An electric hoist does 500 joules of work lifting a crate 2 meters. How

what is the force of gravity attraction between an object with a mass of 0.5 kg and another object has a mass of 0.33 kg and a d

elena55 [62]

Answer:

$from \: newton \: law \: of \: gravitation \\ F = \frac{GMm}{ {r}^{2} } \\ G = 6.67 \times {10}^{ - 11} \\ M = 0.5 \: kg \\ m = 0.33 \: kg \\ r = 0.002 \: m \\ substitute \\ F = \frac{(6.67 \times {10}^{ - 11}) \times (0.5) \times (0.33)) }{ {(0.002)}^{2} } \\ = 2.75 \times {10}^{ - 6} N$

4 0

3 years ago

What is the best description of the constructive interference of light

Verizon [17]

Answer: A. Two waves have a displacement in opposite directions

Explanation:

5 0

3 years ago

Read 2 more answers

Using a horizontal force of 60 N, a wagon is pushed horizontally across the floor a distance of 12 meters at a constant speed of

RideAnS [48]

Answer:

a) The work done by the force is 720 joules, b) The power supplied by the force is 138 watts.

Explanation:

a) Since force is uniform and parallel to the direction of motion, the work ( $W$ ), in joules, done by the force ( $F$ ), in newtons, is defined by this formula:

$W = F\cdot s$ (1)

Where $s$ is the travelled distance, in meters.

If we know that $F = 60\,N$ and $s = 12\,m$ , then the work done by the force is:

$W = F\cdot s$

$W = (60\,N)\cdot (12\,m)$

$W = 720\,J$

The work done by the force is 720 joules.

b) And an expression for the power supplied by the force ( $\dot W$ ), in watts, is concieved by differentiating (1) in time:

$\dot W = F\cdot \dot s$

Where $\dot s$ is the speed of the wagon, in meters per second.

If we know that $F = 60\,N$ and $\dot s = 2.3\,\frac{m}{s}$ , then the power supplied by the force is:

$\dot W = F\cdot \dot s$

$\dot W = (60\,N)\cdot \left(2.3\,\frac{m}{s} \right)$

$\dot W = 138\,W$

The power supplied by the force is 138 watts.

3 0

3 years ago

In a grandfather clock, the second hand moves forward by one second for each half period of the clock’s pendulum.

Maurinko [17]

To solve the problem it is necessary to take into account the concepts related to simple pendulum, i.e., a point mass that is suspended from a weightless string. Such a pendulum moves in a harmonic motion -the oscillations repeat regularly, and kineticenergy is transformed into potntial energy and vice versa.

In the given problem half of the period is equivalent to 1 second so the pendulum period is,

$T= 2s$

From the equations describing the period of a simple pendulum you have to

$T = 2\pi \sqrt{\frac{L}{g}}$

Where

g= gravity

L = Length

T = Period

Re-arrange to find L we have

$L = \frac{gT^2}{4\pi^2}$

Replacing the values,

$L = \frac{(9.8)(2)^2}{4\pi^2}$

$L = 0.99m$

In the case of the reduction of gravity because the pendulum is in another celestial body, as the moon for example would happen that,

$T = 2\pi \sqrt{\frac{L}{g/6}}$

$T = 2\pi\sqrt{\frac{6L}{g}}$

$T = 2\pi \sqrt{\frac{6*0.99}{9.8}}$

$T = 4.89s$

In this way preserving the same length of the rope but decreasing the gravity the Period would increase considerably.

6 0

3 years ago

5. Bill is swimming and drops his swim goggles in the pool. He looks from above the top of the water to see where they fell and

Alexus [3.1K]

The interaction of light that distorted the image of where Bill's goggles appear to be is Refraction.

<h3>What happens to the speed of light when it hits water in a swimming pool?</h3>

The light slows down when it passes from the less dense air into the water, this slow down of the ray of light which causes the ray of light to change direction and then the changes the speed of the light that causes refraction.

The swimming pool appears shallow than it appear to be because the rays of light coming from the bottom of the swimming pool that is refracted by the water making the height of the swimming pool to be above the actual swimming pool bed and there by misleading the observer(Bill).

Get more information about Refraction of light herebrainly.com/question/15838784

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4 0

1 year ago