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ZanzabumX [31]
2 years ago
13

Stars that are not very hot but give off a lot of light are

Physics
2 answers:
melomori [17]2 years ago
6 0

\huge{\textbf{\textsf{{\color{navy}{An}}{\purple{sw}}{\pink{er}} {\color{pink}{:}}}}}

Giants.

  • thanks
  • hope it helps.
MAVERICK [17]2 years ago
3 0

Explanation:

giants are those stars that are not so hot but give a lot of light

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5. What is the amount of force required to accelerate a 20 kg object at a rate of 5 m/sz?
GenaCL600 [577]

Force required is 100 N

<u>Given that;</u>

Rate of acceleration = 5 m/s²

Mass of object = 20kg

<u>Find:</u>

Force required

<u>Computation:</u>

Force = Mass × Acceleration

Force required = Rate of acceleration × Mass of object

Force required = 20 × 5

Force required = 100 N

Learn more:

brainly.com/question/17506203?referrer=searchResults

3 0
2 years ago
A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s
topjm [15]

Answer:

\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

Explanation:

The electrostatic potential energy is given by the following formula

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Now, we will apply this formula to both cases:

U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}

So, the change in the potential energy is

\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

7 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
Compute the density in g/cm^3 of a piece of metal that had mass of 0.485 kg and a volume of 52cm^3
steposvetlana [31]

Answer:

9.3 g/cm³

Explanation:

First, convert kg to g:

0.485 kg × (1000 g / kg) = 485 g

Density is mass divided by volume:

D = (485 g) / (52 cm³)

D = 9.33 g/cm³

Rounding to two significant figures, the density is 9.3 g/cm³.

8 0
3 years ago
An artillery shell of mass 30 kg has a velocity of 250 m/s vertically upward. The shell explodes into two pieces; immediately af
olga_2 [115]

Answer:

9654.34 m

Explanation:

from conservation of momentum

$$\begin{aligned}30 \times 250 &=-10 \times 120+20 \times V \\20 V &=30 \times 250+10 * 120 \\V &=\frac{30 \times 250+10 \times 120}{20}=435 \mathrm{~m} / \mathrm{s}\end{aligned}$$

And from Conservation of Energy

\frac{1}{2} m v^{2}=m g h\\h=\frac{v^{2}}{2 g}\\h=\frac{(435(m/s))^{2}}{2 \times 9.8(m/s^{2} )}\\h=9654.34 (m)

7 0
2 years ago
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