The horizontal force applied is 160 N while the velocity is 2.03 m/s.
<h3>What is the speed of the car?</h3>
The work done by the car is obtained as the product of the force and the distance;
W = F x
F = ?
x = 30.0 m
W = 4,800 J
F = 4,800 J/30.0 m
F = 160 N
But F = ma
a = F/m
a = 160 N/2.30 ✕ 10^3-kg
a= 0.069 m/s
Now;
v^2 = u^2 + 2as
u = 0/ms because the car started from rest
v = √2as
v = √2 * 0.069 * 30
v = 2.03 m/s
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A) 140 degrees
First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is
T = 32 s
So the angular velocity is
Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:
and substituting t = 75 seconds, we find
In degrees, it is
So, the new position is 140 degrees from the initial position at the top.
B) 2.7 m/s
The tangential speed, v, of a point at the egde of the wheel is given by
where we have
r = d/2 = (27 m)/2=13.5 m is the radius of the wheel
Substituting into the equation, we find
<span>This is because Helium
has two valence electrons compared to Hydrogen which has only one. Helium has
more energy levels for an electron to jump thus more spectral lines to occur.
The spectral lines relating to each change of energy level would be more
grouped together and hence the greater chance of them falling in the visible
range.</span>
Answer:
Explanation:
1 )
Here
wave length used that is λ = 580 nm
=580 x 10⁻⁹
distance between slit d = .46 mm
= .46 x 10⁻³
Angular position of first order interference maxima
= λ / d radian
= 580 x 10⁻⁹ / .46 x 10⁻³
= 0.126 x 10⁻² radian
2 )
Angular position of second order interference maxima
2 x 0.126 x 10⁻² radian
= 0.252 x 10⁻² radian
3 )
For intensity distribution the formula is
I = I₀ cos²δ/2 ( δ is phase difference of two lights.
For angular position of θ1
δ = .126 x 10⁻² radian
I = I₀ cos².126x 10⁻²/2
= I₀ X .998
For angular position of θ2
I = I₀ cos².126x2x 10⁻²/2
= I₀ cos².126x 10⁻²
Answer:
Explanation:
la frecuencia = ω/2π, nada cambio
v(max) = ωA → ω2Α = 2ωA duplicara velocidad máxima
a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima
la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía
