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3 years ago

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Physics

1 answer:

andre [41]3 years ago

6 0

Energy is "the ability to do work". Energy is how things change and move. It takes energy to cook food, to drive to school, and to jump in the air. Different forms of Energy. Energy can take a number of different forms.

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A future use of space stations may be to provide hospitals for severely burned persons. It is very painful for a badly burned pe

inessss [21]

Answer:

1.5min

Explanation:

To solve the problem it is necessary to take into account the concepts related to Period and Centripetal Acceleration.

By definition centripetal acceleration is given by

$a_c = \frac{V^2}{r}$

Where,

V = Tangencial velocity

r = radius

With our values we know that

$a_c = \frac{V^2}{r}$

$\frac{V^2}{r} = \frac{1}{10}g$

Therefore solving to find V, we have:

$V = \sqrt{\frac{1}{10}g*r}$

$V = \sqrt{\frac{9.81*200}{10}}$

$V = 14m/s$

For definition we know that the Time to complete are revolution is given by

$t = \frac{Perimeter}{Speed}$

$t = \frac{2\pi R}{V}$

$t = \frac{2\pi * 200}{14}$

$t = 1.5min$

6 0

3 years ago

A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn),

labwork [276]

Answer:

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

Kinetic Frictional Force (fk) : Negative

Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

W = Work Done

f = Force Applied

d = displacement

θ = Angle between force and displacement

FOR FORCE (P):

Since, force P is parallel to the motion of the box. Therefore, θ = 0°

Hence,

W = P d Cos 0°

W = P d(1)

W = Pd

Therefore, work done by force (P) is Positive.

FOR NORMAL FORCE (Fn) AND WEIGHT (W):

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

W = 0

Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.

FOR KINETIC FRICTIONAL FORCE (fk):

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

W = fk d(-1)

W = -fk d

Therefore, work done by Kinetic Frictional Force (fk) is Negative.

8 0

3 years ago

What type of tectonic plate boundary exists along the edge of the North American plate near the coast of Northern California, Or

kobusy [5.1K]

Answer:

-transform plate boundary

- false

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3 years ago

Prisms separate light, such as that from the Sun, by wavelength

SVETLANKA909090 [29]

Sorry I'm so late, but I just took this test and the answer is white (for people who didn't study well ;) )

6 0

3 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

3 years ago