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Svetlanka [38]
4 years ago
15

a toy airplane is flying at a speed of 8 m/s with an acceleration of 0.9 m/s^2. How fast is it flying after 2 seconds?

Physics
1 answer:
Marat540 [252]4 years ago
4 0

Answer:

it  will be flying 1.8 m/s

Explanation:

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Students are completing a lab in which they let a lab cart roll down a ramp. The students record the mass of the cart, the heigh
Dennis_Churaev [7]

Answer:

second column

Explanation:

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3 years ago
State two uses of high specific heat of water​
Nitella [24]
  • It helps in regulating extremes in the environment.
  • It can be used as a car radiators.

<u>Explanation:</u>

  • The amount of heat needed to raise a certain amount of temperature to the water is known as the specific heat of the water. For one gram of water to increase 1 degree celsius, water needs to absorb 4.184 J of heat.

Some of the uses of high specific heat of water are as follows,

  • The temperature of the water will remain the same from day to night in ponds or sea. During day time, the hot air from the sea or pond rises while the cool air from the sea moves to take its place. During the night, the hot air from the land moves to take its place, giving rise to a land breeze and thus maintaining the environment.
  • Another example, When water is circulated throughout an engine, heat will get absorbed. This water is pumped to a radiator and then disposed to the surrounding air.

7 0
4 years ago
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Thepotemich [5.8K]

a) The launch velocity of the rocket is 5.48 m/s

b) The maximum height is 1.53 m

Explanation:

a)

We can solve this part by applying the law of conservation of energy, by considering the kinetic energy and the elastic potential energy only, since there is no change in gravitational potential energy and no friction is involved.

The total energy when the spring is compressed is:

E=KE_i + PE_{si}

with

KE_i = 0 (initial kinetic energy is zero)

PE_{si} = \frac{1}{2}kx^2 is the elastic potential energy stored in the spring, with

k = 450 N/m (spring constant)

x = 0.10 m (compression of the spring)

The total energy when the spring is relased is:

E=KE_f + PE_{sf}

with

KE_f = \frac{1}{2}mv^2 (final kinetic energy), with

m = 0.15 kg (mass of the rocket)

v = velocity of launch of the rocket

PE_{sf} = 0 (elastic potential energy is zero when the spring is released)

Combining the two equations we get

\frac{1}{2}kx^2 = \frac{1}{2}mv^2

And solving for v,

v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(450)(0.10)^2}{0.15}}=5.48 m/s

b)

In this part instead we consider only the kinetic energy and the gravitational potential energy, since the spring is at rest so its energy is now zero.

The total energy at the launch is:

E=KE_i + PE_{gi}

where

KE_i = \frac{1}{2}mv^2 (initial kinetic energy), with

m = 0.15 kg (mass of the rocket)

v = 5.48 m/s (velocity of launch of the rocket)

PE_{gi}=0 (initial gravitational potential energy is zero)

The total energy at the point of maximum height is:

E=KE_f + PE_{gf}

where

KE_f = 0 (kinetic energy is zero since speed is zero)

PE_{gf}=mgh (final gravitational potential energy), with

m = 0.15 kg

g=9.8 m/s^2 (acceleration of gravity)

h = ? (maximum height)

Combining the two equations we find

\frac{1}{2}mv^2 = mgh

And solving for h,

h=\frac{v^2}{2g}=\frac{(5.48)^2}{2(9.8)}=1.53 m

Learn more about potential energy and kinetic energy:

brainly.com/question/1198647

brainly.com/question/10770261

brainly.com/question/6536722

#LearnwithBrainly

3 0
3 years ago
II Force on a tennis ball. The record speed for a tennis ball that is served is 73.14 m/s. During a serve, the ball typically st
AveGali [126]

Answer:

F=248.5W N

Explanation:

Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:

a=\frac{\Delta x}{\Delta t}=\frac{v-v_0}{t-t_0}

Since we start counting at 0s and the ball departs from rest, this is just a=\frac{v}{t}

So we can write:

F=ma=\frac{mv}{t}=\frac{gmv}{gt}

Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:

F=\frac{Wv}{gt}=\frac{v}{gt}W

Substituting our values:

F=\frac{(73.14m/s)}{(9.81m/s^2)(30\times10^{-3}s)}W=248.5W N

Where the average force exerted has been written it terms of the tennis ball's weight W.

8 0
3 years ago
I need help!!!! I don't understand the question
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Mgraphs are proportional ofrelationships
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