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:
D. only briefly while being connected or disconnected.
Explanation:
As we know that transformer works on the principle of mutual inductance
here we know that as per the principle of mutual inductance when flux linked with the primary coil charges then it will induce EMF in secondary coil
So here when AC source is connected with primary coil then it will give output across secondary coil because AC source will have change in flux with time.
Now when we connect DC source across primary coil then it will not induce any EMF across secondary coil because DC source is a constant voltage source in which flux will remain constant always
So here in DC source the EMF will only induce at the time of connection or disconnection when flux will change in it while rest of the time it will give ZERO output
so correct answer will be
D. only briefly while being connected or disconnected.
Answer
D. move a small magnet back and forth within a section of the coiled wire.
Explanation:
i put that for the test and i got it right
1 in=2.54 cm=(2.54 cm)(1 m/100 cm)=0.0254 m
Therefore:
1 in=0.0254 m
1 in³=(0.0254 m)³=1.6387064 x 10⁻⁵ m³
Therefore:
8.06 in³=(8.06 in³)(1.6387064 x 10⁻⁵ m³ / 1 in³)≈1.321 x 10⁻⁴ m³.
Answer: 8.06 in³=1.321 x 10⁻⁴ m³
