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NNADVOKAT [17]
3 years ago
12

A running back with a mass of 70 kg travels down the field with a velocity of 5.0 m s . Calculate the kinetic energy of the foot

ball player.
Physics
2 answers:
Anuta_ua [19.1K]3 years ago
8 0

The kinetic <u>energy</u> of the player is B.) 875 Joules. This is found by multiplying the <u>mass</u> (70 kg) times the <u>velocity</u> (5  m /s) squared, and dividing the <u>product</u> by 2.

Sergio [31]3 years ago
5 0

K.E = 1/2 mv²

    =  1/2 (70kg) (5.0ms)²

    =  1 ˣ ( 1750) / 2

    = 875 kgms²

     

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What net force acting on a 14 KG wagon produces an acceleration of 1.5 MS^2?
valentina_108 [34]

21N

Explanation:

Given parameters:

Mass of wagon = 14kg

Acceleration = 1.5m/s²

Unknown:

Net force on wagon = ?

Solution:

Force is a pull or push on a body that causes a body to change its state. It is expressed as:  

                 Force = mass  x acceleration

  Force on wagon = 14 x 1.5 = 21N

Learn more:

Force brainly.com/question/10470406

#learnwithBrainly

7 0
3 years ago
A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the
Bond [772]

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

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

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

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

Re-arranging for r

r=\frac{Tv}{2\pi}

And substituting into the previous equation

qvB = m \frac{Tv^3}{2\pi}

Solving for T,

T=\frac{2\pi q B}{m v^2}

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

4 0
3 years ago
Which runner has greater kinetic energy: a 45 kg runner moving at a speed of 7 m per second or a 93 kg runner moving at a
bagirrra123 [75]

Kinetic Energy = (1/2) (mass) (speed)

First runner:  KE = (1/2) (45kg) (49 m/s)  =  1,102.5 Joules  

Second runner:  KE = (1/2) (93kg) (9 m/s)  =  418.5 Joules

The <em>first runner </em><em>has 163</em>% more kinetic energy than the second runner has.

7 0
3 years ago
How high would the level be in a gasoline barometer at normal atmospheric pressure?
Sergio [31]

Answer:

h = 13.06 m

Explanation:

Given:

- Specific gravity of gasoline S.G = 0.739

- Density of water p_w = 997 kg/m^3

- The atmosphere pressure P_o = 101.325 KPa

- The change in height of the liquid is h m

Find:

How high would the level be in a gasoline barometer at normal atmospheric pressure?

Solution:

- When we consider a barometer setup. We dip the open mouth of an inverted test tube into a pool of fluid. Due to the pressure acting on the free surface of the pool, the fluid starts to rise into the test-tube to a height h.

- The relation with the pressure acting on the free surface and the height to which the fluid travels depends on the density of the fluid and gravitational acceleration as follows:

                                         P = S.G*p_w*g*h

Where,                              h = P / S.G*p_w*g

- Input the values given:

                                         h = 101.325 KPa / 0.739*9.81*997

                                         h = 13.06 m

- Hence, the gasoline will rise up to the height of 13.06 m under normal atmospheric conditions at sea level.

7 0
3 years ago
Choose the situation below in which the force applied is the greatest.
Gnesinka [82]

Answer:

D

Explanation:

We know the formula for Work to be:

W = f * d

Where W is work done

f is force

d is the distance

A)

Work = 50

Distance = 50

So, Force is:

Force = 50/50 = 1

B)

Work = 400

Distance = 80

Force = 400/80 = 5

C)

Work = 365

Distance = 73

Force = 365/73 = 5

D)

Work = 144

Distance = 16

Force = 144/16 = 9

Hence, D is the situation in which the force applied is the greatest.

6 0
3 years ago
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