Answer:
118.3 J
Explanation:
Givens:
m = 1.4 kg
V = 13 m/s
Formula for kinetic energy:
KE = (1/2)*(m)*(v)^2
KE = .5*(1.4 kg)*(13 m/s)^2
KE 118.3 J
J = Joules
Answer:
Explanation:
- given S = distance from the first = 3.20cm = 0.032m, t = 1.30×10−8 s
- acceleration = 0.032 X 2 /(1.30×10−8)^2
a = 3.79 x 10^14m/s^2
E = ma /q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19
E = magnitude of this electric field. = 2156.3N/C
b) Find the speed of the electron when it strikes the second plate ; V^2 = 2as
= 2 X 3.79 x 10^14 X 0.032
= 4.92 X 10^6m/s
Answer:
θ₀ = 84.78° (OR) 5.22°
Explanation:
This situation can be treated as projectile motion. The parameters of this projectile motion are:
R = Range of Projectile = 150 m
V₀ = Launch Speed of Projectile = 90 m/s
g = 9.8 m/s²
θ₀ = Launch angle (OR) Angle of Elevation = ?
The formula for range of a projectile is given as:
R = V₀² Sin 2θ₀/g
Sin 2θ₀ = Rg/V₀²
Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²
2θ₀ = Sin⁻¹ (0.18)
θ₀ = 10.45°/2
<u>θ₀ = 5.22°</u>
Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:
θ₀ = 90° - 5.22°
<u>θ₀ = 84.78°</u>
Answer:
The new height the ball will reach = (1/4) of the initial height it reached.
Explanation:
The energy stored in any spring material is given as (1/2)kx²
This energy is converted to potential energy, mgH, of the ball at its maximum height.
If the initial height reached is H
And the initial compression of the spring = x
So, mgH = (1/2)kx²
H = kx²/2mg
The new compression, x₁ = x/2
New energy of loaded spring = (1/2)kx₁²
And the new potential energy = mgH₁
mgH₁ = (1/2)kx₁²
But x₁ = x/2
mgH₁ = (1/2)k(x/2)² = kx²/8
H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)
