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Over [174]
3 years ago
15

A car in motion has kinetic energy. What happens to the kinetic energy when the car brakes to a stop? The kinetic energy is tran

sformed into potential energy ready to be released when the car starts back up. The kinetic energy is lost since energy is not conserved. The kinetic energy is transformed into heat energy evident by the warmth of the tires and is dissipated. The kinetic energy is still there, it is just not observable.
Physics
1 answer:
Pie3 years ago
7 0

Answer:

<em>When a moving car brakes to a stop the </em><em>kinetic energy of the car is converted to heat energy. </em>

Explanation:

A moving car has kinetic energy.

It is given by the equation k=\frac{1}{2} mv^2

Where m denotes mass of the car and v denote sits velocity. When the brakes are applied the velocity becomes zero and the car doesn’t possess kinetic energy anymore.

According to law of conservation of energy can neither be created nor be destroyed but can only be transformed from one form to another. On coming to a stop,  the kinetic energy of the car  gets converted to heat. The friction between the tyre and the road heats up the tyre.

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Statement A: Area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.
pochemuha

Explanation:

Formula for calculating the area of a  rectangle A = Length *width

For statement A;

Given area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.

Area of the rectangle = 2.536mm * 1.4mm

Area of the rectangle = 3.5504mm²

The rule of significant figures states that we should always convert the answer to the least number of significant figure amount the given value in question. Since 1.4mm has 2 significant figure, hence we will convert our answer to 2 significant figure.

Area of the rectangle = 3.6mm² (to 2sf)

For statement B;

Given area of a rectangle with measured length = 2.536 mm and width = 1.41 mm.

Area of the rectangle = 2.536mm * 1.41mm

Area of the rectangle = 3.57576mm²

Similarly, Since 1.41mm has 3 significant figure compare to 2.536 that has 4sf, hence we will convert our answer to 3 significant figure.

Area of the rectangle = 3.58mm² (to 3sf)

Based on the conversion, it can be seen that 3.6mm²  is greater than 3.58mm², hence the area of rectangle in statement A is greater than the area of the rectangle in statement B.

7 0
2 years ago
A 57 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. The acceleration of
kari74 [83]

Answer:

Her altitude as she crosses the bar, h₂ is approximately 6.1 m

Explanation:

The given parameters of the motion of the pole vaulter are;

The mass of the pole vaulter, m = 57 kg

The speed with which the pole vaulter is running, u = 11 m/s

The speed of the pole vaulter when she crosses the bar, v = 1.1 m/s

The acceleration due to gravity, g = 9.8 m/s²

From the total mechanical energy, M.E. equation, we have;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy of the motion = m·g·h

K.E. = The kinetic energy of the motion = 1/2·m·v²

By the principle of conservation of energy, we have;

The change (loss) in kinetic energy, ΔK.E. = The change (gain) in potential energy, ΔP.E.

ΔK.E. = 1/2·m·(v² - u²)

ΔP.E. = m·g·(h₂ - h₁)

Where;

h₁ = The ground level = 0 m

h₂ = The altitude with which she crosses the bar

∴ 1/2·m·(v² - u²) = m·g·(h₂ - h₁)

(h₂ - h₁) = (v² - u²)/(2·g) = (11² - 1.1²)/(2·9.8) = 6.11173469388

h₂ = 6.11173469388 + h₁ = 6.11173469388 + 0 = 6.11173469388

h₂ = 6.11173469388

Her altitude as she crosses over the bar, h₂ ≈ 6.1 m.

3 0
3 years ago
He graph below shows the velocity f(t) of a runner during a certain time interval:
Annette [7]
<span>The following that describes the intercepts on the graph is "The initial velocity of the runner was 4 m/s, and the runner stopped after 8 seconds." It is because the starting point of the line is at 4 and then the ending point is at 8.

</span>
5 0
2 years ago
Read 2 more answers
Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be
avanturin [10]

Answer:

The mass rate of the cooling water required is: 1'072988.5\frac{kg}{h}

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

{m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}

Where w refers to the cooling water and s to the steam flow. Reorganizing,

m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}

Write the difference of enthalpy for water as Cp (Tout-Tin):

m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, {h_{s}}^{out}=251.40\frac{kJ}{kg} and {h_{s}}^{in} can be calculated as:

{h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}.

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 \frac{kJ}{kgK}. If the average temperature is actually different, it won't mean a considerable mistake. Also we know that {T_{w}}^{out}-{T_{w}}^{in}\leq 10, so let's work with the limit case, which is {T_{w}}^{out}-{T_{w}}^{in}=10 to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data:

m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}

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5 0
3 years ago
Climatology is a subspecialty of what?
m_a_m_a [10]

 Climatology is a subspecialty of Meterology.
3 0
2 years ago
Read 2 more answers
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