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3 years ago

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An event, a person, a deed, or an object can have a long-lasting or momentary effect. How might you determine if the effect has

the potential to be long-lasting?

1 answer:

Amanda [17]3 years ago

7 0

I found a website called Socratic message me and I will tell you

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The mass of the skier, including his equipment, is 75kg. In the ski race, the total vertical

diamong [38]

The gravitational potential energy

gpe = mgh

$\tt gpe=75\times 9.8\times880=646,800~J$

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2 years ago

Pls help i’ll give brainliest if you give a correct answer!!

arsen [322]

Answer:

C. The distance traveled by an object at a certain velocity.

Explanation:

YW!

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3 years ago

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A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus

babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

$I = m\cdot (\vec{v}_{2} - \vec{v_{1}})$ (1)

Where:

$I$ - Impulse, in kilogram-meters per second.

$m$ - Mass, in kilograms.

$\vec{v_{1}}$ - Initial velocity of the hockey park, in meters per second.

$\vec{v_{2}}$ - Final velocity of the hockey park, in meters per second.

If we know that $m = 0.2\,kg$ , $\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right]$ and $\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right]$ , then the impulse applied by the stick to the park is approximately:

$I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]$

$I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]$

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.

8 0

3 years ago

A 12,000kg. railroad car is traveling at +2m/s when it

ivann1987 [24]

<u>Answer:</u>

The final velocity of the two railroad cars is 1.09 m/s

<u>Explanation:</u>

Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by

$\mathrm{V}=\frac{V 1 M 1+V 2 M 2}{M 1+M 2}$

where

V= Final velocity

M1= mass of the first object in kgs = 12000

M2= mas of the second object in kgs = 10000

V1= initial velocity of the first object in m/s = 2m/s

V2= initial velocity of the second object in m/s = 0 (given at rest)

Substituting the given values in the formula we get

V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s

$\mathrm{V}=\frac{2 \times 1200+0 \times 1000}{12000+10000}=\frac{24000}{22000}=1.09 \mathrm{m} / \mathrm{s}$

Which is the final velocity of the two railroad cars

8 0

3 years ago

A sled of mass 10 kg slides along the ice. it has an initial speed of 2 m/s but stops because of friction. How much work is done

NeX [460]

Answer: The correct answer is option B.

Explanation:

Mass of the sled = 10 kg

Initial speed of the sled = 2 m/s

Kinetic energy of the sled = $\frac{1}{2}mv^2$

$\frac{1}{2}\times 10 kg\times (2 m/s)^2=20 Joules$

Work done by the sled = 20 joules

The work done by the friction will be in opposite direction and equal to the magnitude of the work done of the sled that - 20 J.

Hence, correct answer is option B.

6 0

3 years ago

Read 2 more answers