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4 years ago

8

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity (in m/s) must a basket ball player l

eave the grownd to rist 1.24 m above the floor in an attempt to get the ball?

1 answer:

mamaluj [8]4 years ago

7 0

Answer:4.93 m/s

Explanation:

Given

height to reach is (h )1.24 m

here Let initial velocity is u

using equation of motion

$v^2-u^2=2ah$

here Final Velocity v=0

a=acceleration due to gravity

$0-u^2=2\left ( -g\right )h$

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.81\times 1.24}$

$u=\sqrt{24.328}$

u=4.93 m/s

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A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?

Alex17521 [72]

<h2>Answer:53.63 $ms^{-2}$ </h2>

Explanation:

The equations of motion used in this question is $v=u+at$

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of $9.8ms^{-2}$ .

In X-Direction,

Given that initial velocity= $u_{x}$ = $33.8ms^{-1}$

Using $v=u+at$ ,

$v_{x}=33.8+(0)4.25=33.8ms^{-1}$

In Y-Direction,

Given that initial velocity= $u_{x}$ = $0ms^{-1}$

Using $v=u+at$ ,

$v_{y}=0+(9.8)4.25=41.65ms^{-1}$

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$v=\sqrt{1142.44+1734.72}=\sqrt{2877.163}=53.63ms^{-1}$

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4 years ago

1.<br>What is measurement?<br>

hoa [83]

Answer:

the action of measuring something.

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3 years ago

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with

Misha Larkins [42]

Answer:

Sam's and Abigail speeds before colliding were a. 12.34 m/s and b. 2.86 m/s, respectively. Their total kinetic energy was diminished by c. 1484.42 J, approximately

Explanation:

By conservation of momentum, we have

$\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.$

Writing for each direction at a time,

$m_S v_S = 7m_S \cos{(40)} +9.4 m_A \cos{(18)}\\v_S = 7 \cos{(40)} + 9.4 \frac{m_A}{m_S} \cos{(18)} = 7 \cos{(40)} + 9.4\times\frac{57}{73} \cos{(18)} \approx \mathbf{12.34}\\m_A v_A = 7m_S\sin{(40)} - 9.4m_A\sin{(18)}\\v_A = 7\frac{m_S}{m_A}\sin{(40)} - 9.4\sin{(18)} = 7\times\frac{73}{57}\sin{(40)} - 9.4\sin{(18)} \approx \mathbf{2.86}.$

Their kinetic energy changed by

$K_f - K_i = \left( \frac{1}{2}m_s v_{fs}^2 + \frac{1}{2}m_a v_{fa}^2 \right) - \left( \frac{1}{2}m_s v_{is}^2 + \frac{1}{2}m_a v_{ia}^2 \right) = \left( \frac{1}{2}\times 73 \times 7^2 + \frac{1}{2}\times 57 \times 9.4^2 \right) - \left( \frac{1}{2}\times 73 \times 12.34^2 + \frac{1}{2}\times 57 \times 2.86^2 \right) \approx \mathbf{-1484.418 J}.$

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3 years ago

For the different values given for the radius of curvature RRR and speed vvv, rank the magnitude of the force of the roller-coas

nirvana33 [79]

Explanation:

The force of the roller-coaster track on the cart at the bottom is given by :

$F=\dfrac{mv^2}{R}$ , m is mass of roller coaster

Case 1.

R = 60 m v = 16 m/s

$F=\dfrac{(16)^2m}{60}=4.26m\ N$

Case 2.

R = 15 m v = 8 m/s

$F=\dfrac{(8)^2m}{15}=4.26m\ N$

Case 3.

R = 30 m v = 4 m/s

$F=\dfrac{(4)^2m}{30}=0.54m\ N$

Case 4.

R = 45 m v = 4 m/s

$F=\dfrac{(4)^2m}{45}=0.36m\ N$

Case 5.

R = 30 m v = 16 m/s

$F=\dfrac{(16)^2m}{30}=8.54m\ N$

Case 6.

R = 15 m v =12 m/s

$F=\dfrac{(12)^2m}{15}=9.6m\ N$

Ranking from largest to smallest is given by :

F>E>A=B>C>D

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3 years ago

An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should happen to break

mamaluj [8]

Answer:

Value that the spring constant k = 12Mg / h

Explanation:

According to 2nd law of Newton:

upward force of the spring= F

The weight of the elevator W = mg

F = Mg = M(5g)

==> F =6Mg.

As the spring is compressed to its maximum distance ie s,the maximum upward acceleration comes just , Hence

F =ks = 6Mg

==> s = 6Mg/k

We have gravitational potential energy turning into elastic potential of the spring as the elevator starts at the top some distance h from the spring, and undergoes a total change in height equal to h + s, so:

Mg(h+s) = 1/2ks2

And plugging in our expression for s:

Mg(h+6Mg/k)= 1/2k(6Mg / k)2

gh + 6M2g2/k = 1/2k(36M2g2 /k2)

Mgh +6M2g2/k = 1/2k(36M2g2 /k2)

gh + 6Mg2/k = 18Mg2 / k

gh = 12Mg2 / k

h = 12Mg / k

k = 12Mg / h

7 0

4 years ago

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