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Contact [7]
3 years ago
10

What does the outer part of the disk turn into?

Physics
1 answer:
My name is Ann [436]3 years ago
8 0

Answer:

what does the outer part of the disk turn into

Explanation:

4) it gets sucked into the star

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List out the methods that you can use to separate solid-solid mixtures
kompoz [17]

Handpicking and winowing

5 0
3 years ago
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te
Maksim231197 [3]

Answer:

Tt = 70 + 135e^-0.031t

13 minutes

Explanation:

Given that :

Initial temperature, Ti = 205°

Temperature after 2.5 minutes = 195°

Temperature of room, Ts= 70

Using the relation :

Tt = Ts + Ce^-kt

Temperature after time, t

When freshly poured, t = 0

205 = 70 + Ce^-0k

205 = 70 + C

C = 205 - 70 = 135°

T after 2.5 minutes to find proportionality constant, k

Tt = Ts + Ce^-kt

195 = 70 + 135e^-2.5k

125 = 135e^-2.5k

125 / 135 = e^-2.5k

0.9259 = e^-2.5k

Take In of both sides :

−0.076989 = - 2.5k

k = −0.076989 / - 2.5

k = 0.031

Equation becomes :

Tt = 70 + 135e^-0.031t

t when Tt = 160

160 = 70 + 135e^-0.031k

90 = 135e^-0.031t

90/135 = e^-0.031t

0.6667 = e^-0.031t

In(0.6667) = - 0.031t

−0.405465 = - 0.031t

t = 0.405465/ 0.031

t = 13.071

t = 13 minutes

8 0
2 years ago
A student is throwing a ball. She throws the ball up with her left hand, it passes over her head, and she catches it with her ri
pychu [463]

Answer:

(c) at point 2, the ball is at its highest height do its PE is max. Also at ms height, velocity is zero therefore KE is zero.

8 0
2 years ago
Your office has a 0.025 m^3 cylindrical container of drinking water. The radius of the container is about 13 cm. Required:a. Whe
uranmaximum [27]

Answer:

<em>a) 105935.7 Pa</em>

<em>b) 103630.35 Pa</em>

Explanation:

The volume of the container = 0.025 m^3

The radius of the container = 13 cm = 0.13 m

We have to find the height of the tank

From the equation for finding the volume of the cylinder,

V = \pi r^2h

where

V is the volume of the cylinder

h is the height of the cylinder

substituting values, we have

0.025 = 3.142 x 0.13^2 x h

0.025 = 0.0531h

h = 0.025/0.0531 = 0.47 m

Pressure at the bottom of the tank P = ρgh

where

ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the depth of water which is equal to the height of the tank

substituting values, we have

P = 1000 x 9.81 x 0.47 = 4610.7 Pa

atmospheric pressure = 101325 Pa

therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = <em>105935.7 Pa</em>

b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m

pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa

This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = <em>103630.35 Pa</em>

7 0
3 years ago
Four solid plastic cylinders all have radius 2.41 cm and length 5.94 cm. Find the charge of each cylinder given the following ad
Paladinen [302]

Answer:

Check explanation

Explanation:

QUICK NOTE: THE QUESTION IS NOT COMPLETE. Although it is not, we can make assumptions, since we only need values for the UNIFORM CHARGE DENSITY.

SO, LET US BEGIN;

To solve this question we are to use the equation (1) below;

Charge,Q = uniform charge density,p × Total area of the cylinder,A ------------------------------------------------------------------------(1).

From the question, we are given radius, R to be 2.41 cm and length, L to be 5.94 cm.

Step one: calculate for the total area of the cylinder, A.

Total area of the cylinder, A= area of the top surface + area of the buttom + area of the curved surface of the cylinder.

Hence, total area of the cylinder,A is;

==> πR^2 + πR^2 + 2πRL. -------------------------------------------------------------------------(2).

Then, total area of the cylinder,A is;

==> (L + R)2πR.

Step two: find the charge of each cylinder.

===> For the first cylinder; we have the uniform charge density to be 35 nC/m^2.

Therefore, the combination of equation (1) and (3) gives;

Charge Q= p × (L + R)2πR...----------------------------(4)

Hence, Q= 35 × [(5.94 + 2.41) 2× 3.143 × 5.94].= 10912.615 coulumb.

====> For the second cylinder, we have a uniform charge density of 50 nC/m^2.

Using equation (4), charge,Q= 15,589.45 Coulumb

=====> For THE third cylinder, the uniform charge density is 600, we make use of equation (4);

Charge,Q= 600×311.789.

Charge,Q= 187,073.4 coulumb.

====> For THE fourth cylinder, the uniform charge density is 750 nC/m^2.., we make use of equation (4);

Charge,Q= 233,841.75 coulumb.

7 0
3 years ago
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