Answer:
When frequency increases more wave crests pass a fixed point each second. That means the wavelength shortens. So, as frequency increases, wavelength decreases. The opposite is also true—as frequency decreases, wavelength increases.
Explanation:
Answer:
Explanation:
It is given that,
Mass of the baseball, m = 0.14 kg
It is dropped form a height of 1.8 m above the ground. Let u is the velocity when it hits the ground. Using the conservation of energy as :
h = 1.8 m
u = 5.93 m/s
Let v is the speed of the ball when it rebounds. Again using the conservation of energy to find it :
h' = 1.4 m
v = -5.23 m/s
The change in the momentum of the ball is given by :
So, the change in the ball's momentum occurs when the ball hits the ground is 1.56 kg-m/s. Hence, this is the required solution.
Answer:
12500 V
Explanation:
The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:
where
is the potential difference between the plates
E is the electric field strength
d is the distance between the plates
For the capacitor in this problem, we have
Substituting, we find
Answer:
a. A = 0.1656 m
b. % E = 1.219
Explanation:
Given
mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150 m/s , k = 500 N / m
a.
To find the amplitude of the resulting SHM using conserver energy
ΔKe + ΔUg + ΔUs = 0
¹/₂ * m * v² - ¹/₂ * k * A² = 0
A = √ mB * vₓ² / k
vₓ = mb * u₁ / mb + mB
vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518
A = √ 4.0 kg * (1.852 m/s)² / (500 N / m)
A = 0.1656 m
b.
The percentage of kinetic energy
%E = Es / Ek
Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5
Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N
% E = 13.72 / 1125 = 0.01219 *100
% E = 1.219
F = ma
F 1500N
m = 100 kg
1500 = 100 * a
Solve for a.