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Sonbull [250]
3 years ago
10

A bowling ball rolls 32 meters in 0.8 seconds what is the average speed in m/s of the bowling ball in m/s

Physics
1 answer:
Tanzania [10]3 years ago
7 0

Average speed = (distance rolled) / (time to roll the distance)

Average speed = (32 meters) / (0.8 sec)

Average speed = (32/0.8) (meters/sec)

<em>Average speed = 40 m/s .</em>

<em></em>

I don't know.  That's just a hair under 90 miles per hour.  Is this reasonable for a bowling ball ?  Does anybody roll a bowling ball as fast as a major league pitcher pitches a hardball ? ?

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The distance between two particles is 2 centimeters. If the distance is increased to 4 centimeters, the force will be ?
postnew [5]

Answer:

The new force is 1/4 of the previous force.

Explanation:

Given

Initial\ Distance = 2cm ---- r_1

New\ Distance = 4cm --- r_2

Required

Determine the new force

Let the two particles be q1 and q2.

The initial force F1 is:

F_1 = \frac{kq_1q_2}{r_1^2} --- Coulomb's law

Substitute 2 for r1

F_1 = \frac{kq_1q_2}{2^2}

F_1 = \frac{kq_1q_2}{4}

The new force (F2) is

F_2 = \frac{kq_1q_2}{r_2^2}

Substitute 4 for r2

F_2 = \frac{kq_1q_2}{4^2}

F_2 = \frac{kq_1q_2}{4*4}

F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}

Substitute F_1 = \frac{kq_1q_2}{4}

F_2 = \frac{1}{4}*F_1

F_2 = \frac{F_1}{4}

The new force is 1/4 of the previous force.

3 0
2 years ago
What are possible formulas for impulse? Check all that apply. J = Fdeltat J = StartFraction force over change in time EndFractio
Alex

<u>The possible formulas for impulse are as follows:</u>

J = FΔt

J = mΔv

J = Δp

Answer: Option  A, E and F

<u>Explanation:</u>

The quantity which explains the consequences of a overall force acting on an object (moving force) is known as impulse. It is symbolised as J. When the average overall force acting on an object than such products are formed and in given duration than the start fraction force over change in time end fraction J = FΔt.

The impulse-momentum theorem explains that the variation in momentum of an object is same as the impulse applied to it: J = Δp J = mΔv if mass is constant J = m dv + v dm if mass changes. Logically, the impulse-momentum theorem is equivalent to Newton second laws of motion which is also called as force law.

6 0
3 years ago
How do microwaves use the behavior and characteristics of the electromagnetic wave to function?
serg [7]

Answer:

Radiation moves out of the microwave into waves causing heat.

Explanation:

Inside your microwave oven, electrical energy is transformed into EM energy in the magnetron. When the microwave photons interact with food, food molecules are physically agitated, transforming the EM energy into kinetic energy, or energy of movement   .

Hope it helped!

6 0
3 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

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

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
3 years ago
This outlaw is executed by hanging "in the spring of '25" by
irina [24]
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This is one of the characters in the song, "American Remains", sang by The Highwaymen. The group consisted of </span><span>Johnny Cash, Waylon Jennings, Willie Nelson and Kris Kristofferson. Other characters in the song were a sailor, a dam builder, and a pilot of a starship.
</span>
This is the first stanza of the song:

"I was a highwayman. Along the coach roads I did ride 
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<span>Many a young maid lost her baubles to my trade </span>
<span>Many a soldier shed his lifeblood on my blade </span>
<span>The b*stards hung me in the spring of twenty-five </span>
<span>But I am still alive."</span>
 
8 0
3 years ago
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