from a light source. Then, you back away from the light source until you are 20 meters away from it.

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From my new perspective, the new intensity of the
light source has decreased by a factor of four.

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3 years ago

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Which is the temperature that Fahrenheit and Celsius thermometers would

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If I am not mistaken it’s 0

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Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determin

tia_tia [17]

Answer with Explanation:

We are given that

A=3i-3j m

B=i-4 j m

C=-2i+5j m

a. $D=A+B+C$

$D=3i-3j+i-4j-2i+5j$

$D=2i-2j$

Compare with the vector r=xi+yj

We get x=2 and y=-2

Magnitude= $\mid D\mid=\sqrt{x^2+y^2}=\sqrt{(2)^2+(-2)^2}=2\sqrt 2$ units

By using the formula $\mid r\mid=\sqrt{x^2+y^2}$

Direction: $\theta=tan^{-1}\frac{y}{x}$

By using the formula

Direction of D: $\theta=tan^{-1}(\frac{-2}{2})=tan^{-1}(-1)=tan^{-1}(-tan45^{\circ})=-45^{\circ}$

b.E=-A-B+C

$E=-3i+3j-i+4j-2i+5j$

$E=-6i+12j$

$\mid E\mid=\sqrt{(-6)^2+(12)^2}=13.4$ units

Direction of E= $\theta=tan^{-1}(\frac{12}{-6}=tan^{-1}(-2)=-63.4^{\circ}$

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3 years ago