Let F be the magnitude of the force. The impulse of this force while the ball is in contact with the wall is
Ft = F (0.0210 s)
and this impulse is equal to the change in the ball's momentum,
m ∆v = (1.30 kg) (6.50 m/s - (-10.5 m/s)) = (1.30 kg) (17.0 m/s)
Solve for F :
F (0.0210 s) = (1.30 kg) (17.0 m/s)
F = (1.30 kg) (17.0 m/s) / (0.0210 s)
F ≈ 1050 N
Element formula....
I do hope this helps you Darling
Answer:
Electric flux;
Φ = 30.095 × 10⁴ N.m²/C
Explanation:
We are given;
Charge on plate; q = 17 µC = 17 × 10^(-6) C
Area of the plates; A_p = 180 cm² = 180 × 10^(-4) m²
Angle between the normal of the area and electric field; θ = 4°
Radius;r = 3 cm = 3 × 10^(-2) m = 0.03 m
Permittivity of free space;ε_o = 8.85 × 10^(-12) C²/N.m²
The charge density on the plate is given by the formula;
σ = q/A_p
Thus;
σ = (17 × 10^(-6))/(180 × 10^(-4))
σ = 0.944 × 10^(-3) C/m²
Also, the electric field is given by the formula;
E = σ/ε_o
E = (0.944 × 10^(-3))/(8.85 × 10^(-12))
E = 1.067 × 10^(8) N/C
Now, the formula for electric flux for uniform electric field is given as;
Φ = EAcos θ
Where A = πr² = π × 0.03² = 9π × 10^(-4) m²
Thus;
Φ = 1.067 × 10^(8) × 9π × 10^(-4) × cos 4
Φ = 30.095 × 10⁴ N.m²/C
Answer:
Explanation:
Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.
For faster car on the road,
v = 2v
..........(1)
For the slower car on the road,
............(2)
Equation (1) becomes,
So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.
Answer: 10.3m/s
Explanation:
In theory and for a constant velocity the physics expression states that:
Eq(1): distance = velocity times time <=> d = v*t for v=constant.
If we solve Eq (1) for the velocity (v) we obtain:
Eq(2): velocity = distance divided by time <=> v = d/t
Substituting the known values for t=15s and d=155m we get:
v = 155 / 15 <=> v = 10.3
