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2 years ago

10

Please help me what defines the above collision as being elastic?

1 answer:

Kryger [21]2 years ago

4 0

A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. ... Momentum is conserved in inelastic collisions, but one cannot track the kinetic energy through the collision since some of it is converted to other forms of energy.

I hope this helps

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A charged ball is moving horizontally and perpendicular to a magnetic field of 0.8 Tesla. The ball has a mass of 0.007 kg and ha

goblinko [34]

Answer:

17.15 m/s

Explanation:

Parameters given:

Magnetic field, B = 0.8 T

Mass of ball, m = 0.007 kg

Charge of ball, q = 0.005 C

The magnetic force acting on the charged ball due to the magnetic field is given as:

F = qvBsinθ

where v = velocity of the ball and θ = angle between the horizontal and the magnetic field = 90°

The force of the ball will be in the opposite direction but of equal magnitude:

$F_b$ = -qvBsin(90) = -qvB

To cancel out the effect of gravity, the magnetic force must be equal to the gravitational force acting on the ball:

F = mg

Therefore:

mg = -qvB

Solving for velocity, v, we have:

$v = \frac{mg}{-qB}$

$v = \frac{0.007 * 9.8}{-(-0.005) * 0.8}$

v = 17.15 m/s

The ball must be moving at a velocity of 17.15 m/s.

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3 years ago

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You just found a metallic object in a creek bed. you have found its mass to be 96.5 g, and it displaces 5mL of water. what is th

MaRussiya [10]

Judging by the density (19.3 g/ml) there's a very good chance that it could be gold.

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2 years ago

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the third answer is right.

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An engine performs 6400 j of work on a motorbike the motorbike and the rider have a combined mass of 200 kg if the bike started

spin [16.1K]

The work done by the engine is converted into kinetic energy of the motorbike+rider system:

$W=K=\frac{1}{2}mv^2$

where

W=6400 J is the work

m=200 kg is the mass of the system

v is the speed acquired by the motorbike

Rearranging the equation and substituting the numbers in, we find

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2 years ago

Suppose that the electric field in the Earth's atmosphere is E = 1.16 102 N/C, pointing downward. Determine the electric charge

butalik [34]

Answer:

Electric charge in the earth will be $Q=5.231\times 10^5C$

Explanation:

We have given that E = 116 N/C

Radius of the earth R = 6371 km = 6371000 m

We have to find the electric charge in the earth '

We know that electric field due to charge is given by $E=\frac{1}{4\pi \epsilon _0}\frac{Q}{R^20}=\frac{KQ}{R^2}$ . here K is coulomb's constant

So $116=\frac{9\times 10^9\times Q}{(6371000)^2}$

$Q=5.231\times 10^5C$

So electric charge in the earth will be $Q=5.231\times 10^5C$

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2 years ago