Answer:
20&£+)##&843&()-_££-()&_2+0&&-£_!)
Answer: 0.006in/s
Explanation:
Let the rate at which air is being blown into a spherical balloon be dV/dt which is 1.68in³/s
Also let the rate at which the radius of the balloon is increasing be dr/dt
Given r = 4.7in and Π = 3.14
Applying the chain rule method
dV/dt = dV/dr × dr/dt
If the volume of the sphere is 4/3Πr³
V = 4/3Πr³
dV/dr = 4Πr²
If r = 4.7in
dV/dr = 4Π(4.7)²
dV/dr = 277.45in²
Therefore;
1.68 = 277.45 × dr/dt
dr/dt = 1.68/277.45
dr/dt = 0.006in/s
Answer:
moving the circuit or the magnet gives the same result
Explanation:
The faraday effect establishes that the temporal variation of imaginative flow produces an electric potential
fem = 
dfi / dt
the magnetic flux is
Ф = B. A = B A cos θ
suppose for simplicity that the angle is zero so cos 0 = 1
Φ = B A
By analyzing this expression, the change in magnetic flux can converge while keeping the magnetic field fixed and varying the electric field or keeping the electric field fixed and varying the magnetic field.
Consequently moving the circuit or the magnet gives the same result
Explanation:
w = (4.52 ± 0.02) cm, x = ( 2.0 ± 0.2) cm, y = (3.0 ± 0.6) cm. Find z = x + y - w and its uncertainty.
z = x + y - w = 2.0 + 3.0 - 4.5 = 0.5 cm
Dz = Dx + Dy + Dw = 0.2 + 0.6 + 0.02 = 0.82 rounding to 0.8 cm
So z = (0.5 ± 0.8) cm
Solution with standard deviations, Eq. 1b, Dz = 0.633 cm
z = (0.5 ± 0.6) cm
Notice that we round the uncertainty to one significant figure and round the answer to match.
1. The rocket would experience a downwards force due to gravity from the moon. Also, the rocket would experience an upwards force from the acceleration of the rocket thrusters. Assuming you are neglecting the force of air friction, there would be 2 forces. If you are considering air friction as well, there would be also a downwards force of kinetic air friction acting down on the rocket as it flies, making it 3 forces. If the rocket has already taken off, then there will be no normal force. However, if the rocket is still on the surface of the moon, the rocket would experience a normal force from the moon. Since the question isn't specific I can't provide a concrete answer, but it would range from 2-4 forces depending on the above factors.
2. Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.
F = ma
F = (15.3)(-9.8)
F = -150N
Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object. So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.
3. Same idea as question 2. First determine the weight of the object:
F = ma
F = (30)(-9.8)
F = -294N
The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.
-294 + 150N + x = 0
x = 144N
So the person is exerting 144 N. (Which I hope is a choice that you forgot to type out on the question)
5. Image 3 has the least amount of tension since both tension forces exerted on the cables are shared amongst the two cables and they are parallel to the downwards force of gravity that acts on the object and against them.
