A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at -1.2 m/s2, which force is closest to the net force a
1 answer:
Answer:
F= 1320 N
Explanation:
Given that
mass ,m= 1100 kg
Acceleration ,a= 1.2 m/s²
We know that ,According to the second law of Newton's
Force = Mass x acceleration
F= m a
F=Force ,a=acceleration
m=mass
Now by putting the values in the above equation
F= 1320 N
Therefore the force on the vehicle will be 1320 N.
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Answer:
T = 295.57 s
Explanation:
given,
mass of the rocket = 200 Kg
mass of the fuel = 100 Kg
acceleration = 35 m/s²
time, t = 35 s
time taken by the rocket to hit the ground, = ?
Final speed of the rocket when fuel is empty
using equation of motion
v = u + a t
v = 0 + 35 x 35
v = 1225 m/s
height of the rocket where fuel is empty
v² = u² + 2 a s
1225² = 0 + 2 x 35 x h₁
h₁ = 21437.5 m
After 35 s the rocket will be moving upward till the final velocity becomes zero.
Now, using equation of motion to find the height after 35 s
v² = u² + 2 g h₂
0² = 1225² + 2 x (-9.8) h₂
h₂ = 76562.5 m
total height = h₁ + h₂
= 76562.5 m + 21437.5 m = 98000 m
now, time taken by before the rocket hit the ground
using equation of motion
negative sign is used because the distance travel by the rocket is downward.
4.9 t² - 1225 t - 13500 = 0
t = 260.57 s
neglecting the negative sign
total time the rocket was in air
T = t₁ + t₂
T = 35 + 260.57
T = 295.57 s
Time for which rocket was in air is equal to 295.57 s.
Answer:
a. Fab = 3000 N tension
Fac = 1500 N compression
Fbc = 3000 N compression
b. Aluminum for AB, titanium for AC and BC.
c. Fc = 4000 N
Explanation:
a. Draw free body diagrams of the pins at A and B, assuming that all internal forces are in tension (pulling away from the pin).
Sum of the forces on A in the x direction:
∑F = ma
H − Fab sin 60° = 0
Fab = H / sin 60°
Fab = 2600 / sin 60°
Fab = 3000 N
Sum of the forces on A in the y direction:
∑F = ma
-Fab cos 60° − Fac = 0
Fac = -Fab cos 60°
Fac = -H / tan 60°
Fac = -2600 / tan 60°
Fac = -1500 N (it's negative, so it's actually in compression).
Sum of the forces on B in the x direction:
∑F = ma
Fab cos 30° + Fbc cos 30° = 0
Fbc = -Fab
Fbc = -H / sin 60°
Fbc = -3000 N (it's negative, so it's in compression)
b. AB is in tension, so it should be aluminum.
AC and BC are in compression, so they should be titanium.
c. Draw a free body diagram of the entire triangle ABC. Assume there is both a horizontal and vertical reaction force at C, and assume they point in the +x and +y directions.
Sum of the forces in the x direction:
∑F = ma
Fcx + H = 0
Fcx = -H
Fcx = -2600 N
Sum of the forces in the y direction:
∑F = ma
Fcy − V = 0
Fcy = V
Fcy = 3000 N
The magnitude of the net force is:
Fc² = Fcx² + Fcy²
Fc² = (-2600 N)² + (3000 N)²
Fc = 4000 N
