Answer:
The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.
Explanation:
Answer: E = 1.62*10^-3 v
Explanation:
Given
L = 1.9 m
V = 19 m/s
B = 4.5*10^-5 T
Induced emf can be found by using the formula for induced emf in a moving conductor.
e = blv, where
B is the component of the earth's magnetic field.
L is the width of the car in m
V is the velocity of the car in m/s
E is the induced emf in volts.
E = 4.5*10^-5 * 1.9 * 19
E = 1.62*10^-3 v
The drivers side is more positive, so the charge would remain positive
The force between two system is inversely proportional to the square the distance between them. The indirect proportion between these items may be expressed as,
F = k / d²
where k is the constant proportionality. Substituting the known values,
0.68 N = k / (1 m)² ; k = 0.68
For the second case,
F = 0.68 / (0.5 m)² = 2.72 N
Thus, the answer is 2.72 N.
Answer:
10.5 m/s
Explanation:
For the first chestnut:
y₀ = 10 m
v₀ = 0 m/s
a = -9.8 m/s²
y = y₀ + v₀ t + ½ at²
y = 10 + (0) t + ½ (-9.8) t²
y = 10 − 4.9t²
When y = 7.5:
7.5 = 10 − 4.9t²
t = 5/7
When y = 0:
0 = 10 − 4.9t²
t = 10/7
For the second chestnut:
y₀ = 10 m
y = 0 m
a = -9.8 m/s²
t = 10/7 s − 5/7 s = 5/7 s
y = y₀ + v₀ t + ½ at²
0 = 10 + v₀ (5/7) + ½ (-9.8) (5/7)²
0 = 10 + 5/7 v₀ − 2.5
v₀ = -10.5
The second chestnut must be thrown downwards at 10.5 m/s.
That relationship is as DIRECT as you can get.