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Anika [276]
2 years ago
6

a 1. You found that the MCB was tied with a thread and the thread was fixed with a nail on the wall in your friend's house. i. I

s it good idea to do this? ii. What could be the possible hazard of this? iii. What should have done to keep the circuit safe?​
Physics
1 answer:
Katen [24]2 years ago
5 0

The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.

<h3>What is a miniature circuit breaker?</h3>

A miniature circuit breaker is a circuit breaker that is used in homes as a means of guarding against damage to appliances due to a very high current.

This miniature circuit breaker is also harzardous in the sense that it could lead to an electrical fault related fire outbreak especially when it is being blown freely by wind as you tie it with a thread. Doing this a very bad idea because of the risk of a fire hazard.

The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.

Learn more about miniature circuit breaker: brainly.com/question/1298425

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Two balls, one twice as massive as the other, are dropped from the roof of a building (freefall). Just before hitting the ground
Katena32 [7]

KE= 1/2 mv^2

Kinetic Energy is equal to 1/2 x mass x velocity squared

The mass of the larger ball has TWICE
the kinetic energy. KE is directly proportional to the mass.

6 0
3 years ago
I throw a ball upward at an initial speed of 20 m/s. How much time does it take before the ball slows to a speed of 0 m/s
masha68 [24]
20/9.8 = 2.0 seconds. The ball stops after 2 seconds.
5 0
3 years ago
GETS BRAINILIST!!!!Which of the following will increase the gravitational force exerted by one object on another? Increasing the
jeka94

Answer:

C. Decreasing the distance between the objects.

Explanation:

Gravitational force depends on mass and distance. The farther the objects are apart, the less gravitational force/pull they have.

5 0
3 years ago
One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet
Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

               \frac{G M m}{r^{2}}=m \omega^{2} r

The orbit’s period is given by,

               T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}

Where,

T_{e} = Earth’s period

T_{p} = planet’s period

M_{s} = sun’s mass

r_{e} = earth’s radius

Now,

             T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}

As, planet mass is equal to 0.7 times the sun mass, so

            T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}

Taking the ratios of both equation, we get,

             \frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}

            \frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}

           \frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}

Given T_{p}=9.5 \text { days } and T_{e}=365 \text { days }

          \frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}

         r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}

7 0
3 years ago
True or false:The potential energy of an object is affected by its speed
Verdich [7]
True because the speed of a mass can change it's gravitational potential energy in the gravitational field of another massive object.
7 0
3 years ago
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