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Pepsi [2]
3 years ago
10

Powerful sports cars can go from zero to 25 m/s (about 60 mph) in 6 seconds. What is the magnitude of the acceleration, includin

g units
Physics
1 answer:
Vikentia [17]3 years ago
7 0

Answer:

The magnitude of the acceleration is 4.2 m/s²

Explanation:

Given;

initial velocity of the powerful sports car, u = 0

final velocity of the powerful sports car, v = 25 m/s

time of motion, t = 6 seconds

Acceleration = Δv / Δt

Δv is change in velocity

Δt is change in time

The magnitude of the acceleration of the powerful sports car during the motion is given by;

a = \frac{v-u}{t}\\\\a = \frac{25-0}{6}\\\\a = 4.2 \ m/s^2

Therefore, the magnitude of the acceleration is 4.2 m/s²

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A(n) 131 g ball is dropped from a height
larisa [96]

Answer:

26.59 N/m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 131 g

Extention (e) = 4.82755 cm

Acceleration due to gravity (g) = 9.8 m/s²

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Next, we shall convert 131 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

131 g = 131 g × 1 Kg / 1000 g

131 g = 0.131 Kg

Thus, 131 g is equivalent to 0.131 Kg.

Next, we shall the force exerted by the ball on the spring. This can be obtained as follow:

Mass (m) = 0.131 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = ma

F = 0.131 × 9.8

F = 1.2838 N

Next, we shall convert 4.82755 cm to metre (m)

This can be obtained as follow:

100 cm = 1 m

Therefore,

4.82755 cm = 4.82755 cm × 1 m / 100 cm

4.82755 cm = 0.0482755 m

Thus, 4.82755 cm is equivalent to 0.0482755 m

Finally, we shall determine the spring constant as follow:

Force (F) = 1.2838 N

Extention (e) = 0.0482755 m

Spring constant (K) =?

F = Ke

1.2838 = K × 0.0482755

Divide both side by 0.0482755

K = 1.2838 / 0.0482755

K = 26.59 N/m

Thus the spring constant is 26.59 N/m

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