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Ksivusya [100]
3 years ago
10

A 5kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 8 N opposing the motion. Draw a fo

rce diagram and determine the net force. Then calculate the acceleration of the object.

Physics
1 answer:
Taya2010 [7]3 years ago
3 0

Answer:

Net force on the block is 32 N.

Acceleration of the object is 6.4 m/s².

Explanation:

Let the acceleration of the object be a m/s².

Given:

Mass of the block is, m=5\ kg

Force of pull is, F=40\ N

Frictional force on the block is, f=8\ N

The free body diagram of the object is shown below.

From the figure, the net force in the forward direction is given as:

F_{net}=F-f=40-8=32\ N

Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,

F_{net}=ma\\32=5a\\a=\frac{32}{5}=6.4\ m/s^2

Therefore, the acceleration of the object in the forward direction is 6.4 m/s².

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3 years ago
2 strings both vibrate at exactly 220 Hz. The tension in one of them is then decreased sightly. As a result, 3 beats per second
MAVERICK [17]

Explanation:

Given that,

2 strings both vibrate at exactly 220 Hz. The frequency of sound wave depends on the tension in the strings.

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So, the new frequency of the string is 217 Hz. Hence, this is the required solution.

3 0
3 years ago
A bicycle rim has a diameter of 0.65 m and a moment of inertia, measured about its center, of 0.21 kg⋅m2What is the mass of the
stepan [7]

Answer:

m = 1.99 kg = 2 kg

Explanation:

The moment of inertia of a bicycle rim about it's center is given by the following formula:

I = mr^{2}\\

where,

I = Moment of Inertia of the Bicycle Rim = 0.21 kg.m²

r = Radius of the Bicycle Rim = Diameter of the Bicycle Rim/2

r = 0.65 m/2 = 0.325 m

m = Mass of the Bicycle Rim = ?

Therefore,

0.21\ kg.m^{2} = m(0.325\ m)^{2}\\m = \frac{0.21\ kg.m^{2}}{(0.325\ m)^{2}}\\

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3 0
3 years ago
The most common units for expressing the density of a substance are the g/cm3.
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True.

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Which calculates the intensity of an electric field at a point where a 0.50 C charge experiences a force of 20. N?
tester [92]

Answer: 40\ N/C

Explanation:

Given

Magnitude of charge is q=0.5\ C

Force experienced is F=20\ N

Electric field intensity is the electrostatic force per unit charge

\therefore E=\dfrac{F}{q}\\\\\Rightarrow E=\dfrac{20}{0.50}\\\\\Rightarrow E=40\ N/C

Thus, the electric field intensity is 40\ N/C

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3 years ago
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