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vaieri [72.5K]
1 year ago
14

An object is pulled northward with a force of 10 n and southward with a force of 15 n. the magnitude of the net force on the obj

ect is:________
a. 0 n.
b. 5 n.
c. 10 n.
d. 15 n.
Physics
1 answer:
Anastaziya [24]1 year ago
6 0

The correct option is (b) 5n

As a result, there is a net downward force of 5N operating on the object.

The resultant force is the force that results from adding the vector sums of all the forces operating on an item. The combined action of all the acting forces on the object produces the same effect as the resulting force. When determining the resulting force, the direction of the forces must be taken into account.

Given;

The northward force is Fn = 10N

The southward force is Fs = 15N

Required;

The net force on the mobile phone is Fnet = ?N

The object's weight exerts downward pressure, and upward resistance exerts upward pressure. The vector sum of these two forces will be the net force.

Fnet = Fs - Fn (Considering the direction downward as positive)

Fnet= 15N - 10N

Fnet = 5N

As a result, there is a net downward force of 5 N operating on the object.

Learn more about the Force with the help of the given link:

brainly.com/question/7362815

#SPJ4

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(a) 1200 rad/s

The angular acceleration of the rotor is given by:

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where we have

\alpha = -80.0 rad/s^2 is the angular acceleration (negative since the rotor is slowing down)

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Solving for \omega_f, we find the final angular speed after 10.0 s:

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We can calculate the time needed for the rotor to come to rest, by using again the same formula:

\alpha = \frac{\omega_f - \omega_i}{t}

If we re-arrange it for t, we get:

t = \frac{\omega_f - \omega_i}{\alpha}

where here we have

\omega_i = 2000 rad/s is the initial angular speed

\omega_f=0 is the final angular speed

\alpha = -80.0 rad/s^2 is the angular acceleration

Solving the equation,

t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s

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