The purpose of a fuse or circuit breaker is _____.to speed up the current
to slow down the current
to stop current that is flowing too fast
to stop current that is flowing too slow

Answers

Answer 1

Answer:

Answer:

to stop current that is flowing too fast

Explanation:

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When astronauts travel to the moon, their bodies experience a lower gravitational pull than on Earth. Which type of pull are they experiencing?

Answers

Answer:

The pull the astronauts feel in the moon will be gravitational too.

Explanation:

The pull will be gravitational too because there's a gravitational force exerted by the moon over the astronaut, this force keeps pulling the astronaut towards the moon surface but it's weaker than in earth because the mass of the moon is significantly less that the mass of the earth.

They are experiencing the pull of leaving the atmosphere

Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.(a) If the string is stationary and the yo-yo accelerates away from it at a rate of 1.80 m/s2, what is the angular acceleration of the yo-yo in rad/s2? rad/s2.
(b) What is the angular velocity in rad/s after 0.750 s if it starts from rest? rad/s.
(c) The outside radius of the yo-yo is 3.10 cm.What is the tangential acceleration in m/s2 of a point on its edge? m/s2

Answers

Answer:

Part a)

$\alpha = 782.6 rad/s^2$

Part B)

$\omega = 587 rad/s$

Part c)

$a_t = 24.3 m/s^2$

Explanation:

Part a)

As we know that

$a = R \alpha$

so we will have

$a = 1.80 m/s^2$

$R = 0.230 cm$

$\alpha = (a)/(R)$

$\alpha = (1.80)/(0.230 * 10^(-2))$

$\alpha = 782.6 rad/s^2$

Part B)

Angular speed of the yo-yo

$\omega = \alpha t$

so we have

$\omega = 782.6 * 0.750$

$\omega = 587 rad/s$

Part c)

Tangential acceleration is given as

$a_t = R \alpha$

$a_t = (3.10 * 10^(-2))(782.6)$

$a_t = 24.3 m/s^2$

To practice Problem-Solving Strategy 3.2 Relative velocity. An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward. The speed of the plane relative to the air (called the "air speed") as measured by instruments aboard the plane is 310 km/hour . In which direction should the pilot head?

Answers

The pilot should head 13.6° north of west.

Why?

We can solve the problem by using trigonometric relations. Since there is a right triangle formed between the direction that the pilot wants to fly to and the wind's speed, we can use the following formula:

$Tan(\alpha )=(WindSpeed)/(PlaneSpeed)\n\n\alpha =Tan((WindSpeed)/(PlaneSpeed))^(-1)=ArcTan((WindSpeed)/(PlaneSpeed))\n\n\alpha =ArcTan((WindSpeed)/(PlaneSpeed))$