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3 years ago

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To catch a fast-moving ball, you extend your hand forward before contact with the ball and let it ride backward in the direction

of the ball's motion. Doing this reduces the force of contact on your hand principally because the Group of answer choices force of contact is reduced relative velocity is less time of contact is increased time of contact is decreased none of these

1 answer:

const2013 [10]3 years ago

7 0

Answer:

C: time of contact is increased

Explanation:

When a ball is falling from height, it maintains the same mass but velocity may change depending on the conditions in that environment.

Now, to catch the ball on your palm, you need to find a way to reduce the momentum of the ball upon hitting your hand.

Now, to reduce the momentum, we need to apply impulse.

This is because we know that;

Impulse(I) = F × t = change in momentum (m(v - u))

Thus;

I = Ft

F = I/t

We are told that extending the hand reduces the force of contact on your hand.

Thus;

For F which is the force to be reduced, the denominator (t) has to be increased.

Thus, the correct answer is that time of contact is increased

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A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s

strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

$m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v$

Where:

$m_{M}$ , $m_{S}$ - Masses of the small meteorite and the communication satellite, measured in kilograms.

$v_{M}$ , $v_{S}$ - Speeds of the small meteorite and the communication satellite, measured in meters per second.

$v$ - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

$v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}$

If $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ and $v_{S} = 0\,\frac{m}{s}$ , the final speed is now calculated:

$v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}$

$v = 0.1\,\frac{m}{s}$

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

$K_{S} + K_{M} - K - Q_{disp} = 0$

$Q_{disp} = K_{S}+K_{M}-K$

Where:

$K_{S}$ , $K_{M}$ - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

$K$ - Kinetic energy of the satellite-meteorite system, measured in joules.

$Q_{disp}$ - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

$Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]$

Given that $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ , $v_{S} = 0\,\frac{m}{s}$ and $v = 0.1\,\frac{m}{s}$ , the dissipated heat is:

$Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]$ $Q_{disp} = 200000\,J$

$Q_{disp} = 200\,kJ$

The energy coverted to heat is 200 kilojoules.

4 0

3 years ago

1 question 20points<br> How is frequency related to the sound we here

pishuonlain [190]

Frequency is the vibration of noise and the vibration determines the pitch, which we depend on to be a pitch or frequency we can hear. If it's too high or too low our ears can't hear it

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3 years ago

Before Collision Consider a system to be one train car moving toward another train car at rest When the train cars collide, the

erma4kov [3.2K]

Answer:

2,400kg * m/s

Explanation:

You are missing some information in the question but the rest could be found some where else.

The question gives the masses and starting velocity of each car.

Car 1: m = 600kg and sv = 4m/s

Car 2: m 400kg and sv = 0m/s

Find the momentum of both cars.

Car 1: 600 * 4 = 2400

Car 2: 400 * 0 = 0

Add both.

2400 + 0 = 2400

Best of Luck!

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3 years ago

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The images show two different types of Galapagos tortoises that scientists believe descended from the same species. The first ty

vichka [17]

Answer:

C. turtles with genes for long necks had a better chance of surviving to reach reproductive age.

Explanation:

The turtles that had long necks were more fit to the environmnet in which they were lovated and were able to grow larger and have more reproductive time because of their ability to feed on grass and small shrubs, this helped them always haev food available, and made them the dominant gene eventually.

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4 years ago

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A bullet fired vertically at a velocity of 36m/s .after 45 the bullet hit the top of a bulid how height is a bulid?

zzz [600]

Answer:

The height of the building is 8,302.5 m

Explanation:

Given;

velocity of the projectile, u = 36 m/s

time of motion, t = 45 s

Let the upward direction of the bullet be negative,

The height of the building is calculated as;

$h = ut - \frac{1}{2} gt^2\\\\h = (36\times 45) - (\frac{1}{2} \times 9.8 \times 45^2)\\\\h = 1620 - 9922.5\\\\h = -8,302.5 \ m\\\\The \ height \ of \ the \ building \ is \ 8,302.5 \ m$

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3 years ago