To solve this problem we will use the definition of the kinematic equations of centrifugal motion, using the constants of the gravitational acceleration of the moon and the radius of this star.
Centrifugal acceleration is determined by
Where,
v = Velocity
r = Radius
From the given data of the moon we know that gravity there is equivalent to
While the radius of the moon is given by
If we rearrange the function to find the speed we will have to
The speed for this to happen is 1.7km/s
<h3>
Answer:</h3>
5.5 N
<h3>
Explanation:</h3>
<u>We are given;</u>
- Real weight of an object in air as 10 N
- Apparent weight in water as 7 N
- Relative density of a liquid is 1.5
We need to calculate the apparent weight when in liquid .
- First we calculate upthrust
Upthrust = Real weight - Apparent weight
= 10 N - 7 N
= 3 N
- Then calculate the upthrust in the liquid.
we need to know that;
Relative density of a liquid = Upthrust in liquid/Upthrust in water
Therefore;
1.5 = U ÷ 3 N
Upthrust in liquid = 3 N × 1.5
= 4.5 N
- Therefore, the upthrust of the object in the liquid is 4.5 N
But, Upthrust = Real weight - Apparent weight
Therefore;
Apparent weight in Liquid = Real weight - Upthrust
= 10 N - 4.5 N
= 5.5 N
Thus, the weight when the object is immersed in the liquid is 5.5 N
Answer:
80 Joules.
Explanation:
Given that a 20 kg sled stars at the top of a hill which is 10 m above the bottom and slides a distance of 50 m, ending at the bottom of the hill with a speed of 8 m/s. During the slide, the work done by gravity is about ?
The parameters given are:
Mass M = 20 kg
Height h = 10 m
Velocity V = 8 m/s
The workdone by gravity will be equal to the total energy
At the bottom of the hill, the total energy will be equal to the maximum kinetic energy.
Maximum K.E = 1/2mv^2
K.E = 1/2 × 20 × 8
K.E = 80J
Therefore, the workdone by the gravity is equal to 80 Joules.
Answer:
<h3>0.445</h3>
Explanation:
In friction, the coefficient of friction formula is expressed as;
Ff is the frictional force = Wsinθ
R is the reaction = Wcosθ
Substitute inti the equation;
Given
θ = 24°
Hence the coefficient of kinetic friction between the box and the ramp is 0.445
