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
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>
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
Answer:
1.41 m/s^2
Explanation:
First of all, let's convert the two speeds from km/h to m/s:


Now we find the centripetal acceleration which is given by

where
v = 12.8 m/s is the speed
r = 140 m is the radius of the curve
Substituting values, we find

we also have a tangential acceleration, which is given by

where
t = 17.0 s
Substituting values,

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

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.