Answer:
Magnetic field = 0.534 T
Explanation:
The solving is on the attach document.
The springs of a 1700-kg car compress 5.0 mm when its 66-kg driver gets into the driver seat. If the car goes over a bump, what
1 answer:
Answer:
The frequency of oscillations is 7 Hz
Explanation:
Given;
mass of car, = 1700 kg
mass of driver, = 66 kg
compression of the spring, x = 5mm = 0.005 m
The frequency of the oscillation is given as;
where;
k is force constant
m is the total mass of the car and the driver
m = 1700 kg + 66 kg = 1766 kg
Weight of the car and the driver;
W = mg
W = 1766 x 9.8
W = 17306.8 N
Apply hook's law, to determine the force constant;
F = kx
W = F
Thus, k = W/x
k = 17306.8 / 0.005
k = 3461360 N/m
Now, calculate the frequency
Therefore, the frequency of oscillations is 7 Hz
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Answer:
0.82 mm
Explanation:
The formula for calculation an bright fringe from the central maxima is given as:
so for the distance of the second-order fringe when wavelength = 745-nm can be calculated as:
where;
n = 2
= 745-nm
D = 1.0 m
d = 0.54 mm
substituting the parameters in the above equation; we have:
= 0.00276 m
= 2.76 × 10 ⁻³ m
The distance of the second order fringe when the wavelength = 660-nm is as follows:
= 1.94 × 10 ⁻³ m
So, the distance apart the two fringe can now be calculated as:
= 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m
= 10 ⁻³ (2.76 - 1.94)
= 10 ⁻³ (0.82)
= 0.82 × 10 ⁻³ m
= 0.82 × 10 ⁻³ m
= 0.82 mm
Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm
Answer:
0.25m/s
Explanation:
Given parameters
m₁ = 5kg
v₁ = 1.0m/s
m₂ = 15kg
v₂ = 0m/s
Unknown:
velocity after collision = ?
Solution:
Momentum before collision and after collision will be the same. For inelastic collision;
m₁v₁ + m₂v₂ = v(m₁ + m₂)
Insert parameters and solve for v;
5 x 1 + 15 x 0 = v (5 + 15 )
5 = 20v
v = = 0.25m/s