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Zina [86]
3 years ago
9

A 878-kg (1940 lb) dragster, starting from rest, attains a speed of 25.9 m/s (57.9 mph) in 0.62 s. (a) Find the average accelera

tion of the dragster during this time interval?
Physics
1 answer:
salantis [7]3 years ago
7 0

Answer:

41.8m/s^2

Explanation:

Since the dragster starts from rest, initial velocity (u) = 0m/s, final velocity (v) = 25.9m/s, time (t) = 0.62s

From the equations of motion, v = u + at

a = (v - u)/t = (25.9 - 0)/0.62 = 25.9/0.62 = 41.8m/s^2

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stich3 [128]
We have: v i (initial velocity) = 6 m/sv = 1.1 m/sa = - 9.8 m/s²v = v i + a · t1.1 m/s = 6 m/s - 9.8 m/s² t9.8 t = 6 - 1.19.8 t = 4.9t = 4.9 : 9.8t = 0.5 sThen the replacement:x = xi + vi · t + a t² / 2( xi = 0 )x = 6 · 0.5 - 9.8 · 0.25 / 2x = 3 - 1.225Answer:
x = 1.775 m

4 0
3 years ago
Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 22oC and exits at 2.0
KonstantinChe [14]

Answer:

The  the quality of the refrigerant at the exit of the expansion valve is 0.179.

Explanation:

Given that,

Initial pressure = 10 bar

Temperature = 22°C

Final pressure = 2.0 bar

We using the value of h

h = 293.4\ kJ/kg

The refrigerant during expansion undergoes a throttling process

Therefore, h_{1}=h_{2}

We need to calculate the quality of the refrigerant at the exit of the expansion valve

At 2.0 bar,

The property of ammonia

h_{f}=47.8 kJ/kg

h_{g}=1417.7 kJ/kg

Using formula

h_{2}=h_{f}+x(h_{g}-h_{f})

Put the value into the formula

293.4 =47.8+x(1417.7-47.8)

x=\dfrac{293.4-47.8}{1417.7-47.8}

x=0.179

Hence, The  the quality of the refrigerant at the exit of the expansion valve is 0.179.

6 0
3 years ago
Reference frame definitely changes when what also changes?
kompoz [17]

Explanation is in the file

tinyurl.com/wpazsebu

8 0
2 years ago
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A 230-km-long high-voltage transmission line 2.0 cm in diameter carries a steady current of 1,100 A. If the conductor is copper
Molodets [167]

Answer:

28.23 years

Explanation:

I = 1100 A

L = 230 km = 230, 000 m

diameter = 2 cm

radius, r = 1 cm = 0.01 m

Area, A = 3.14 x 0.01 x 0.01 = 3.14 x 10^-4 m^2

n = 8.5 x 10^28 per cubic metre

Use the relation

I = n e A vd

vd = I / n e A

vd = 1100 / (8.5 x 10^28 x 1.6 x 10^-19 x 3.14 x 10^-4)

vd = 2.58 x 10^-4 m/s

Let time taken is t.

Distance = velocity x time

t = distance / velocity = L / vd

t = 230000 / (2.58 x 10^-4) = 8.91 x 10^8 second

t = 28.23 years

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3 years ago
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What is the definition of energy ​
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Energy (in Physics) is the ability to do work.

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