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3 years ago

When the velocity of an object is doubled, by what factor is its momentum changed? By what factor is its kinetic energy changed?

Physics

1 answer:

anzhelika [568]3 years ago

4 0

Answer:

Kinetic energy would increase by a factor of 4 where as momentum would increase by a factor of 2.

Explanation:

Kinetic Energy is given by 0.5*mass*velocity^2. Kinetic Energy is proportional to Velocity^2.

Momentum is given by mass*velocity. Momentum is proportional to Velocity.

If the velocity of an object is doubled, Kinetic energy would increase by a factor of 2^2 i.e 4 times. Momentum would increase by a factor of 2.

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Simplify -2/3 - 3/5 A) -1/15 B) -19/15 C) -3/5 D) -5/8

Leto [7]

I got B,when you subtract 3/5 from NEGATIVE 2/3 it creates a negative 19 over a positive 15.

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3 years ago

Sunglasses that reduce glare take advantage of which kind of wave

marysya [2.9K]

Answer:

It should be option B polarization

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3 years ago

Read 2 more answers

Your friend is constructing a balancing display for an art project. She has one rock on the left ( ms=2.25 kgms=2.25 kg ) and th

Firdavs [7]

Answer:

Torque = 35.60 N.m (rounded off to 3 significant figures.

Explanation:

Given details:

The mass of the rock on the left, ms = 2.25 kg

The total mass of the rocks, mp = 10.1 kg

The distance from the fulcrum to the center of the pile of rocks, rp = 0.360 m

(a) The torque produced by the pile of rock, T = F*rp = m*g*rp

Torque = 9.8*0.360*10.1 = 35.6328

Torque = 35.60 N.m (rounded off to 3 significant figures).

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3 years ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

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3 years ago

Table C: Average Speeds for Lower Racetrack

Blababa [14]

Answer:

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= (circumference/time)^2/r

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= ((2³.14*50/14.3)²)/50

= 9.64 m/s²

brainlist?

Explanation:

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3 years ago