Answer:
0 N
Explanation:
Applying,
F = qvBsin∅................. Equation 1
Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.
From the question,
Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°
Substitute these values into equation 1
F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)
Since sin0° = 0,
Therefore,
F = 0 N
Answer:
Explanation:
s = s₀ + v₀t + ½at²
s = 0 + 0(15) + ½(6)(15²)
s = 675 m
Not sure what the free fall acceleration is needed for, but if the object is dropped from a high enough point, it will travel in 15 seconds
s = ½10(15²) = 2250 m if air resistance is ignored
Answer:
Explanation:
We are given that
Mass,
Radius,r=0.8 m
Height,h=2.9 m
We have to find the angular acceleration of the cylinder.
According to question
Where
Substitute the value
Where 
Angular acceleration,
Answer:
Average speed is 6m/s
Average velocity is 5.859m/s
Explanation:
Average speed, s = d/t
d is distant
t is time
speed north is 800m and 110s
speed due south is 400m and 90s
Average speed = (800+400) / (110+90)
= 1200/200
=6m/s
Average velocity V = (v + u)/2
v = final velocity
u = initial velocity
V = (7.273 + 4.444)/2 = 11.717/2
V = 5.859m/s
Answer:
Approximately 
(rounded down,) assuming that 
.
The number of repetitions would increase if efficiency increases.
Explanation:
Ensure that all quantities involved are in standard units:
Energy from the cookie (should be in joules, 
):
.
Height of the weight (should be in meters, 
):
.
Energy required to lift the weight by 
without acceleration:
.
At an efficiency of 
, the actual amount of energy required to raise this weight to that height would be:
.
Divide 
by 
to find the number of times this weight could be lifted up within that energy budget:
.
Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.
