Are ultrasound waves high energy waves or low energy waves?

When the core of a star like the sun uses up its supply of hydrogen for fusion, the core begins to ________?

pav-90 [236]

When the core of a star like the sun uses up its supply of hydrogen for fusion, the core begins to contract in size. The gravitational forces becomes strong and cause the star to reduce in size. as the star contracts, the temperature of the star also rises since the heat is distributed inside smaller Volume now.

4 0

3 years ago

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The force generated by a single muscle fiber can be increased by increasing is called

mars1129 [50]

Answer:

The force generated by a single muscle fiber can be increased by increasing the frequency of action potentials

Explanation:

The force generated by a muscle fiber is the result of the shortening of the skeletal muscle, and this force is also know as muscle tension. The larger motor units shorten along with the smaller units to produce the muscle force. The time lapsed between the beginning of the action potential in the muscle and the beginning of the contraction is the latent period. Action potential is the result of the difference electrical potential as a result of passage of an impulse along the membrane of a muscle or nerve cell.

4 0

2 years ago

What is the amount of charge a capacitor can store per volt of potential difference?

sladkih [1.3K]

The amount of charge a capacitor can store per volt of potential difference is called the capacitance of the capacitor.

<h3>What is the capacitance?</h3>

The capacitance has to do with the amount of charge that is stored by a capacitor. We know that a capacitor is a device that can be used to store electric charges. We have in it, two capacitors separated by a dielectric material.

Hence, the amount of charge a capacitor can store per volt of potential difference is called the capacitance of the capacitor.

Learn more about capacitor:brainly.com/question/17176550

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1 year ago

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A 95 kg clock initially at rest on a horizontal floor

Nataly [62]

Answer:

You are given that the mass of the clock M is 95 kg.

This is true whether the clock is in motion or not.

Fs is the frictional force required to keep the clock from moving.

Thus Fk = uk W = uk M g the force required to move clock at constant speed. (the kinetic frictional force)

uk = 560 N / 931 N = .644 since the weight of the clock is 931 N (95 * 9.8)

us is the frictional force requited to start the clock moving

us = static frictional force = 650 / 931 -= .698

4 0

2 years ago

A ship sets sail from Rotterdam, The Netherlands, heading due north at 7.00 m/s relative to the water. The local ocean current i

lisov135 [29]

Answer:

8.07 m/s, 81.7º NE.

Explanation:

The ship, due to the local ocean current, will be deviated from its original due north bearing.
In order to find the magnitude of the velocity of the ship, we need to convert a vector equation, in an algebraic one.
If we choose two axes coincident with the N-S and W-E directions, we can find the components of the velocity along these directions.
Clearly, the velocity of the ship, relative to water, is only due north, so it has no component along the W-E axis.
The local ocean current, as it is directed at an angle between both axes, has components along these axes.
These components can be found from the projections of the velocity vector along these axes, as follows:

$vocx = voc* cos 40 = 1.53 m/s * 0.766 = 1.17 m/s\\vocy = voc* sin 40 = 1.53 m/s * 0.643 = 0.98 m/s$

The component along the N-S axis (y-axis) of the velocity of the ship will be the sum of the velocity relative to water, plus the component of the ocean current along this same axis:

$vshy = vsw + vwy = 7.00 m/s + 0.98 m/s = 7.98 m/s$

The component along the W-E axis, is just the component of the local ocean current in this direction:

vshx = 1.17 m/s

We can find the magnitude of the velocity vector, applying the Pythagorean theorem, as follows:

$v = \sqrt{vshx^{2} + vshy^{2} } =\sqrt{(7.98m/s)^{2} +(1.17m/s)^{2} } =8.07 m/s$

The direction of the vector relative to the W-E axis (measured in counterclockwise direction) is given by the relative magnitude of the x and y components, as follows:

$tg \theta = \frac{vshy}{vshx} = \frac{7.98}{1.17} = 6.82 \\ \theta = tg^{-1} (6.82)\\ \theta= 81.7\deg$

The velocity of the ship, relative to Earth, is 8.07 m/s, 81.7º North of East.

7 0

2 years ago