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krok68 [10]
3 years ago
14

What is the momentum of an 18-kg object moving at 0.5 m/s?

Physics
1 answer:
Otrada [13]3 years ago
6 0

sorry Idk the answer ..???

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A sprinter starts from rest and reaches a speed of 15 m/s in 4.25 s. Find his acceleration
kakasveta [241]

Answer:

a=3.53 m/s^2

Explanation:

Vo=0 m/s (because he is not moving at the start)

V1=15 m/s

t= 4.25 s

a = (V1-Vo) / t = 15/4.25 = 3.53 m/s^2

5 0
3 years ago
A 50.0 kg woman climbs a flight of stairs 6.00 m high in 15.0 s. How much power does she use.
WITCHER [35]

Her weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

Work = (weight) · (height) = (50kg) · (9.8 m/s²) · (6 m)

Power = (work) / (time) = (50kg) · (9.8 m/s²) · (6 m) / (15 s)

Power = (50 · 9.8 · 6 / 15) · (kg · m² / s³)

Power = 196 (kg · m / s²) · (m) / s

Power = 196 Newton-meter/second

<em>Power = 196 watts</em>

6 0
3 years ago
You stand on a merry-go-round which is spinning at f = 0.25 revolutions per second. You are R = 200 cm from the center. (a) Find
wariber [46]

Answer:

a) ω = 9.86 rad/s

b) ac = 194. 4 m/s²

c) minimum coefficient of static friction, µs = 19.8

Explanation:

a) angular speed, ω = 2πf, where f is frequency of revolution

1 rps = 6.283 rad/s, π = 3.142

ω = 2 * 3.14 * 0.25 * 6.28

ω = 9.86 rad/s

b) centripetal acceleration, a = rω²

where r is radius in meters; r = 200 cm or 2 m

a = 2 * 9.86²

a = 194. 4 m/s²

c) µs = frictional force/ normal force

frictional force = centripetal force = ma; where a is centripetal acceleration

normal force = mg; where g = 9.8 m/s²

µs = ma/mg = a/g

µs = 194.4 ms⁻²/9.8 ms⁻²

c) minimum coefficient of static friction, µs = 19.8

5 0
2 years ago
A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round
Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s

v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s

Now we find the centripetal acceleration which is given by

a_c=\frac{v^2}{r}

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2

we also have a tangential acceleration, which is given by

a_t = \frac{v-u}{t}

where

t = 17.0 s

Substituting values,

a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2

6 0
3 years ago
Read 2 more answers
A proton is moving in a circular orbit of radius 12 cm in a uniform 0.31-T magnetic field perpendicular to the velocity of the p
PIT_PIT [208]

Answer:

Explanation:

A proton of charge

q=+1.609×10^-19C

Orbit a radius of 12cm

r=0.12m

Magnetic Field of 0.31T

Angle between velocity and field is 90°

a. Because the magnetic force F supplies the centripetal force Fc.

The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by

F = qvB sin θ

And the centripetal force is given as

Fc=mv²/r

Where m is mass of proton

m=1.673×10^-27kg

Then, F=Fc

qvB sin θ=mv²/r

qBSin90=mv/r

rqB=mv

Then, v=rqB/m

v=0.12×1.609×10^-19×0.31/1.673×10^-23

v=3577692.78m/s

v=3.58×10^6m/s

b. Since,

F=qVBSin90

F=1.609×10^-19×3.58×10^6×0.31

F=1.785×10^-13 N.

6 0
3 years ago
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