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inysia [295]
2 years ago
14

A mover does 422 J of work pushing a crate 8.39 m. How much force did he exert?

Physics
2 answers:
ratelena [41]2 years ago
8 0

Answer:

heartbroken

Explanation:

just failed my test cause i typed my answers in Caps. Acellus a scam . leave a heart if you feel for me :(  (Answers above were right btw)

kicyunya [14]2 years ago
4 0

Answer:

50.3N

Explanation:

Work done = force x distance

422J. = force x 8.39m

÷8.39 both side to get force

Force is 50.3N to 1 d.p.

Check:

50.3 x 8.39=422.017J

Same as 422J to 1 d.p

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What is the refractive index of air if light travels through it at 3.0 108 m/s?
olchik [2.2K]

Answer:

n = c/v = (3.00 x 108 m/s)/(2.76 x 108 m/s) = 1.09. This does not equal any of the indices of refraction listed in the table.

4 0
2 years ago
I really need help please just answer at least one
yuradex [85]

Answer:

9) a = 25 [m/s^2], t = 4 [s]

10) a = 0.0875 [m/s^2], t = 34.3 [s]

11) t = 32 [s]

Explanation:

To solve this problem we must use kinematics equations. In this way we have:

9)

a)

v_{f}^{2} = v_{i}^{2}-(2*a*x)\\

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = acceleration [m/s^2]

x = distance = 200 [m]

Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.

0 = (100)^2 - (2*a*200)

a = 25 [m/s^2]

b)

Now using the following equation:

v_{f} =v_{i} - (a*t)

0 = 100 - (25*t)

t = 4 [s]

10)

a)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

Note:  The positive sign of the equation means that the car increases his speed.

5^2 = 2^2 + 2*a*(125 - 5)

25 - 4 = 2*a* (120)

a = 0.0875 [m/s^2]

b)

Now using the following equation:

v_{f}= v_{i}+a*t\\

5 = 2 + 0.0875*t

3 = 0.0875*t

t = 34.3 [s]

11)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

10^2 = 2^2 + 2*a*(200 - 10)

100 - 4 = 2*a* (190)

a = 0.25 [m/s^2]

Now using the following equation:

v_{f}= v_{i}+a*t\\

10 = 2 + 0.25*t

8 = 0.25*t

t = 32 [s]

4 0
3 years ago
What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?
viktelen [127]

Answer: 27.21 V

Explanation:

The <u>electric potential</u> V_{E} due to a point charge is expressed as:

V_{E}=k\frac{q}{r}

Where:

k=9(10)^{9}\frac{Nm^{2}}{C^{2}} is the <u>electric constant</u>

q=1.6(10)^{-19}C is the <u>electric charge of the hydrogen nucleus</u>, which is positive

r=5.292(10)^{-11}m is the <u>distance</u>

Rewritting the equation with the known values:

V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}

Finally:

V_{E}=27.21 V

5 0
3 years ago
EASY MULTIPLE CHOICE QUESTIONS! PLEASE HELP!
skad [1K]
1.A and 2.B there the answers
8 0
3 years ago
One of the world's largest Ferris wheels, the Cosmo Clock 21 with a radius of 50.0 m is located in Yokohama City, Japan. Each of
STatiana [176]

Answer:

a = 0.55 m / s²

Explanation:

The centripetal acceleration is given by the relation

         a = v² / r

angular and linear velocities are related

         v = w r

we substitute

          a = w² r

In the exercise they indicate the angular velocity w = 1 rev/min, let's reduce to the SI system

          w = 1 rev / min (2pi rad / 1rev) (1min / 60s) = 0.105 rad/ s

let's calculate

          a = 0.105² 50.0

          a = 0.55 m / s²

4 0
3 years ago
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