The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from
1 answer:
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The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.
For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.
As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.
i.e.
P.E = K.E + R.K.E
Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
Learn more about angular velocity here:
Answer:
The corresponding magnetic field is
Explanation:
From the question we are told that
The electric field amplitude is
Generally the magnetic field amplitude is mathematically represented as
Where c is the speed of light with a constant value
So
Since 1 T is equivalent to
Answer:
a) 6 times farther. b) 6 times longer.
Explanation:
Once released, in the horizontal direction, no other forces act on the ball, so it continues moving at the same initial velocity, which is given by the projection of the velocity vector in the horizontal direction, as follows:
vₓ = v* cos (25º) = 23 m/s * 0.906 = 20.8 m/s
In the vertical direction, the initial velocity is the projection of the velocity vector along the vertical axis, as follows:
vy = v* sin (25º) = 23 m/s * 0.422 = 9.72 m/s
Assuming that the acceleration is constant, and equal to 1/6*g, we can calculate the total time of flight, with the following kinematic equation for the vertical displacement:
If the total displacement in the vertical direction is 0 (which means that the time if the total time of flight), we can solve for t, as follows:
On earth, this time could be calculated in the same way:
As the time is defined by the vertical movement, we can find the horizontal distance travelled on the moon, as follows:
Δx = v₀ₓ * t = 20.8 m/s * 11. 9 s = 248.1 m
On earth, the distance travelled had been as follows:
Δx = v₀ₓ * t = 20.8 m/s * 1.98 s = 41.3 m
⇒ Δx(moon) / Δx(earth) = 248.1 / 41.3 = 6.00
b) As we have just found, the time of flight on the moon and on the earth are as follows:
tmoon = 11. 9 s
tearth = 1.98 s
⇒ t(moon) / t(earth) = 11.9 / 1.98 = 6.0