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2 years ago

8. A 740 kg car traveling 19 m/s comes to a complete stop in 2.0 s. What is the force exerted on the car during this stop?

2 answers:

6 0

Force = mass * acceleration

acceleration = change_in_velocity / time

so:

force = 740 kg * (19 m/s - 0 m/s) / 2.0 s
= 740 * 19 / 2 kg m per second^{2}
= 7030 kg m per second^{2}
= 7030 newtons of force

svet-max [94.6K]2 years ago

3 0

Answer:

Force exerted on the car is 7030 N.

Explanation:

It is given that,

Mass of the car, m = 740 kg

Initial speed of the car, u = 19 m/s

Final speed of the car, v = 0

Time taken, t = 2 s

Let F is the force exerted on the car during this stop. We know that it is equal to the product of force and acceleration. Mathematically, it is given as :

$F=m\times \dfrac{v-u}{t}$

$F=740\times \dfrac{0-19}{2}$

F = -7030 N

So, the force exerted on the car during this stop is 7030 N. Hence, this is the required solution.

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A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o

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Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

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g is the gravitational constant whose value is 9.8 m/s^2
h is the height which is to be calculated

$P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m$

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a-3

By the relation of volume and density

$M=\rho \times V$

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Part b

Pressure variation is given as

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Now corrected pressure is as

$P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa$

Finding the value of height for this corrected pressure as

$P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m$

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Answer:

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b) 2.3 m/s

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Now, when the boat is on the river, the velocity of the boat will be equal to the velocity of the boat in still water plus the velocity of the river.

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