Answer:
Taking forces along the plane
F cos θ - M g sin θ -100 = M a net of forces along the plane
F = (M a + M g * .5 + 100) / .866 solving for F
F = (80 * 1.5 + 80 * 9.8 * .5 + 100) / .866 = 707 N
F = 707 N acting along the plane
Fn = F sin θ + M g cos θ forces acting perpendicular to plane
Fn = 707 * 1/2 + 80 * 9.8 * .866 = 1030 Newtons forces normal to plane
(this would give a coefficient of friction of 100 / 1030 = .097 = Fn)
Answer:
the final velocity of the car is 59.33 m/s [N]
Explanation:
Given;
acceleration of the car, a = 13 m/s²
initial velocity of the car, u = 120 km/h = 33.33 m/s
duration of the car motion, t = 2 s
The final velocity of the car in the same direction is calculated as follows;
v = u + at
where;
v is the final velocity of the car
v = 33.33 + 13 x 2
v = 59.33 m/s [N]
Therefore, the final velocity of the car is 59.33 m/s [N]
Answer:
True
Explanation:
East, up, and left all define as a direction.
Answer:
Time period of the osculation will be 0.0671 sec
Explanation:
It is given a vertical spring is stretched by 4 cm
So change in length of the spring x = 4 cm = 0.04 m
Mass which is hung from it m = 12 gram = 0.012 kg
Sprig force will be equal to weight of the mass
So 
k = 244.7 N/m
Now new mass is m = 28 gram = 0.028 kg
So time period with new mass will be
Let volume of empty boat be = 100% = 1V
and mass of boat be M
In water 10%, 0.1V of the volume is submerged.
Mass, m of 1200kg increases the submerging from 10%, 0.1V to 70%, 0.7V
M leads to 0.1V boat submerging
boat submerging.
M + 1200kg leads to 0.7V boat submerging.
This is 60%, 0.6 V increase
By comparison
(M+1200kg) * 0.1V = 0.7V * M
0.1M + 120kg = 0.7M
120kg = 0.7M - 0.1M
120kg = 0.6M
M = (120/0.6)kg
M = 200kg.
The mass of the boat is 200kg.
