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beks73 [17]
4 years ago
11

The car has a constant deceleration of 4.20 m/s^2. If its initial velocity was 24.0 m/s, how long does it take to come to a stop

? Answer in s
Physics
1 answer:
nalin [4]4 years ago
5 0

Answer:

The time is 5.71 sec.

Explanation:

Given that,

Acceleration a= -4.20 m/s^2

Initial velocity = 24.0 m/s

We need to calculate the time

Using equation of motion

v = u+at[/tex]

Where, v = final velocity

u = inital velocity

t = time

a = acceleration

Put the value into the formula

0 =24.0 +(-4.20)\times t

t = \dfrac{-24.0}{-4.20}

t=5.71\ sec

Hence, The time is 5.71 sec.

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The anawer is D hope this helps!
4 0
3 years ago
In gases heat is transferred by?​
Sedaia [141]

Answer:

Convection

Convection transfers heat energy through gases and liquids. As air is heated, the particles gain heat energy allowing them to move faster and further apart, carrying the heat energy with them. Warm air is less dense than cold air and will rise.

4 0
3 years ago
It is proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of foil. W
Anvisha [2.4K]

Answer:

962291.57928 m²

Explanation:

P_r = Pressure = 2\dfrac{I}{c}  (full reflection)

I = Intensity = \dfrac{P}{A}=\dfrac{P}{4\pi r^2}

P = Power = 3.9\times 10^{26}\ W

c = Speed of light = 3\times 10^8\ m/s

M = Mass of Sun = 1.99\times 10^{30}\ kg

m = Mass of ship = 1500 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Force of radiation is given by

F_r=P_rA\\\Rightarrow F_r=2\dfrac{I}{c}\times A\\\Rightarrow F_r=2\dfrac{P}{4\pi r^2c} A

This force will balance the gravitational force as stated in the question

\dfrac{GMm}{r^2}=2\dfrac{P}{4\pi r^2c} A\\\Rightarrow A=\dfrac{4\pi cGMm}{2P}\\\Rightarrow A=\dfrac{4\times \pi\times 3\times 10^8\times 6.67\times 10^{-11}\times 1.99\times 10^{30}\times 1500}{2\times 3.9\times 10^{26}}\\\Rightarrow A=962291.57928\ m^2

The area of the must be 962291.57928 m²

3 0
3 years ago
a 2.5-kg object is dropped from a height of 1000 m. what is the force of air resistance on the object when it reaches terminal v
Sergio [31]

Answer:

24.5 N

Explanation:

The falling object experiences its weight acting downwards and the air resistance in the opposite direction.

The air resistance increases with velocity so there may come a point, depending on the shape of the object and if there is sufficient height, where these 2 forces are equal.

Since the object has no net forces acting on it it will, according to Newton, no longer accelerate but continue with a constant velocity.

This is called Terminal Velocity.

So:

Air resistance = weight

R = m g

R = 2.5 × 9.8  = 24.5N

3 0
4 years ago
Saturated steam at 125 kpa is compressed adiabatically in a centrifugal compressor to 700 kpa at the rate of 2.5 kg⋅s−1. the com
Tpy6a [65]
M° = 2.5 kg/sec
For saturated steam tables
at p₁ = 125Kpa
hg = h₁ = 2685.2 KJ/kg
SQ = s₁ = 7.2847 KJ/kg-k
for isotopic compression
S₁ = S₂ = 7.2847 KJ/kg-k
at 700Kpa steam with S = 7.2847
h₂ 3051.3 KJ/kg
Compressor efficiency
h =  0.78
0.78 = h₂ - h₁/h₂-h₁
0.78 = h₂-h₁ → 0.78 = 3051.3 - 2685.2/h₂ - 2685.2
h₂ = 3154.6KJ/kg
at 700Kpa with 3154.6 KJ/kg
enthalpy gives
entropy S₂ = 7.4586 KJ/kg-k
Work = m(h₂ - h₁) = 2.5(3154.6 - 2685.2
W = 1173.5KW
5 0
4 years ago
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