Answer:
The tension of the string is 41.876 N
Explanation:
Given;
length of the string, L = 2.11 m
mass of the string, m = 19.5 g = 0.0195 kg
frequency of the wave, f = 440 Hz
wavelength, λ = 15.3 cm = 0.153 m
The velocity of the wave is given by;
v = fλ
v = 440 x 0.153
v = 67.32 m/s
Also the velocity of the wave is given by
where;
μ is mass per unit length = 0.0195 / 2.11 = 0.00924 kg/m
T is the tension of the string
T = v²μ
T = (67.32)²(0.00924)
T = 41.876 N
Therefore, the tension of the string is 41.876 N
Either 175 N or 157 N depending upon how the value of 48° was measured from.
You didn't mention if the angle of 48° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.
Since we need a torque of 55 N·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 N·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.
Ideally, the force will have a 0° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".
If the angle of 48° is from the normal to the lug wrench, the usefully applied power will be:
U = F*cos(48)
where
U = Useful force
F = Force applied
So solving for F and calculating gives:
U = F*cos(48)
U/cos(48) = F
117 N/0.669130606 = F
174.8537563 N = F
So 175 Newtons of force is required in this situation.
If the 48° is from the lug wrench itself, that means that the force is 90° - 48° = 42° from the normal. So doing the calculation again (this time from where we started plugging in values) we get
U/cos(42) = F
117/0.743144825 = F
157.4390294 = F
Or 157 Newtons is required for this case.
Answer:
he kinetic energy increases on the descent, being maximum at the lowest point of the trajectory.
Explanation:
In these semicircular sections the skaters slide from one side to the other, in the downward path their kinetic energy increases and their potential energy decreases; When it leaves the ramp and is in the air, the kinetic energy decreases rapidly, up to the point of maximum height where the kinetic energy is zero.
Consequently, the kinetic energy increases on the descent, being maximum at the lowest point of the trajectory.
Answer: 4.27 *10^6 N/C
Explanation: In order to calculate the electric field along the axis of charged ring we have to use the following expression:
E=k*x/(a^2+x^2)^3/2 where a is the ring radius and x the distance to the point measured from the center of the ring.
Replacing the data we have:
E= (9* 10^9* 0.3* 50 * 10^-6)/(0.1^2+0.3^2)^3/2
then
E=4.27 * 10^6 N/C
