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ki77a [65]
4 years ago
9

An acrobat of mass mA jumps upwards off a trampoline with an initial speed v0. At a height h0, the acrobat grabs a clown of mass

mB. Assume that the time the acrobat takes to grab the clown is negligibly small. What is the maximum height hf reached by the acrobat and clown
Physics
1 answer:
SpyIntel [72]4 years ago
8 0

Answer:

Explanation:

Initial speed u = v₀

Final speed v at height h₀

v² = u² -2gh₀

v = \sqrt (v_0^2 -2gh_0)}

Applying conservation of momentum at height h₀

Velocity of combined mass upwards  V = m_A v / m_A+m_B

if H is the maximum height attained

H = V² /2g = m_A^2 (V_0^2-2gh_0) / 2g( m_A+m_B)^2

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True the only way to emit ur proton is if there are ground electrical
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Suppose a car is moving in a straight line and steadily increases its speed. It moves from 35 km/h to 40 km/h the first second a
snow_lady [41]

Answer:

c

Explanation:

the first 35 to 40 acceleration will be 5 same as 40 to 45 ,

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7 0
3 years ago
A racing car has a mass of 1525kg. What is its kinetic energy if it has a speed of 108km/h
Reptile [31]
E = 1/2 m v^2

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3 years ago
in a _system supply and demand forces affect the production and consumption decisions. There is little to no _control in such a
Alchen [17]

Answer:

in a free market system supply and demand forces affect the production and consumption decisions. There is little to no government control in such a system .

Explanation:

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5 0
4 years ago
A 22kg box slides down a 30.0 degree incline. If there is no friction, what is the acceleration of the box?
Rashid [163]

Answer:

Acceleration of the box is 4.9 m/s².

Explanation:

The free body diagram is shown below.

The weight of the body can be resolved into two mutually perpendicular components as shown. The force responsible for sliding motion of the box along the incline is due to mg\sin \theta acting along the incline.

Therefore, as per Newton's second law of motion,

Net force = mass\times acceleration

Now, net force acting along the incline is only mg\sin \theta

Therefore,

mg\sin \theta=ma\\a=g\sin \theta

Now, g=9.8\textrm{ and } \theta = 30

Therefore, the acceleration of the box is given as:

a=9.8\times \sin(30)=9.8\times 0.5=4.9\textrm{ }m/s^{2}

Acceleration of the box is 4.9 m/s².

6 0
4 years ago
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