A ball is thrown straight up in the air and just before it lands it is travelling -47.5 m/s. How long was the ball in the air?

Why do you think it might be useful to know the distance between a gun fired and it's target?

inysia [295]

Distance plays an incredible factor, especially when taking into consideration especially long distances in terms of shooting a gun. You may be familiar with the Coriolis Effect, and how a bullet's trajectory will change in relation to the rotation of the Earth. The farther the distance, the greater the change in trajectory.

4 0

3 years ago

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As relay runner A enters the 65-ft-long exchange zone with a speed of 30 ft/s, he begins to slow down. He hands the baton to run

Hoochie [10]

Answer:

a_a = -3.2 ft/s^2 , a_b = 3.723 ft/s^2

t = 5.909 s

Explanation:

Given:

- Initial velocity of A v_i,a = 30 ft /s

- Initial distance s_o = 0

- Length of the exchange zone s_f = 65 ft

- Time taken t = 2.5 s

Start out by A's velocity as he gets to the end of the exchange zone.

Part a

Runner A decelerates at a uniform rate, so you can use the equation:

s_f = s_o + v_i,a*t + 0.5a_a*t^2

65 = 0 + 30*2.5 + 0.5*a_a*2.5^2

a_a = -20 / 2.5^2

a _a= -3.2 ft/s^2

Runner B accelerates at v_f,a as final velocity at a uniform rate, so you can use the equation:

v_f,b^2 - v_i,b^2 = 2*a_b*s

a_b = (30 -3.2*2.5)^2 /2*65

a_b = 3.723 ft/s^2

Part b

When Runner B should begin running:

t = (v_f,b - v_i,b) / a_b

t = (30 - 3.2*2.5) / 3.723

t = 5.909 s

5 0

3 years ago

The distance between two cities is 185miles.Express the distance in kilometers.

murzikaleks [220]

297.729km :-) hope this helps

8 0

3 years ago

A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

Olegator [25]

Answer:

366.90149 m/s

923.821735 J

324.734 J

Initial Kinetic energy > Final kinetic energy

Explanation:

$m_1$ = Mass of block = 0.072 kg

$m_2$ = Mass of bullet = 4.67 g

$u_1$ = Initial Velocity of block = 0

$u_2$ = Initial Velocity of bullet = 629 m/s

$v_1$ = Final Velocity of block = 17 m/s

$v_2$ = Final Velocity of bullet

In this system the linear momentum is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow v_2=\frac{m_{1}u_{1}+m_{2}u_{2}-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.072\times 0+4.67\times 10^{-3}\times 629-0.072\times 17}{4.67\times 10^{-3}}\\\Rightarrow v_2=366.90149\ m/s$

Final Velocity of bullet is 366.90149 m/s

The initial kinetic energy

$K_i=\frac{1}{2}m_2u_2^2\\\Rightarrow K_i=\frac{1}{2}4.67\times 10^{-3}\times 629^2\\\Rightarrow K_i=923.821735\ J$

The final kinetic energy

$K_f=\frac{1}{2}m_2v_2^2+\frac{1}{2}m_1v_1^2\\\Rightarrow K_f=\frac{1}{2}4.67\times 10^{-3}\times 366.90149^2+\frac{1}{2}0.072\times 17^2\\\Rightarrow K_f=324.734\ J$

Initial Kinetic energy > Final kinetic energy

3 0

3 years ago

Physical activity and exercise typically lead to

Dmitry [639]

Answer:

The answer is C (Higher energy levels)

Explanation:

Because it's C

8 0

3 years ago

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