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ozzi
3 years ago
10

Two identical balls move directly toward each other with equal speeds. how will the balls move if they collide and stick togethe

r'
Physics
1 answer:
Citrus2011 [14]3 years ago
7 0
The momentum of both the identical balls would eventually be transferred to one another when it comes to a point wherein they will collide. In addition, the phenomenon is called an elastic collision wherein both the momentum and energy of the system would considered to be conserved.
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Is an electrical current which comes from a battery
Oxana [17]
Hrdudikdodidbshshsjjsksks
6 0
3 years ago
An electroscope is a fork-shaped device commonly used to detect the presence of charge. The tin leaves of an electroscope will s
velikii [3]

Answer:

All these is caused by the repulsion force.

Explanation:

The electroscope produces a series of electric charges that produce a repulsion force when is putted in contact with a electric charged object.

As the physics law mentions, two different forces are repealed, the electrocospe is charged negatively and the object positively, causing a repulsion force that avoids that both objects touch the other.

7 0
3 years ago
Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that
Leno4ka [110]

Answer:

8.829 m/s²

Explanation:

M = Mass of Earth

m = Mass of Exoplanet

g_e = Acceleration due to gravity on Earth = 9.81 m/s²

g = Acceleration due to gravity on Exoplanet

m=M-0.1M\\\Rightarrow m=0.9M

g_e=G\frac{M}{r^2}

g=G\frac{0.9M}{r^2}

Dividing the equations we get

\frac{g}{g_e}=\frac{G\frac{0.9M}{r^2}}{G\frac{M}{r^2}}\\\Rightarrow \frac{g}{g_e}=0.9\\\Rightarrow g=0.9g_e\\\Rightarrow g=0.9\times 9.81\\\Rightarrow g=8.829\ m/s^2

Acceleration due to gravity on the surface of the Exoplanet is 8.829 m/s²

3 0
3 years ago
Scientists use what scale to compare the hardness of minerals?
frozen [14]
Mohs hardness scale. hope this helps
4 0
3 years ago
The cross section of a copper strip is 1.2 mmthick and 20 mm wide. There is a 25-A current through this cross section, with the
Naily [24]

To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

V_d = \frac{I}{nAq}

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

I = 25 A

A= 1.2*20 *10^{-6} m^2

q= 1.6*10^{-19}C

N = 8.47*10^{19} mm^{-3}

V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}

V_d = 7.68*10^{-5}m/s

The hall voltage is given by

V=\frac{IB}{ned}

Where

B= Magnetic field

n = number of free electrons

d = distance

e = charge of electron

Then using the formula and replacing,

V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}

V = 3.84*10^{-6}V

5 0
2 years ago
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