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.
It is an example of liquid. if thats what you are asking for...
The Earth's gravity keeps the Moon orbiting us. It keeps changing the direction of the Moon's velocity. This means gravity makes the Moon accelerate all the time, even though its speed remains constant.
Answer:
The answer to the question is as follows
The acceleration due to gravity for low for orbit is 9.231 m/s²
Explanation:
The gravitational force is given as

Where
= Gravitational force
G = Gravitational constant = 6.67×10⁻¹¹
m₁ = mEarth = mass of Earth = 6×10²⁴ kg
m₂ = The other mass which is acted upon by
and = 1 kg
rEarth = The distance between the two masses = 6.40 x 10⁶ m
therefore at a height of 400 km above the erth we have
r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m
and
=
= 9.231 N
Therefore the acceleration due to gravity =
/mass
9.231/1 or 9.231 m/s²
Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²
To determine the centroid of the object first moment of area is used.
To predict the resistance of a shape to bending and deflection which are directly proportional, second moment of area is used.