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3 years ago

When two or more forces act together on an object what do they combine to form

1 answer:

7 0

Without knowing anything about their magnitudes or directions, the only thing you can always say about thier combination is that it's the "net force" on the object.

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What change will most likely increase the strength of a magnetic field produced by an electromagnet? a. Reduce the number of tur

Anettt [7]

Answer:

Option D

Explanation:

When another battery is added to the circuit, the power supplied through the coil and to the magnet becomes greater leading to stronger magnetic field lines being produced.

7 0

3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$

v is the speed of cat, $v=\dfrac{2\pi r}{t}$

$\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29$

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0

3 years ago

In the figure below the pulley is a solid disk of mass M and radius R with rotational inertia MR 2/2. Two blocks one of mass m a

matrenka [14]

Assuming you are looking for the acceleration a:

1. $m_1a = T_1 -m_1g$
2. $m_2a = m_2g - T_2$
where T is the tension and a is the acceleration of the blocks. The acceleration of the two blocks and the acceleration of the pulley must be equal.

The torque on the pulley is given by:
3. $\tau = \overrightarrow r \times \overrightarrow F = (T_2 - T_1)R = I\alpha = \frac{1}{2} MR^2 \frac{a}{R}$
where $I = \frac{1}{2} mR^2$ and $a = \alpha R$ .

Combining the three equations:
$T_2 - T_1 = \frac{1}{2} Ma \\ m_2g - m_2a -m_1g - m_1a = (m_2-m_1)g - (m_1 + m_2)a = \frac{1}2}Ma \\ \\ a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2}M }$

6 0

3 years ago

A small plane flies at a speed of 102 km/h in still air. Suppose the wind blows out from the west (with the air moving east) at

Semmy [17]

Answer:

2.68 hours

Explanation:

A.) Suppose the wind blows out from the west (with the air moving east). The pilot should then head her plane to northwest direction to move directly north.

B.) Given that plane flies at a speed of 102 km/h in still air. And the wind blows out from the west (with the air moving east) at a speed of 46 km/h.

The plan resultant speed can be calculated by using pythagorean theorem.

Resultant Speed = Sqrt( 102^2 + 46^2 )

Resultant Speed = Sqrt( 12520)

Resultant speed = 111.89 km/h

From the definition of speed,

Speed = distance/time

Where distance = 300 km

Substitute the resultant speed and the distance into the formula.

111.89 = 300/time

Time = 300/111.89

Time = 2.68 hours

Therefore, it take her 2.68 hours to reach a point 300 km directly north of her srarting point

7 0

3 years ago

An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert

cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, $k=4.2\times 10^5\ N/m$

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

$mgh=\dfrac{1}{2}kx^2$

x is the compression in spring

$x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m$

So, the compression in the spring is 5.88 meters.

6 0

3 years ago