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Allushta [10]
3 years ago
11

During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of

72.0 kg and its horizontal component of velocity is 6.50 m/s just before the 65.0 kg performer catches it. If the performer is initially motionless on nearly frictionless roller skates, what is his speed immediately after catching the cannon ball
Physics
1 answer:
olasank [31]3 years ago
8 0

Answer:

3.416 m/s

Explanation:

Given that:

mass of cannonball m_A = 72.0 kg

mass of performer m_B = 65.0 kg

The horizontal component of the ball initially \mu_{xA} = 6.50 m/s

the final velocity of the combined system v = ????

By applying the linear momentum of conservation:

m_A \mu_{xA}+m_B \mu_{xB} = (m_A+m_B) v

72.0 \ kg \times 6.50 \ m/s+65.0 \ kg \times 0 = (72.0 \ kg+65.0 \ kg) v

468 kg m/s + 0 = (137 kg)v

v = \dfrac{468\  kg m/s }{137 \ kg}

v = 3.416 m/s

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<span>At an instant when the displacement is equal to a/2, Potential energy U = 1/2ka(square) where a is displacement. when a= a/2 U = 1/4ka(square) U = E/4 Potential Energy = 1/4 Total energy</span>
4 0
3 years ago
You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then lo
IgorC [24]

ANSWER:

F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.

STEP-BY-STEP EXPLANATION:

F(h) is Horizontal Force = 200 N

V is Speed = 2.4 m/s

The total weight increase by 42%

coefficient of rolling friction decrease by 19%

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

F(h) = F(f)

F(h) = mg* u

m is mass

g is gravitational acceleration = 9.8 m/s^2

200 = mg*u

Since weight increases by 42% and friction coefficient decreases by 19%

New weight = 1+0.42 = 1.42 = (1.42*m*g)

New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u

F(h) = (0.81μ) (1.42 m g)

       = (0.81) (1.42) (μ m g)

       = (0.81) (1.42) (200)

       = 230 N

4 0
3 years ago
a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f
harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = \frac{d}{x}

⇒ \theta = tan^{-1} \frac{d}{x}

Now for the \DeltaBAC:

tan\theta = \frac{d + h}{x}

⇒ \theta = tan^{-1} \frac{d + h}{x}

Now, differentiating w.r.t x:

\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}]

For maximum angle, \frac{d\theta }{dx} = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

0 = \frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}

\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}

After solving the above eqn, we get

x = \sqrt{\frac{d}{d + h}}

The observer should stand at a distance equal to x = \sqrt{\frac{d}{d + h}}

4 0
3 years ago
A body is dropped from the roof of a 20 m high building by how much:
USPshnik [31]

Answer:

t = 2.01 s

Vf = 19.7 m/s

Explanation:

It's know through the International System that the earth's gravity is 9.8 m/s², then we have;

Data:

  • Height (h) = 20 m
  • Gravity (g) = 9.8 m/s²
  • Time (t) = ?
  • Final Velocity (Vf) = ?

==================================================================

Time

Use formula:

  • \boxed{t=\sqrt{\frac{2*h}{g}}}

Replace:

  • \boxed{t=\sqrt{\frac{2*20m}{9.8\frac{m}{s^{2}}}}}

Everything inside the root is solved first. So, we solve the multiplication of the numerator:

  • \boxed{t=\sqrt{\frac{40m}{9.8\frac{m}{s^{2}}}}}

It divides:

  • \boxed{t=\sqrt{4.08s}}

The square root is performed:

  • \boxed{t=2.01s}

==================================================================

Final Velocity

use formula:

  • Vf = g * t

Replace:

  • Vf = 9.8 m/s² * 2.01 s

Multiply:

  • Vf = 19.7 m/s

==================================================================

How long does it take to reach the ground?

Takes time to reach the ground in <u>2.01 seconds.</u>

How fast does it hit the ground?

Hits the ground with a speed of <u>19.7 meters per seconds.</u>

7 0
3 years ago
13 points and brainlyest if possible. Thanks.
nikdorinn [45]
Most likely it would be C not completely sure 
3 0
3 years ago
Read 2 more answers
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