Answer:
D). energy resulting from the attraction between two masses.
Explanation:
Explanation:
Momentum is mass times speed.
p = mv
a) p = (1500 kg) (25.0 m/s) = 37,500 kg m/s
b) p = (40,000 kg) (1.00 m/s) = 40,000 kg m/s
The truck has more linear momentum.
Momentum in the y direction:
pᵧ = (1500 kg) (25.0 m/s) = 37,500 kg m/s
Momentum in the x direction:
pₓ = (1500 kg) (15.0 m/s) = 22,500 kg m/s
Total linear momentum:
p² = pₓ² + pᵧ²
p² = (22,500 kg m/s)² + (37,500 kg m/s)²
p = 43,700 kg m/s
Answer:
The separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)
Explanation:
The relationship between energy and wavelength is expressed below:
E = hc/λ
λ = hc/EK - EL
Considering the condition of Bragg's law:
2dsinθ = mλ
For the first order Bragg's law of reflection:
2dsinθ = (1)λ
2dsinθ = hc/EK - EL
d = hc/2sinθ(EK - EL)
Where 'd' is the separation distance between the parallel planes of an atom, 'h' is the Planck's constant, 'c' is the velocity of light, θ is the angle of reflection, 'EK' is the energy of the K shell and 'EL' is the energy of the K shell.
Therefore, the separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)
Answer:
23 m/s downward
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<em>Taking the downward direction as positive</em>
<u>We are given:</u>
Initial velocity of the marble (u) = 0 m/s
Time interval (t) = 2.3 seconds
Final velocity (v) = x m/s
<u>Solving for the Final velocity:</u>
<u>Acceleration of the Marble:</u>
We know that gravity will make the marble accelerate at a constant acceleration of 10 m/s
<u>Final velocity:</u>
v = u + at [First equation of motion]
x = 0 + (10)(2.3) [replacing the given values]
x = 23 m/s
Hence, after 2.3 seconds, the marble will move at a velocity of 23 m/s in the downward direction
Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
