a rocket with a lift of mass 20 000kg is blasted upwards with an initial acceleration of 5.0m/s². calculate the initial thurst(f
1 answer:
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Answer:
The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.
Explanation:
Given that,
Mass of object = 5 kg
Speed = 3 m/s
Mass of stationary object = 3 kg
Moving object deflected = 30°
Stationary object deflected = 31°
We need to calculate the velocity of each ball after collision
Using conservation of momentum
Along x-axis
Put the value into the fomrula
....(I)
Along y -axis
Put the value into the formula
...(II)
From equation (I) and (II)
Put the value of v₁ in equation (I)
Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.
Answer:
a) 29062.125 N·m
b) 0 N·m
c)
d) T = 11186.02 N
Explanation:
We are given
Beam mass = 1975 kg
Beam length = 3 m
Cable angle = 60° above horizontal
a) We have the formula for torque given as follows;
Torque about the pin = Force × Perpendicular distance of force from pin
Where the force = Force due to gravity or weight, we have
Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N
Point of action of force = Midpoint for a uniform beam = length/2
∴ Point of action of force = 3/2 = 1.5 m
Torque due to gravity = 19374.75 N × 1.5 m = 29062.125 N·m
b) Torque about the pinned end due to the contact forces between the pin and the beam is given by the following relation;
Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m
c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows
Sum of moments about p is given as follows
∑M = 0 gives;
T·sin(θ) × L= M×L/2×g
Therefore torque due to tension is given by the following expression
d) Plugging in the values in the torque due to tension equation, we have;
Therefore, we make the tension force, T the subject of the formula hence