<h3>
Answer: 130 newtons</h3>
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Explanation:
We'll need the acceleration first.
- The initial speed (let's call that Vi) is 8.0 m/s
- The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
- This happens over a duration of t = 4.0 seconds
The acceleration is equal to the change in speed over change in time
a = acceleration
a = (change in speed)/(change in time)
a = (Vf - Vi)/(4 seconds)
a = (0 - 8.0)/4
a = -8/4
a = -2
The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.
Here's a further break down of Sam's speeds at the four points of interest
- At 0 seconds, he's going 8 m/s
- At the 1 second mark, he's slowing down to 8-2 = 6 m/s
- At the 2 second mark, he's now at 6-2 = 4 m/s
- At the 3 second mark, he's at 4-2 = 2 m/s
- Finally, at the 4 second mark, he's at 2-2 = 0 m/s
Next, we'll apply Newton's Second Law of motion
F = m*a
where,
- F = force applied
- m = mass
- a = acceleration
We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg
So the force applied must be:
F = m*a
F = 65*(-2)
F = -130 newtons
This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.
The magnitude of this force is |F| = |-130| = 130 newtons
Answer:
B)
The magnitude of induced emf in the conducting loop is 0.99 mV.
Explanation:
Rate of increase in magnetic field per unit time = 0.090 T/s
Area of the conducting loop = 110 cm^2 = 0.0110 m^2
Electromagnetic induction is the production of an emf or voltage in a coil of wire due to a changing magnetic field through the coil.
Induced e.m.f is given as:
EMF = (-N*change in magnetic field/time)*Area
EMF = rate of change of magnetic field per unit time * Area
EMF = 0.090 * 0.0110
EMF = 0.00099 V
EMF = 0.99 mV
Answer:
The neutron can be found in the nucleus of the atom with the proton.
Answer:
V = 576 V
Explanation:
Given:
- The area of the two plates A = 0.070 m^2
- The space between the two plates d = 6.3 mm
- Te energy density u = 0.037 J /m^3
Find:
- What must the potential difference between the plates V?
Solution:
- The energy density of the capacitor with capacitance C and potential difference V is given as:
u = 0.5*ε*E^2
- Where the Electric field strength E between capacitor plates is given by:
E = V / d
Hence,
u = 0.5*ε*(V/d)^2
Where, ε = 8.854 * 10^-12
V^2 = 2*u*d^2 / ε
V = d*sqrt ( 2*u / ε )
Plug in values:
V = 0.0063*sqrt ( 2 * 0.037 / (8.854 * 10^-12) )
V = 576 V