What does the path of an object look like when it is in uniform motion ?

HELP! What are 5 changes that a fidget spinner can be?

noname [10]

Answer:

I would hope they can change this question

6 0

3 years ago

The diagram shows the top view of a 65-kg student at point A on an amusement park ride. The ride spins the student in a horizont

Elza [17]

Answer:

1923 N

Explanation:

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

Mass (m) = 65 Kg

Radius (r) = 2.5 m

Velocity (v) = 8.6 m/s

Centripetal force (F) =?

The centripetal force, F, can be obtained by using the following formula:

F = mv²/r

F = 65 × 8.6² / 2.5

F = 65 × 73.96 / 2.5

F = 4807.4 / 2.5

F = 1922.96 ≈ 1923 N

Thus, the magnitude of the centripetal's force acting on the student is approximately 1923 N

3 0

3 years ago

How does your Liver work with your Bladder in your body?

andreyandreev [35.5K]

Answer:

The liver filters the blood of toxins before it continues through the body. It also removes excess sugar from the arteries. Your kidney then further filters the blood for water-soluble waste and toxins to be excreted and finally, your bladder gets read of that water-soluble waste.

6 0

2 years ago

The buoyant force on an object placed in a liquid is (a) always equal to the volume of the liquid displaced. (b) always equal to

Nana76 [90]

Answer:

(c) always equal to the weight of the liquid displaced.

Explanation:

Archimedes principle (also called physical law of buoyancy) states that when an object is completely or partially immersed in a fluid (liquid, e.t.c), it experiences an upthrust (or buoyant force) whose magnitude is equal to the weight of the fluid displaced by that object.

Therefore, from this principle the best option is C - always equal to the weight of the liquid displaced.

7 0

3 years ago

A hollow non-conducting spherical shell has inner radius R1 = 5 cm and outer radius R2 = 19 cm. A charge Q = -35 nC lies at the

Gnom [1K]

Answer:

a. E = -13.8 kN/C

b. E = +8.51 kN/C

Explanation:

We will apply Gauss' Law to the regions where the electric field is asked.

Gauss' Law states that if you draw an imaginary surface enclosing a charge distribution, then the electric field through the imaginary surface is equal to the total charge enclosed by this surface divided by electric permittivity.

$\int\vec{E}d\vec{a} = \frac{Q_{\rm enc}}{\epsilon_0}$

a. For this case, we will draw the imaginary surface between the inner and outer shell of the sphere. The total charge enclosed by this surface will be equal to the sum of the charge Q at the center and charge of the shell within the volume from R1 and r.

Here, r = 0.5(R1+R2) = 12 cm.

$E4\pi r^2 = \frac{Q_{\rm enc}}{\epsilon_0}\\Q_{\rm enc} = Q + \rho V_{\rm enc} = Q + (Ar) (\frac{4}{3}\pi (r^3 - R_1^3)) = (-35\times 10^{-9}) + (16\times 10^{-6})(12\times 10^{-2})(\frac{4}{3}\pi((12\times 10^{-2})^3 - (5\times 10^{-2})^3)) = -2.21\times 10^{-8}~C$

$E = \frac{-2.21\times 10^{-8}}{4\pi (12\times10^{-2})^2 \epsilon_0} = -1.38\times 10^4~N/C\\E = -13.8~kN/C$

b. For this case, we will draw the imaginary surface on the outside of the shell.

The total charge enclosed by this surface will be equal to the sum of the charge at the center and the total charge of the shell.

$Q_{\rm enc} = Q + \rho V = Q + (Ar)[\frac{4}{3}\pi (R_2^3 - R_1^3)]\\Q_{\rm enc} = (-35\times 10^{-9}) + [(16\times 10^{-6})(38\times 10^{-2})][\frac{4}{3}\pi((19\times 10^{-2})^3 - (5\times 10^{-2})^3)]\\Q_{\rm enc} = 1.36\times 10^{-7}~C$

$E = \frac{1.36\times 10^{-7}}{4\pi (38\times10^{-2})^2 \epsilon_0} = 8.51\times 10^3~N/C\\E = 8.51~kN/C$

7 0

3 years ago