Answer:
t = 4.08 s
R = 40.8 m
Explanation:
The question is asking us to solve for the time of flight and the range of the rock.
Let's start by finding the total time it takes for the rock to land on the ground. We can use this constant acceleration kinematic equation to solve for the displacement in the y-direction:
We have these known variables:
- (v_0)_y = 0 m/s
- a_y = -9.8 m/s²
- Δx_y = -20 m
And we are trying to solve for t (time). Therefore, we can plug these values into the equation and solve for t.
- -20 = 0t + 1/2(-9.8)t²
- -20 = 1/2(-9.8)t²
- -20 = -4.9t²
- t = 4.08 sec
The time it takes for the rock to reach the ground is 4.08 seconds.
Now we can use this time in order to solve for the displacement in the x-direction. We will be using the same equation, but this time it will be in terms of the x-direction.
List out known variables:
- v_0 = 10 m/s
- t = 4.08 s
- a_x = 0 m/s
We are trying to solve for:
By using the same equation, we can plug these known values into it and solve for Δx.
- Δx = 10 * 4.08 + 1/2(0)(4.08)²
- Δx = 10 * 4.08
- Δx = 40.8 m
The rock lands 40.8 m from the base of the cliff.
Answer:
(
)=1913.31 N/m^2
Explanation:
given:
=0.85
=90 m/s
γ∞=1.23 kg/m^3
solution:
since outside pressure is atm pressure vaccum can be defined by ()
=√2()/γ∞[
-1]
()=1913.31 N/m^2
Answer:
21870.3156 N
Explanation:
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 1.6 m/s²
Equation of motion
The acceleration of the craft should be 1.02234 m/s²
Weight of the craft
Thrust
The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N
Answer:
v = 24 m/s, rightwards
Explanation:
Given that,
The mass of TBT explosive = 5 kg
It explodes into two pieces.
One of the pieces weighing 2.0 kg flies off to the left at 36 m/s. Let left be negative and right be positive.
The law of conservation of momentum holds here. Let v be the final speed of the remaining piece. So,
So, the final speed of the remaining piece is 24 m/s and it is in the right direction.
Let us examine the given situations one at a time.
Case a. A 200-pound barbell is held over your head.
The barbell is in static equilibrium because it is not moving.
Answer: STATIC EQUILIBRIUM
Case b. A girder is being lifted at a constant speed by a crane.
The girder is moving, but not accelerating. It is in dynamic equilibrium.
Answer: DYNAMIC EQUILIBRIUM
Case c: A jet plane has reached its cruising speed at an altitude.
The plane is moving at cruising speed, but not accelerating. It is in dynamic equilibrium.
Answer: DYNAMIC EQUILIBRIUM
Case d: A box in the back of a truck doesn't slide as the truck stops.
The box does not slide because the frictional force between the box and the floor of the truck balances out the inertial force. The box is in static equilibrium.
Answer: STATIC EQUILIBRIUM
