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Crazy boy [7]
3 years ago
13

A box of mass 64 kg is at rest on a horizontal frictionless surface. A constant horizontal force F~ then acts on the box and acc

elerates it to the right. It is observed that it takes the box 5.4 seconds to travel 32 meters. What is the magnitude of the force?
Physics
1 answer:
iren2701 [21]3 years ago
3 0

Answer:

The correct answer is;

The magnitude of the force is 35.12 N

Explanation:

To solve the question, we note that the friction is zero and the force causes motion of a stationary mass

One of the equations of motion is required such as

v² = u² + 2× a× s

Where

v = Final velocity = 5.93 m/s

u = Initial velocity = 0 m/s , object at rest

a = acceleration

s = distance moved = 32 meters

But v = Distance/Time = 32 m /5.4 s = 5.93 m/s

Therefore

5.93² = 2×a×32

or a = 35.12/ 64 = 0.55 m/s²

Therefore Force F = Mass m × Acceleration a

Where mass m = 64 kg

Therefore F = 64 kg×0.55 m/s² = 35.12 N

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Answer:

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A Pitot-static tube is mounted on a 2.5 cm pipe where oil (???? = 860 kg/m3, ???? = 0.0103 kg/m·s) is flowing. The Pitot tube is
emmasim [6.3K]

Answer:

\dot{V}=0.0733 \,m^3.s^{-1}

Explanation:

Given:

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diameter of the pipe, d=2.5\times 10^{-2}m

pressure difference, \Delta P=95.8\times 10^{5}\,Pa

In case of  pitot tube, the velocity is given by:

v=\sqrt{\frac{2.\Delta P}{\rho} }

v=\sqrt{\frac{2\times 95.8\times 10^{5}}{860} }

v=149.26\,m.s^{-1}

Now we know that volumetric flow rate is given as:

\dot{V}=a.v

where :

a= cross sectional area of the pipe

v= velocity of flow

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3 years ago
What is the primary source of a sound?
brilliants [131]

Answer:

A vibration is the primary source of a sound.

8 0
4 years ago
49W
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Answer:

16

Explanation:

The magnitude of the electrostatic force between two charged particles is given by

F=\frac{kq_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the charges of the two particles

r is the separation between the particles

In this problem, the initial force between the particles is F.

Later, the distance between the two particles is increased by four, so

r' = 4r

So, the new force between the particles will be

F'=\frac{kq_1 q_2}{(4r)^2}=\frac{kq_1 q_2}{16r^2}=\frac{1}{16}F

So, the new force decreases by a factor of 16.

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Answer:

1.5 m/s

Explanation:

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