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4 years ago

8

A mineral deposit along a strip of length 8 cm has density s(x) = 0.01x(8 − x) g/cm for 0 ≤ x ≤ 8. Calculate the total mass of t

he deposit. (Round your answer to three decimal places.)

1 answer:

vazorg [7]4 years ago

3 0

Answer:

8z

Explanation:

It is 8z

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A large mass collides with a stationary, smaller mass. How will the masses behave if the collision is inelastic?

iragen [17]

Logically both masses will collide and well make a reaction. first of all depending on the small mass it will either merge or unite with the big mass or it will bounce away from it . if this happen it will make a reaction that will affect both masses. Hope this helps if it is incorrect please let me know :)

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4 years ago

The electric field everywhere on the surface of a thin, spherical shell of radius 0.710 m is of magnitude 936 N/C and points rad

kaheart [24]

Explanation:

Given that,

Radius of a spherical shell, r = 0.71 m

Electric field that points radially toward the center of the sphere is 936 N/C

(a) Let q is the net charge within the sphere's surface. Using Gauss's law, we can find it :

$\dfrac{1}{q}\times \epsilon_0=EA\\\\q=\dfrac{\epsilon_0}{EA}\\\\q=\dfrac{8.85\times 10^{-12}}{936\times \pi \times (0.71)^2}\\\\q=5.97\times 10^{-15}\ C$

(b) Gauss's law is used to find the amount of charge enclosed within a surface. It doesn't say anything about the distribution of the charge inside the spherical shell.

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4 years ago

Whats is the conclusion of the mass on the period of spring?

Alchen [17]

Answer:

Explanation:

The period of a spring-mass system is proportional to the square root of the mass and inversely proportional to the square root of the spring constant.

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3 years ago

3. Consider the motion of the object whose velocity-time graph is given

notka56 [123]

Answer:

16 m.

Explanation:

From the question given above, we obtained the following data:

Initial velocity (u) = 0 m/s

Final velocity (v) = 8 m/s

Initial time (t1) = 0 sec

Final time (t2) = 2 secs

Net Displacement (ΔD) =?

Velocity is defined as the rate of change of the displacement of an object with time. Mathematically, it is expressed as:

Change in velocity (Δv) = change in displacement (ΔD) / change in time (Δt)

Δv = ΔD / Δt

Next we shall determine the change in velocity and time. This is illustrated below below:

Initial velocity (u) = 0 m/s

Final velocity (v) = 8 m/s

Change in velocity (Δv) =?

Change in velocity (Δv) = v – u

Change in velocity (Δv) = 8 – 0

Change in velocity (Δv) = 8 m/s

Initial time (t1) = 0 sec

Final time (t2) = 2 secs

change in time (Δt) =?

change in time (Δt) = t2 – t1

change in time (Δt) = 2 – 0

change in time (Δt) = 2 secs.

Finally, we shall determine the net displacement of the object as follow:

Change in velocity (Δv) = 8 m/s

change in time (Δt) = 2 secs.

Net Displacement (ΔD) =?

Δv = ΔD / Δt

8 = ΔD/2

Cross multiply

ΔD = 8 × 2

ΔD = 16 m

Therefore, the net displacement of the object is 16 m.

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4 years ago

An electrostatic paint sprayer contains a metal sphere at an electric potential of 25.0 kV with respect to an electrically groun

ira [324]

Answer:

The charge is $Q = 2.177 *10^{-9} \ C$

Explanation:

From the question we are told that

The electric potential is $V = 25.0 \ kV = 25.0 *10^{3}\ V$

The mass of the drop is $m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg$

The speed is $v = 18.8 \ m/s$

Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question is mathematically represented as

$Q = \frac{m v^2 }{ 2 * V }$

Substituting values

$Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }$

$Q = 2.177 *10^{-9} \ C$

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3 years ago