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2 years ago

How does a push or pull affect motion?

Physics

1 answer:

STatiana [176]2 years ago

8 0

Answer:

The motion of an object acted on by a force depends partly on the strength of the push or pull. The stronger the push or pull, the faster the object will move.

Explanation:

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A thin spherical shell has a radius of 0.70 m. An applied torque of 860 N m gives the shell an angular acceleration of 4.70 rad/

Artyom0805 [142]

Answer:

$I=182.97\ kg-m^2$

Explanation:

Given that,

Radius of a spherical shell, r = 0.7 m

Torque acting on the shell, $\tau=860\ N$

Angular acceleration of the shell, $\alpha =4.7\ m/s^2$

We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

$\tau=I\alpha$

I is the rotational inertia of the shell

$I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{860}{4.7}\\\\I=182.97\ kg-m^2$

So, the rotational inertia of the shell is $182.97\ kg-m^2$ .

7 0

3 years ago

Below is the data from a gas law experiment comparing the pressure and the volume of a gas at a given temperature.

Wewaii [24]

Answer:

The combined gas equation relates three variables pressure, temperature and volume when the number of moles is constant.

The equation is PV / T = constant. Which is valid for a fixed number of moles of the gas.

You can derive the combined gas equation from the combination of Bolye's law, Charles' law and Gay-Lussac's law, which needs some algebra.

Explanation:

9 0

3 years ago

Read 2 more answers

What is the magnitude of the electric force of attraction between two point charges + 2.8 mc and -1.2 mc if the distance

Anit [1.1K]

Answer:

Explanation:

it just a

8 0

3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago

Which force can affect an object without direct physical contact?

denis-greek [22]

Gravity affet everything and it touches nothing.
Hope this helps!

8 0

4 years ago

Read 2 more answers