Answer:
The maximum compression of the spring after the collision is 0.15 m
Explanation:
Given data
Mass of the block (m) = 0.80 kg
Initial velocity (v) = 1.2 m/s
Spring constant (k) = 50 N/m
Find the maximum compression of the spring (x) after compression
Potential energy of the spring = Kinetic energy of the block
Kinetic energy of the block = 0.5 × (mv)²
Kinetic energy of the block = 0.5 × (0.80 × 1.2)²
Kinetic energy of the block =0.5 × 0.9216
Kinetic energy of the block = 0.4608 ---------->(1)
Potential energy of the spring = 0.5 × k × x²
Potential energy of the spring = 0.5 × 50 × x²
Potential energy of the spring = 25 x² ---------> (2)
Equate (1) and (2)
25 x² = 0.4608
x² = 0.018432 m²
x =0.1357 = 0.15 m
Therefore the maximum compression of the spring after collision is 0.15 m
Answer:
t= 24080 s
Explanation:
Given that
Current in the wire ,I = 4 A
The charge ,q = 6.02 x 10²³ e C
We know that
I=Current
q=Charge
t=time
Now by putting the values in the above equation we get'
t= 24080 s
Answer:
Option A is correct.
Explanation:
In the given triangle , x is the base and 14 is the hypotenuse for angle 33°.
cos 33 = base / hypotenuse
Putting the given values
cos 33 = x / 14
.84 = x / 14
x = .84 x 14
= 11.7
Option A is correct.
C.) a magnetic field is the correct answer…
Answer:
3317.44401 kg/m³
Explanation:
= Actual density of cube = 1800 kg/m³
= Density change due to motion
v = Velocity of cube = 0.84c
c = Speed of light =
Relativistic density is given by
The cube's density as measured by an experimenter in the laboratory is 3317.44401 kg/m³