Answer:
18 radians
Explanation:
The computation is shown below:
As we know that
Torque = Force × Moment arm
= 1N × 1M
= 1N-M
Torque = 

Now

Here t = 1 minutes = 60 seconds
The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is
1/2 • (10.0 m/s) • (4.0 s) = 20.00 m
Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as
<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>
and under constant acceleration,
<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2
According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so
∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2
∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2
∆<em>x</em> = 20.00 m
Answer:
See Explanation
Explanation:
a) We know that;
v = λf
Where;
λ = wavelength of the wave
f = frequency of the wave
v = velocity of the wave
So;
T = 2 * 2.10 s = 4.2 s
Hence f = 1/4.2 s
f = 0.24 Hz
The wavelength = 6.5 m
Hence;
v = 6.5 m * 0.24 Hz
v = 1.56 m/s
b)The amplitude of the wave is;
A = 0.600 m/2 = 0.300 m
c) Since the wave speed does not depend on the amplitude of the wave then the answer in (a) above remains the same
Where d = 0.30 m
A = 0.30 m/2 = 0.15 m
Explanation:
Terminal velocity is given by:

Here, m is the mass of the falling object, g is the gravitational acceleration,
is the drag coefficient,
is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is
.

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:
