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2 years ago

1. An 800-gram bowling ball is rolling in a straight line toward you. If its momentum is 16 kg•m/sec, how fast is it traveling?

Physics

1 answer:

ziro4ka [17]2 years ago

3 0

Answer:

The ball is rolling at a speed of 0.02 meter per second.

Step by step explanation:

We are given that there is a 800 gram bowling ball rolling in a straight line. If its momentum is given to be 16 kg.m/sec, we are to find its velocity.

For this, we will use the formula of momentum.

Momentum = mass × velocity

16 = 800 × velocity

Velocity = 16/800 = 0.02 meter per second

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Paolo says that his bicycle is hard to pedal.Mia looks at the bicycle and tells him he needs to oil the chain.explain why oiling

laila [671]

Answer:

Less friction

Explanation:

Paolo's bike is too difficult to pedal because there is too much friction in the mechanisms of the bike. To reduce friction, Paolo must oil the chain. This will make the bike run much more smoothly and allow for easier pedalling.

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2 years ago

The rectangular plates in a parallel-plate capacitor are 0.063 m x 5.4 m. A distance of 3.5 x 10^-5m separate the plates. The pl

Aleks04 [339]

The capacitance of the capacitor is $1.8\cdot 10^{-9}F$

Explanation:

The capacitance of a parallel-plate capacitor is given by the equation

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the medium

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the capacitor in this problem, we have:

k = 2.1 is the dielectric constant

$d=3.5\cdot 10^{-5}m$ is the separation between the plates

$A=0.063m \cdot 0.054m = 0.0034 m^2$ (I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)

Therefore, the capacitance of the capacitor is

$C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F$

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3 years ago

As capacitor was discharging, what did you observe about q on its plates and the motion of charges in the external circuit?

liubo4ka [24]

As capacitor was discharging, The charge on the plate got reversed and the motion of charge is opposite to the flow of current.

The charging contemporary asymptotically processes 0 as the capacitor becomes charged up to the battery voltage.

The capacitor is completely charged when the voltage of the electricity supply is equal to that at the capacitor terminals. that is referred to as capacitor charging; and the charging segment is over when modern-day stops flowing thru the electrical circuit.

A capacitor can be slowly charged to the important voltage and then discharged quick to provide the power wanted. it's far even viable to charge several capacitors to a positive voltage and then discharge them in any such way as to get extra voltage out of the gadget than became installed.

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1 year ago

A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =

Mila [183]

Answer:

$v=6.65m/sec$