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Goryan [66]
3 years ago
5

You lower a 2.50 kg textbook (remember when textbooks used to be made out of paper instead of being digital?) from a height of 1

.85 m to 1.50 m. What is its change in potential energy?
Physics
1 answer:
aev [14]3 years ago
6 0

Answer:

Change in potential energy = 8.58 J

Explanation:

Potential energy = Mass x Acceleration due to gravity x Height

Potential energy = mgh

m = 2.50 kg

g = 9.81 m/s²

Initial height, h₁ = 1.85 m

Final height, h₂ = 1.50 m

Change in potential energy = mgh₁  - mgh₂

Change in potential energy = 2.5 x 9.81 x 1.85 - 2.5 x 9.81 x 1.50

Change in potential energy = 8.58 J

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What does wave speed have to do with all SONAR AND RADAR technologies?
Elena L [17]

Explanation:

1) Radar uses radio waves, which are a type of electromagnetic energy. Sonar uses the echo principle by sending out sound waves underwater or through the human body to locate objects. Sound waves are a type of acoustic energy. Because of the different type of energy used in radar and sonar, each has its own applications.

2)Radar systems operate using radio waves primarily in air, while sonar systems operate using sound waves primarily in water (Minkoff, 1991). Despite the difference in medium, similarities in the principles of radar and sonar can frequently result in technological convergence.

4 0
2 years ago
Read 2 more answers
If a high jumper needs to make his center of gravity rise 1.50 m, how fast must he be able to sprint? Assume all of his kinetic
sergiy2304 [10]

Answer:

v = 5.42 m/s

Explanation:

given,

height of the jumper = 1.5 m

velocity of sprinter = ?

kinetic energy can be transformed into potential energy

m g h = \dfrac{1}{2}mv^2

g h = \dfrac{1}{2}v^2

v =\sqrt{2gh}

v =\sqrt{2\times 9.8 \times 1.5}

v = 5.42 m/s

Speed of the sprinter is equal to v = 5.42 m/s

7 0
3 years ago
At a certain elevation, the ________ , the air becomes saturated and water-vapor molecules ________.
andriy [413]

At certain altitude, the temperature of air decrease, The air becomes saturated and water vapour molecules starts condensing.


As the altitude of air increase, the atmospheric pressure decrease due to which the temperature of the air decrease. The water molecules in the atmosphere start condensing, which saturate the air (that is air can no hold water molecules), due to which the water vapour molecules starts condensing and falls on the earth in the form of rain.


4 0
3 years ago
A 36.3 kg cart has a velocity of 3 m/s. How much kinetic energy does the object have?
uysha [10]

Answer:

163.35

__________________________________________________________

<u>We are given:</u>

Mass of the object (m) = 36.3 kg

Velocity of the object (v) = 3 m/s

<u>Kinetic Energy of the object:</u>

We know that:

Kinetic Energy = 1/2(mv²)

KE = 1/2(36.3)(3)²            [replacing the variables with the given values]

KE = 18.15 * 9

KE = 163.35 Joules

Hence, the cart has a Kinetic Energy of 163.35 Joules

7 0
2 years ago
As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i
11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

Explanation:

Given that,

Height = 1.94  m

Bounced height = 1.48 m

Time interval t=14.86\times10^{-3}\ s

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

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

Put the value into the formula

9.8\times1.94=\dfrac{1}{2}\times v^2

v=\sqrt{2\times9.8\times1.94}

v=6.166\ m/s

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2'

Put the value into the formula

9.8\times1.48=\dfrac{1}{2}\times v^2'

v'=\sqrt{2\times9.8\times1.48}

v'=5.38\ m/s

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

a=\dfrac{v-v'}{t}

Put the value into the formula

a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}

a=776.98\ m/s^2

a=0.77\times10^{3}\ m/s^2

Hence,The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

6 0
3 years ago
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