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3 years ago

What is kinetic energy of a 7.26kg bowling ball that is rolling at a speed of 2m/s?

1 answer:

8 0

Kinetic energy is the energy possessed by an object when that object is moving in space. The higher the mass of an object or higher the speed of an object the higher the kinetic energy will be.

So to calculate the Kinetic Energy we can use the following formula

K.E=(1/2)*m*v^2

Inserting the values in formula gives:

K.E=1/2*7.26*2^2

14.52J

This is the final answer which gives the kinetic energy of the ball.

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standing side by side, you and a friend step off a bridge and fall for 1.6s to the water below. your friend goes first, and you

Andrew [12]

(a) The distance will be more than 2.0 meters.

In fact, you starts your fall after your friend has already fallen 2.0 meters. This means that your friend has already accelerated for a while, therefore his velocity will be greater than yours. But this statement will be actually true for the entire fall, since you has some delay, therefore when your friend will hit the water, the separation between you and him will be greater than the initial separation of 2.0 meters.

b) First of all we need to calculate the height of the bridge with respect to the water. We know that you take 1.6 s to fall down, therefore we can use the following equation:

$S=\frac{1}{2}gt^2=\frac{1}{2}(9.81 m/s^2)(1.6s)^2=12.56 m$

We know that your friend will take 1.6 s to falls down. Instead, you start your jump after he has already fallen 2.0 m, therefore after a time given by the equation:

$S=\frac{1}{2}gt^2$

Using S=2.0 m,

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(2.0 m)}{9.81 m/s^2}}=0.64 s$

So we know that you start your fall 0.64 s after your friend. Therefore, now we can find how much did you fall between the moment you started your fall (0.64 s) and the moment your friend hits the water (1.6 s). Using

$t=1.6 s-0.64 s=0.96 s$

we find

$S=\frac{1}{2}gt^2=\frac{1}{2}(9.81 m/s^2)(0.96 s)^2 =4.52 m$

So, when your friend hits the water, you just covered 4.52 m, while he already covered 12.56 m. Therefore, the separation between you and your friend is more than 2 meters.

8 0

3 years ago

An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and

Kobotan [32]

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is $F_{ma}=BqV---(i)$

The given formula to find the electric force is $F_{el}=qE---(ii)$

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

$Bqv=qE\\\\v=\frac{E}{B}$

$v=\frac{187500}{0.125} \\\\v=15\times10^5m/s$

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

$F_{ce}=\frac{mv^2}{r}---(iii)$

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

$Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg$