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motikmotik
3 years ago
11

Name the apparatus used to measure friction force? PLEASE HELP AS SOON AS POSSIBLE

Physics
1 answer:
melisa1 [442]3 years ago
8 0
There is no apparatus for it. It is either use something like ruler and table and rub together or rub our hands, and friction force will be showed.
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Which region of the early universe was most likely to become a galaxy?
kramer

Answer:

This is likely possible for a region whose matter density is higher than the normal average.

Explanation:

A galaxy is a collection of lumps in space which are clumped together and interact with each other. There are a lot of speculations on how galaxies were birthed. some believe its formed by a collection of massive gas, dust which eventually collapsed under their own gravitational pull. others says its formed by the combination of large lumps of matter which accumulated forming thee galaxies. The possibility of a galaxy forming is dependent on how massive the matter in the region of the universe is.

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3 years ago
Consider the four masses. Describe the motion of the masses if you used THE SAME amount of force to move each one over THE SAME
Illusion [34]
The answer is c. if itis heavier, u have to push hardier or it to move the same distance. make sense??
4 0
3 years ago
Read 2 more answers
if a 60 kg person was standing on a platform at the surface of saturn and they jumped, they would have to push with a force grea
lidiya [134]

Answer:

A 60 kg person standing on a platform at the surface of Saturn and they jumped, they would have to push with a force greater than 540 N

Explanation:

The gravitational attraction between an object on the surface of a planet and the planet is given by the weight of the object

Therefore the force needed to be applied for an object to lift off the surface of a planet = The weight of the object

The weight of the object on the surface of a planet = m × g

Where;

m = The mass of the object

g = The strength of gravity on the planet's surface in N/kg

The given parameters are;

The mass of the person standing on a platform at the surface of Saturn, m = 60 kg

The strength of gravity on the surface of Saturn = 9 N/kg

Therefore, we have;

The weight of the person = The force greater than which the person would have to push on the surface of Saturn so as to Jump = The weight of the person on the surface of Saturn = 60 kg × 9 N/kg = 540 N

Therefore, for a 60 kg person standing on a platform at the surface of Saturn and they jumped, they would have to push with a force greater than 540 N.

4 0
3 years ago
Julietta and Jackson are playing miniature golf. Julietta's ball rolls into a long. Straight upward incline with a speed of 2.95
Verdich [7]

Answer:

The length of the incline is 3.504 meters.

Explanation:

Let suppose that Julietta's ball decelerates uniformly, then we determine the length of the incline is determined by the following equation of motion:

\Delta s = v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (Eq. 1)

Where:

\Delta s - Length of the incline, measured in meters.

v_{o} - Initial speed of the ball, measured in meters per second.

a - Aceleration of the ball, measured in meters per square second.

t - Time, measured in second.

If we know that v_{o} = 2.95\,\frac{m}{s}, t = 1.54\,s and a = -0.876\,\frac{m}{s^{2}}, then the length of the incline is:

\Delta s = \left(2.95\,\frac{m}{s} \right)\cdot (1.54\,s)+\frac{1}{2}\cdot \left(-0.876\,\frac{m}{s^{2}} \right) \cdot (1.54\,s)^{2}

\Delta s = 3.504\,m

The length of the incline is 3.504 meters.

6 0
2 years ago
A flat loop of wire consisting of a single turn ofcross-sectional area 7.90 cm2is perpendicular to a magnetic field that increas
Westkost [7]

To develop this problem, it is necessary to apply the concepts related to Faraday's law and Magnetic Flow, which is defined as the change that the magnetic field has in a given area. In other words

\Phi = BA Cos\theta

Where

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A = Area

\theta = Angle between magnetic field lines and normal to the area

The differentiation of this value allows us to obtain in turn the induced emf or electromotive force.

In this case we have that the flat loop of wire is perpendicular to the magnetic field, therefore the angle is 0 degrees, since its magnetic field acts parallel to the area:

\theta = 0 then our expression can be written as

\Phi = BA

From the same value of the electromotive force we have to

\epsilon = -\frac{d\Phi}{dt}

Replacing we have

\epsilon = -A\frac{B}{dt}

Replacing with our values we have that

\epsilon = -(7.9*10^{-4}m)\frac{(3.5-0.5)}{0.1}

\epsilon = -0.0237V

Therefore the magnitude of the induced emf in the loop is 0.0237V

On the other hand we have that the current by Ohm's Law can be defined as

I = \frac{\epsilon}{R}

For the given value of the resistance and the previously found potential we have to

I = \frac{0.237}{1.3}

I= 0.0182A

6 0
3 years ago
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