Static friction is the friction that exists between a stationary object and the surface on which it's resting.
frictional force occurs when you try to push an object alongside a surface.
I attached the full question.
We know that for a parallel-plate capacitor the surface charge density is given by the following formula:
Where V is the voltage between the plates and d is separation.
Voltage is by definition:
Voltage is analog to the mechanical work done by the force.
Above formula is correct only If the field is constant, and we can assume that it is since no function has been given.
The charge density would then be:
Please note that elecric permittivity of air is very close to elecric permittivity of vacum, it is common to use them <span>interchangeably</span>.
The answer is D. Electric resistance increases with an increase in the length of a wire and as a result current flow decreases. There is a direct relationship between the length of the wire and the resistance. The longer the wire, the more resistance there will be. Additionally, from Ohm's Law, current is inversely proportional to resistance. This means as the current increases, resistance decreases or vice versa.
Answer:
x component 3.88 y- component 14.488
Explanation:
We have given a vector A which has a magnitude of 15 m/sec which is at 75° counter-clock wise ( anti-clock wise) from x -axis which is clearly shown in bellow figure
Now x-component will be 15 cos75°=3.8822 ( as it makes an angle of 75° with x-axis )
y- component will be 15 sin 75°=14.488
For verification the resultant of x and y component should be equal to 15
So
Answer:
1.86 m
Explanation:
First, find the time it takes to travel the horizontal distance. Given:
Δx = 52 m
v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s
a = 0 m/s²
Find: t
Δx = v₀ t + ½ at²
52 m = (22.2 m/s) t + ½ (0 m/s²) t²
t = 2.35 s
Next, find the vertical displacement. Given:
v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s
a = -9.8 m/s²
t = 2.35 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²
Δy = 4.91 m
The distance between the ball and the crossbar is:
4.91 m − 3.05 m = 1.86 m