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Answer:
a) time taken to catch up with speeding car is 12.25 secs
b) the police car will travel 273.8 m to catch up with the speeding car
Explanation:
Given that;
speed of car = 50 mi/hr = 22.352 m/s
acceleration of police car = 10 mi/hr = 4.47 m/s²
= 70 mi/hr = 31.29 m/s
Now time taken to reach maximum speed is t₁
so
=
+ at₁
we substitute
31.29 = 0 + 4.47t₁
t₁ = 31.29 / 4.47
t₁ = 7 sec
now
d₁ = 0 + 1/2 × at₁²
d₁ = 0 + 1/2 × 0 + 4.47×(7)²
d₁ = 109.5 m
so distance travelled by the speeding car in time t₁ will be
=
× t₁
we substitute
= 22.352 × 7
= 156.46 m
now distance between polive car and speeding car
Δd = - d₁
Δd = 156.46 - 109.5
Δd = 46.96 m
time taken to cover Δd will be
t₂ = Δd / ( -
)
t₂ = 46.96 / ( 31.29 - 22.352 )
t₂ = 46.96 / 8.938
t₂ = 5.25 sec
distance travelled by the police in time t₂ will be
d₂ = × t₂
d₂ = 31.29 × 5.25
d₂ = 164.3 m
a) How long will it take before the officer catches up to the speeding car;
time taken to catch up with speeding car;
t = t₁ + t₂
t = 7 + 5.25
t = 12.25 secs
Therefore, time taken to catch up with speeding car is 12.25 secs
b) how far will it have travelled in order to do so;
distance = d₁ + d₂
distance = 109.5 + 164.3
distance = 273.8 m
Therefore, the police car will travel 273.8 m to catch up with the speeding car
a) E = 0
b)
Explanation:
a)
We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:
where
E is the electric field
q is the charge contained by the Gaussian surface
is the vacuum permittivity
Here we want to find the electric field at a distance of
r = 12 cm = 0.12 m
Here we are between the inner radius and the outer radius of the shell:
However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.
This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:
q = 0
Therefore, the magnitude of the electric field is also zero:
E = 0
b)
Here we want to find the magnitude of the electric field at a distance of
r = 20 cm = 0.20 m
from the centre of the shell.
Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.
Therefore, it is given by:
where in this problem:
is the charge on the shell
is the distance from the centre of the shell
Substituting, we find:
To solve this problem, it is necessary to apply the concepts related to force described in Newton's second law, so that
F = ma
Where,
m = mass
a = Acceleration (Gravitational acceleration when there is action over the object of the earth)
Torque, as we know, is the force applied at a certain distance, that is,
Where
F= Force
d = Distance
Our values are given as,
Since the system is in equilibrium the difference of the torques is the result of the total Torque applied, that is to say
Therefore the magnitude of the frictional torque at the axle of the pulley if the system remains at rest when the balls are released is