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2 years ago

6

A racing car has a mass of 1720 kg. the acceleration of gravity is 9.8 m/s 2 . what is its kinetic energy if it has a speed of 1

29 km/h?

1 answer:

lesantik [10]2 years ago

3 0

Is this to late to answer?

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What unit can be used to measure the mass of a grain of salt

Anna007 [38]

milligrams

Explanation:

The best unit to measure the mass of a grain of salt is the milligram.

This is about; a thousandth of a gram.

A grain of salt is a very small particle size that can still be visible with the eye.

It has a very low and small mass.

For substances like this, we use the milligram:

1000milligram = 1g

learn more:

Mass brainly.com/question/4736384

#learnwithBrainly

5 0

2 years ago

A pair of toy freight cars, one twice the mass of the other, fly apart when a compressed spring that joins them is released. Acc

saul85 [17]

Answer:

greater acceleration is experienced by the car with lower mass

Explanation:

Since both the toys are connected by same spring so the force due to spring on both the toys will be same and it is given as

$F = kx$

now we know by Newton's II law

$F = ma$

so here we have

$a = \frac{F}{m}$

here we have same force on both the blocks

so acceleration will be more if mass is less

so greater acceleration is experienced by the car with lower mass

8 0

3 years ago

what is the property of an object due to its mass by which it resists any change in its position unless overcome by force ?

UkoKoshka [18]

Answer:

Mass as a Measure of the Amount of Inertia

All objects resist changes in their state of motion. All objects have this tendency - they have inertia.

Explanation:

hope this helps

5 0

2 years ago

A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the

BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

$F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}$ I'm going to do the math on the top and then on the bottom and divide at the end.

$F_g=\frac{2.4012*10^{40}}{3.721*10^{23}}$ and now when I divide I will express my answer to the correct number of sig dig's:

$Fg=$ 6.45 × 10¹⁶ N

8 0

2 years ago

A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c

Nady [450]

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

$I= \Delta p = m \Delta v = m (v-u)$

where

m is the mass of the object

$\Delta v$ is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

$I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s$

(b) 155 N

The impulse can also be rewritten as

$I=F \Delta t$

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

$\Delta t$ is the duration of the collision

In this situation, we have

$\Delta t = 0.06 s$

So we can re-arrange the equation to find the magnitude of the average force:

$F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N$

6 0

2 years ago