Answer:
The magnitude of the electric field is 
Explanation:
Given:
Charge of electron
C
Separation between two charges
m
For finding the magnitude of the electric field,

Where 


Therefore, the magnitude of the electric field is 
Answer:
I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.
Explanation:
In order to increase the horizontal distance covered by the ball, we need to examine the variables involved in the formula of range of projectile. The formula for the range of projectile is given as follows:
R = V₀² Sin 2θ/g
where, g is a constant on earth (acceleration due to gravity) and θ is the angle of ball with ground at the time of launching. The value of θ should be 45° for maximum range. In this case we do not know the angle so, we can not tell if we should change it or not.
The only parameter here which we can increase to increase the range is launch velocity (V₀). The formula for V₀ in terms of horizontal and vertical components is as follows:
V₀ = √(V₀ₓ² + V₀y²)
where,
V₀ₓ = Horizontal Velocity
V₀y = Vertical Velocity
Hence, it is clear from the formula that we can increase both the horizontal and vertical velocity to increase the initial speed which in turn increases the horizontal distance covered by the ball.
<u>Therefore, I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.</u>
Answer:
a) t = 0.75 s, b) t = 2.25 s
Explanation:
The speed of sound is constant in a material medium
v = 340 m / s
we can use the relations of uniform motion to find the time
v = x / t
t = x / v
In the exercise, the observer's distance to the wall is indicated d = 510 m, it also indicates that the shot is fired at the midpoint
x = d / 2
a) direct sound the distance from the observer to the screen is
x = 510/2 = 255 m
t = 255/340
t = 0.75 s
b) echo sound.
In this case the sound reaches the wall bounces, the distance is
x = d / 2 + d
x = 3/2 d
x = 3/2 510
x = 765 m
the time is
t = x / v
t = 765/340
t = 2.25 s
Answer:
v₁ = 4 [m/s].
Explanation:
This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.

where:
P = linear momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]

where:
m₁ = mass of the tank = 500 [kg]
v₁ = velocity of the tank after firing the missile [m/s]
m₂ = mass of the missile = 20 [kg]
v₂ = velocity of the missile after firing = 100 [m/s]
![(500*v_{1})=(20*100)\\v_{1}=2000/500\\v_{1}=4[m/s]](https://tex.z-dn.net/?f=%28500%2Av_%7B1%7D%29%3D%2820%2A100%29%5C%5Cv_%7B1%7D%3D2000%2F500%5C%5Cv_%7B1%7D%3D4%5Bm%2Fs%5D)
Answer:
Vector Quantity
Explanation:
A Vector quantity is a quantity has both magnitude and direction while scalar quantity has only magnitude.