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erastova [34]
2 years ago
9

PLEASE HELP ME PLS PLS Imagine an object held at some height above the ground. It is released and falls toward the ground. Ignor

ing air resistance or friction, which of the following must be true?
a
Before it falls, all of its energy must be potential energy
b
At the end of its fall, all of its energy must be converted to kinetic energy
c
During its fall, the combination of its kinetic and potential energy must equal the amount of potential energy with which it started
d
Total energy = Kinetic Energy + Potential Energy
e
All of the above
Physics
2 answers:
Stels [109]2 years ago
8 0

Option e is true. The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

<h3>What is the law of conservation of energy?</h3>

According to the Law of conservation of energy. Although energy cannot be generated or destroyed, it may be transferred from one form to another.

The following statements are true;

a)All of its energy must be potential energy before it falls.

b)At the conclusion of its fall, all of its energy must be transformed to kinetic energy.

c)During its fall, the sum of its kinetic and potential energy must match the initial quantity of potential energy.

d)Total energy = Kinetic Energy + Potential Energy.

Hence, option e is correct.

To learn more about the law of conservation of energy, refer ;

brainly.com/question/2137260

#SPJ2

iragen [17]2 years ago
8 0
I think it’s E not due
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TEA [102]

Answer:

4 m/s or 4 meters per second.

Explanation:

In order to calculate the speed of wave, you multiply the wavelength in meters and the frequency of the Wave in Hertz. 2 times 2 equals 4. The wave speed is always in m/s considering that the wavelength is also in meters.

7 0
3 years ago
What property of light waves does the michelson-morley interferometer directly demonstrate?
yan [13]
The wave nature of light, due to the experiment having bright and dark bands corresponding to places where you have constructive and destructive interference.
8 0
3 years ago
The driver of a car slams on the brakes, causing the car to slow down at a rate of
sdas [7]

Answer:

A. The time taken for the car to stop is 3.14 secs

B. The initial velocity is 81.64 ft/s

Explanation:

Data obtained from the question include:

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Final velocity (V) = 0

Time (t) =?

Initial velocity (U) =?

A. Determination of the time taken for the car to stop.

Let us obtain an express for time (t)

Acceleration (a) = Velocity (V)/time(t)

a = V/t

Velocity (V) = distance (s) /time (t)

V = s/t

a = s/t^2

Cross multiply

a x t^2 = s

Divide both side by a

t^2 = s/a

Take the square root of both side

t = √(s/a)

Now we can obtain the time as follow

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Time (t) =..?

t = √(s/a)

t = √(256/26)

t = 3.14 secs

Therefore, the time taken for the car to stop is 3.14 secs

B. Determination of the initial speed of the car.

V = U + at

Final velocity (V) = 0

Deceleration (a) = –26ft/s2

Time (t) = 3.14 sec

Initial velocity (U) =.?

0 = U – 26x3.14

0 = U – 81.64

Collect like terms

U = 81.64 ft/s

Therefore, the initial velocity is 81.64 ft/s

7 0
3 years ago
what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe
ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2

Hence, this is the required solution.

8 0
3 years ago
Read 2 more answers
A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle
Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of <em>θ</em> = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height <em>y</em> at time <em>t</em> would be

<em>y</em> = (40 m/s) <em>t</em> - 1/2 <em>g t</em> ²

Set <em>y</em> = 160 m, and you'll find that there is no (real) solution for<em> t</em>, so the ball never attains the given maximum height.

From another perspective: recall that

<em>v </em>² - <em>v</em>₀² = 2<em>a </em>∆<em>y</em>

where

• <em>v</em>₀ = initial velocity

• <em>v</em> = final velocity

• <em>a</em> = acceleration

• ∆<em>y</em> = displacement

At its maximum height, the ball has zero vertical velocity, and ∆<em>y</em> = maximum height = 160 m. The ball is in free fall once it's launched, so <em>a</em> = -<em>g</em>.

So we have

0² - (40 m/s)² = -2<em>g </em>(160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0
3 years ago
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