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3 years ago

A bowling ball with a mass of 8 kg is moving at a speed of 5 m/s. What is its kinetic energy?

Physics

1 answer:

natali 33 [55]3 years ago

5 0

Answer:

100J

Explanation:

Kinetic energy=1/2mv^2

Kinetic energy=(1/2 x 8)x5^2

Kinetic energy=4x25

Kinetic energy=100

100J

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The diagram illustrates the movement of sound waves between an observer and a race car. As the race car drives away from the obs

Andrei [34K]

I believe the answer is d

6 0

3 years ago

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be obser

Alona [7]

Answer:

<h2><em>6000 counts per second</em></h2>

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>

5 0

3 years ago

Comparativo entre a máquina a vapor de Neu carmen e James Watt!

Bas_tet [7]

Answer:

jame watt

Explanation:

James Watt tiña máis importancia que outro home

3 0

3 years ago

Identify the law, write the equation and calculate the answer to the problem below.

lyudmila [28]

Find refractive index first

$\\ \rm\Rrightarrow \mu=\dfrac{c}{v}$

$\\ \rm\Rrightarrow \mu=\dfrac{1.0003}{1.33}$

$\\ \rm\Rrightarrow \mu =0.75$

Now

$\\ \rm\Rrightarrow \dfrac{sini}{sinr}=\mu$

$\\ \rm\Rrightarrow \dfrac{sin45}{sinr}=0.75$

$\\ \rm\Rrightarrow \dfrac{sin45}{0.75}=sinr$

$\\ \rm\Rrightarrow sinr=0.94$

$\\ \rm\Rrightarrow r=sin^{-1}(0.94)$

$\\ \rm\Rrightarrow r=70^{\circ}$

5 0

2 years ago

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

lutik1710 [3]

Answer:

speed of electrons = 3.25 × $10^{7}$ m/s

acceleration in term g is 3.9 × $10^{17}$ g.

radius of circular orbit is 2.76 × $10^{-4}$ m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

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

v = $\sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}$

v = 3.25 × $10^{7}$ m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = $\frac{Bqv}{m}$

a = $\frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}$

a = 3.82 × $10^{18}$ m/s²

and acceleration in term g

a = $\frac{3.82\times 10^{18}}{9.81}$

a = 3.9 × $10^{17}$ g

acceleration in term g is 3.9 × $10^{17}$ g.

and

electron moving in circular orbit has centripetal force

F = $\frac{mv^2}{r}$

Bqv = $\frac{mv^2}{r}$

r = $\frac{mv}{Bq}$

r = $\frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}$

r = 2.76 × $10^{-4}$ m

radius of circular orbit is 2.76 × $10^{-4}$ m

8 0

3 years ago