Answer:
θ=π/2
Explanation:
The definition of work is W = → F ⋅ → d = q E c o s θ d W=F→⋅d→=qEcosθd. So if no work is done, the displacement must be in the direction perpendicular to the force ie c o s θ = 0 → θ = π / 2 cosθ=0→θ=π/2
A charged particle can be displaced without any external work done on it in a uniform electric field when its movement is perpendicular to the direction of the electric field.
In a uniform electric field, the electric force is the same in every direction. Therefore, if a charge were to be displaced perpendicular to the original direction of the electric field (i.e., in the y or z direction), it would not encounter any extra electric forces. This means there would be no external work being done on the charge. When a charge is moved perpendicular to an electric field, the field does not affect it, and hence, no work is done by the field.
In other words, a charge can be displaced in this field without any external work being done on it when it is moved in a direction perpendicular to the uniform electric field, either in y-axis or z-axis, assuming the electric field is constant in the x-axis direction.
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Velocity vs. Time
1400
1300
1200
1100
1000
4 45
3
1 (s)
At what point is the car the fastest?
A. t = 1.0 s
B. t = 4.2 s
C. t = 3.0 s
D. t = 4.5 s
From the graph, it is clear that, the velocity is at a time of 1 s is highest. The velocity at 1 second corresponds to 1250 km/hr. Then it decreases with time.
The velocity - time graph shows the change in velocity with respect to time. The velocity is placed in y -axis and time is given in x - axis. The slope of the curve in velocity - time graph gives the acceleration of the object.
Similarly, the position of the object in meter after a t seconds can be determined from the velocity - time graph. It is the rate of change in velocity of the object.
From the graph, it is clear that, the curve has its peak at 1 second. After that the peak descends down. Hence, the maximum velocity of the car is at a time of 1 second at which the velocity is 1250 km/hr.
Find more on velocity - time graph :
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Answer:
Explanation:
Given that,
Frequency of the radio signal,
It is detected at a pint 2.1 km from the transmitter tower, x = 2.1 km
The amplitude of the electric field is, E = 800 mV/m
Let I is the intensity of the radio signal at that point. Mathematically, it is given by :
is the rms value of electric field,
So, the intensity of the radio signal at that point is . Hence, this is the required solution.
At constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.
According to Newton's second law of motion, the force applied to a an object is directly proportional to the product of mass and acceleration of the object.
F = ma
Acceleration is the change in the velocity of an object per change in time of motion.
Thus, we can conclude that at constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.
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Answer:
Explanation:
When the puck is sliding on the ice, there is no force being exerted on the puck to keep it moving forward. Instead, inertia keeps the puck moving forward. Friction between the puck and the ice gradually slows the puck down. You hit a hockey puck and it slides across the ice at nearly a constant speed
So lets fill out what we have first:
Vi or initial velocity = 20 m/s
Acceleration or a = 4 m/s^2
Time for the motion = 10s
Now, using the four main kinematic equations we can deduce that the best kinematic equation to use in these terms is:
Δx = Vi(t) + 0.5at²
Plug all of our information in:
Δx = (20)(10) + (0.5)(4)(100)
Δx = 400 m
Answer:
400 m
Explanation:
answer on ed
a). The magnitude along with the direction of the electric field releasing westward force of × N would be:
× N/C is Eastward Direction
b). The magnitude along with the force of the direction that this field releases on proton would be:
× in Eastward Direction
a). Given that,
Force × N
As we know,
Force Charge × Electric Field
So,
∵ Electric Field
× ) ×
×
The direction of the field would be opposite i.e. Eastward direction due to the field carrying a -ve charge.
b). The magnitude carried by the force working on the proton would be the same with an opposite direction due to +ve charge.
∵ Force × N in Eastward direction.
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Explanation:
(a) E = F/q
E = 4.8×10^-17/1.6×10^-19
E = 300 N/C
(b) same magnitude of electric field is exerted on proton