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Elenna [48]
3 years ago
6

Plzz answer this questions will mark as brainlist ​

Physics
1 answer:
EastWind [94]3 years ago
4 0

Answer:

1. 19.28 secs

2. 154.22 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

Force (F) = 1000 N

Mass (m) = 1200 Kg

Time (t) =..?

Distance (s) =...?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Force (F) = 1000 N

Mass (m) = 1200 Kg

Acceleration (a) =.?

Force (F) = mass (m) x acceleration (a)

F = ma

1000 = 1200 x a

Divide both side by 1200

a = 1000/1200

a = 0.83 m/s²

Since the car is coming to rest, it means it is decelerating. Therefore, the acceleration is – 0.83 m/s²

1. Determination of time taken for the car to halt i.e stop. This can be obtained as follow:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

acceleration (a) = – 0.83 m/s²

Time (t) =.?

v = u + at

0 = 16 + (–0.83 x t)

0 = 16 – 0.83t

Rearrange

0.83t = 16

Divide both side by 0.83

t = 16/0.83

t = 19.28 secs.

Therefore, the time taken for the car to halt is 19.28 secs.

2. Determination of the distance travelled by the car before coming to rest. This can be obtained as follow:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

acceleration (a) = – 0.83 m/s²

Distance (s) =..?

v² = u² + 2as

0 = 16² + (2 x –0.83 x s)

0 = 256 – 1.66s

Rearrange

1.66s = 256

Divide both side by 1.66

s = 256/1.66

s = 154.22 m

Therefore, the distance travelled by the car before coming to rest is 154.22 m.

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250Nm

Explanation:

Given parameters:

Length of the long pry bar  = 1m

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Angle  = 90°

Unknown:

Amount of torque applied  = ?

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Torque is the turning force on a body that causes the rotation of the body.

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Answer:

W = 5701 KW

Explanation:

From the question let inlet be labelled as point 1 and exit as point 2, for the fluid steam, we can get the following;

Inlet (1): P1 = 40 bar ; T1 = 500°C and V1 = 200 m/s

Exit(2) : At saturated vapour; P2 = 0.8 bar and V2 = 150 m/s

Volumetric flow rate = 15 m^(3)/s

Now, to solve this question, we assume constant average values, steeady flow and adiabatic flow.

Specific volume for steam at P2 = 0.8 bar in the saturated vapour state can be gotten from saturated steam tables(find a sample of the table attached to this answer).

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v2 = 2.087 m^(3)/kg

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We also know energy equation could be defined as;

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Since the flow is adiabatic, potential energy can be taken to be zero. Therefore, we get;

-W = m[(h2 - h1) + {(V2(^2) - (V1(^2)} /2)}

From, table 2, i attached , at P1 = 40 bar and T1 = 500°C; specific enthalpy was calculated to be h1 = 3445.3 KJ/Kg

Likewise, at P2 = 0.8 bar; from the table, we get specific enthalpy as;

h2 = 2665.8 KJ/Kg

So we now calculate power developed;

W = - 7.17 [(2665.8 - 3445.3) + {(150^(2) - 200^(2))/2000 = 5701KW

Since the sign is not negative but positive, it means that the power is developed from the system.

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