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2 years ago

A police officer draws a sketch of the scene of an accident, as shown.

Physics

2 answers:

iren2701 [21]2 years ago

5 0

I would have to say that it is Y

larisa86 [58]2 years ago

4 0

Answer:

Arrow Y

Explanation:

It is given that, the car was rounding the curve when it hit an oil slick and went off the road.

It is clear that the car is moving in a circular track. The force acting on a circular path is centripetal force. Also, the speed remains constant and the velocity changes in the case of circular motion. The direction of motion when the car leaves circular motion will be tangential.

Arrow Y shows the direction that the car moved in after hitting the oil slick.

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Roberto creates an animation titled “Flying through an Atom.” The animation makes it seem as if the viewer passes all the way th

vladimir2022 [97]

<span>The viewer travels through the electron cloud, then the nucleus, and then the electron cloud again.</span>

8 0

2 years ago

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Nathan accelerates his skateboard uniformly along a straight path from rest to 12.5 m/s in 2.5 s.

kicyunya [14]

Answer:

<h2>a) Nathan's acceleration is 5 m/s² </h2><h2>b) Nathan's displacement during this time interval is 15.625 m</h2><h2>c) Nathan's average velocity during this time interval is 6.25 m/s</h2>

Explanation:

a) We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Final velocity, v = 12.5 m/s

Time, t = 2.5 s

Substituting

v = u + at

1.25 = 0 + a x 2.5

a = 5 m/s²

Nathan's acceleration is 5 m/s²

b) We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 5 m/s²

Time, t = 2.5 s

Substituting

s = ut + 0.5 at²

s = 0 x 2.5 + 0.5 x 5 x 2.5²

s = 15.625 m

Nathan's displacement during this time interval is 15.625 m

c) Displacement = 15.625 m

Time = 2.5 s

We have

Displacement = Time x Average velocity

15.625 = 2.5 x Average velocity

Average velocity = 6.25 m/s

Nathan's average velocity during this time interval is 6.25 m/s

5 0

2 years ago

A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold

tino4ka555 [31]

$q_{c}$ = Heat released to cold reservoir

$q_{h}$ = Heat released to hot reservoir

$W_{max}$ = maximum amount of work

$t_{c}$ = temperature of cold reservoir

$t_{h}$ = temperature of hot reservoir

we know that

$\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}$

$q_{h} = (\frac{t_{h}}{t_{c}})q_{c}$ eq-1

maximum work is given as

$W_{max}$ = $q_{h}$ - $q_{c}$

using eq-1

$W_{max}$ = $(\frac{t_{h}}{t_{c}})q_{c}$ - $q_{c}$

6 0

2 years ago

160 students sit in an auditorium listening to a physics lecture. Because they are thinking hard, each is using 125 W of metabol

anastassius [24]

Answer:

minimum power should be used to operate the air conditioner is 4000 W

Explanation:

given data

students n = 160

power p = 125 W

COP = 5.0

to find out

what minimum power should be used

solution

we know the COP formula that is given below

COP = students × power / minimum power

minimum power = n × p / COP

put all value

minimum power = n × p / COP

minimum power = 160 × 125 / 5

minimum power = 4000 W

minimum power should be used to operate the air conditioner is 4000 W

8 0

2 years ago

Arrange the objects in order from least to greatest potential energy. Assume that gravity is constant. (Use PE = m × g × h.)

yawa3891 [41]

Answer:The objects in order from least to greatest potential energy:

3<1<4<2

Explanation:

1. Potential energy of a 15-kilogram stone found at a height of 3 meters:

$mgh=15 kg\times 9.8 m/s^2\times 3 m=441 J$

2. Potential energy 10 kilograms of water stored at a height of 9 meters:

$mgh=10 kg\times 9.8 m/s^2\times 9 m=882 J$

3. Potential energy 1-kilogram ball located 20 meters in the air:

$mgh=1 kg\times 9.8 m/s^2\times 20 m=196 J$

4. Potential energy box of books weighing 25 kilograms placed on a shelf 2 meters high:

$mgh=25 kg\times 9.8 m/s^2\times 2 m=490 J$

The objects in order from least to greatest potential energy:

3<1<4<2

8 0

2 years ago

Read 2 more answers