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2 years ago

newton's 3rd law is known as the law of______? is it different objects or the same objects or objects in motion?

Physics

1 answer:

Levart [38]2 years ago

5 0

Answer:

th law of action and reaction

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It appears as though the moon disappears and the sun comes out in the daytime. Based on what you know, explain why this is happe

zimovet [89]

Answer:

The Earth is toward the sun

Explanation:

DONT LISTEN TO ME I AM A CHILD AND I JUST GUESSED

6 0

2 years ago

Read 2 more answers

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

2 years ago

21. Calculate the acceleration of the bus from point D to E. Show your work.

Marat540 [252]

21) Acceleration from D to E: $1 m/s^2$

22) The acceleration of the bus from D to E is $1 m/s^2$

Explanation:

21)

The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time elapsed

In this problem, we want to measure the acceleration of the bus from point D to point E. We have:

- Initial velocity at point D: u = 0

- Final velocity at point E: v = 5 m/s

- Time elapsed from D to E: t = 21 - 16 = 5 s

Therefore, the acceleration between D and E is

$a=\frac{5-0}{5}=1 m/s^2$

22) This question is the same as 21), so the result is the same.

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

4 0

2 years ago

A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it

Savatey [412]

The charge of the object must be $1.11 \times e^{-5} \text { coulomb }$

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

$Electric field, E=\frac{\text { Force }(F)}{q}$

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

$q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }$

6 0

2 years ago

A girl throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounc

erma4kov [3.2K]

Answer:

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Applying the impulse-momentum equation;

Impulse = change in momentum

Ft = m∆v

F = (m∆v)/t

Where;

F = force

t = time

m = mass

∆v = v2 - v1 = change in velocity

Given;

m = 0.80 kg

t = 0.050 s

The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.

v2 = 25 m/s

v1 = -25 m/s

∆v = v2 - v1 = 25 - (-25) m/s = 25 +25 = 50 m/s

Substituting the values;

F = (m∆v)/t

F = (0.80×50)/0.05

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

4 0

2 years ago