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2 years ago

If you kick a ball and the ball accelerates through the air, the forces involved must be

Physics

2 answers:

sineoko [7]2 years ago

6 0

Unbalanced is likely the answer your teacher wants

Liono4ka [1.6K]2 years ago

3 0

It would be kinetic and potential energy, potential from the force of the kick, and kinetic from energy exerted.

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A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in

lukranit [14]

Explanation:

Value of the cross-sectional area is as follows.

A = $1.5 \times 2.30$

= 3.45 $in^{2}$

The given data is as follows.

Allowable stress = 14,500 psi

Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

$P_{max} = \sigma_{a}A$

= $14500 \times 3.45$

= 50025 lb

Now, maximum load from shear stress is as follows.

$P_{max} = 2 \times \tau_{a} \times A$

= $2 \times 7100 \times 3.45$

= 48990 lb

Hence, $P_{max}$ will be calculated as follows.

$P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})$

= 48990 lb

Thus, we can conclude that the maximum permissible load $P_{max}$ is 48990 lb.

4 0

3 years ago

You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh

lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

$F=\frac{kQq}{r^2}$

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

$\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}$

And the force on the charge due to the charge on the north is,

$\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}$

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of $45^\circ$ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

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1 year ago

What is the time required for an object starting from rest.

sashaice [31]

Time required for what?

6 0

1 year ago

If an ideal gas does not exist then why laws were stated?

mihalych1998 [28]

That is because it is impossible to create a law for the behavior of every single different gas, so creating laws for an ideal gas helps us understand the basic nature of gasses which might or might not differ slightly or a lot. By understanding how an ideal gas works, we can understand how a normal gas works.

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2 years ago

Two factor that affect gravitational force

Gnoma [55]

When dealing with the force of gravity between two objects, there are only two things that are important – mass, and distance. The force of gravity depends directly upon the masses of the two objects, and inversely on the square of the distance between them.

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2 years ago