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3 years ago

How is a crowbar used to remove the top of a box change the direction of the force

1 answer:

7 0

The crowbar is placed under a small edge under the top of the box. The crowbar will send the box lid flying from the opposite direction that the crowbar is placed. Please give Brainlest if you understand.

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Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0

3 years ago

In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e

Zanzabum

Answer:

The velocity is $v_t = 0.02175 \ m/s$

Explanation:

From the question we are told that

The mass of the bullet is $m_b = 0.024 \ kg$

The initial speed of the bullet is $u_b = 1200 \ m/s$

The mass of the target is $m_t = 320 \ kg$

The initial velocity of target is $u_t = 0 \ m/s$

The final velocity of the bullet is is $v_b = 910 \ m/s$

Generally according to the law of momentum conservation we have that

$m_b * u_b + m_t * u_t = m_b * v_b + m_t * v_t$

=> $0.024 * 1200 + 320 * 0 = 0.024 * 910 + 320 * v_t$

=> $v_t = 0.02175 \ m/s$

3 0

3 years ago

A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm. How

Sati [7]

0.36 J of work is done in stretching the spring from 15 cm to 18 cm.

To find the correct answer, we need to know about the work done to strech a string.

<h3>What is the work required to strech a string?</h3>

Mathematically, the work done to strech a string is given as 1/2 ×K×x².
K is the spring constant.

<h3>What will be the spring constant, if 40N force is required to hold a 10 cm to 15 cm streched spring?</h3>

The force experienced by a streched spring is given as Kx. x is the length of the spring streched from its natural length.
Then K = Force / x.
Here x = 15 - 10 = 5 cm = 0.05 m
K = 40/0.05 = 800N/m.

<h3>What will be the work required to strech that spring from 15 cm to 18 cm?</h3>

Work done = 1/2×k×x²
Here x= 18-15=3cm or 0.03 m
So, W= 1/2×800×0.03² = 0.36 J.

Thus, we can conclude that the work done is 0.36 J.

Learn more about the spring force here:

brainly.com/question/14970750

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6 0

2 years ago

Please help due today

Leya [2.2K]

Answer:

Explanation:

(8√2)² = x² + x²

8² × √2² = 2x²

64 × 2 = 2x²

128 = 2x²

64 = x²

x = 8

give me brainliest please

5 0

3 years ago

When a positively charged conductor touches a neutral conductor, the neutral conductor will:

frutty [35]

Answer:

Lose electrons

Explanation:

When a positively charged conductor touches a neutral conductor, the neutral conductor will lose electrons. Only electrons can move from one conductor to another, so if the neutral conductor ended up with a positive charge it means it lost electrons. The conductor touching and the neutral conductor both end up being charged positively.

6 0

4 years ago