Answer:
The mass of the object is 24.5 kg and weight of the object on Mars is 91.14 N.
Explanation:
Weight of the object on the surface of Earth, W = 245 N
On the surface of Earth, acceleration due to gravity, g = 10 m/s²
Weight of an object is given by :
W = mg
m is mass
So, the mass of the object is 24.5 kg
Acceleration due to gravity on Mars, g' = 3.72 m/s²
Weight of the object on Mars,
W' =mg'
W' = 24.5 kg × 3.72 m/s²
= 91.14 N
So, the weight of the object on Mars is 91.14 N.
Answer:
Explanation:
Given:
dimension of uniform plate,
mass of plate,
Now we find the moment of inertia about the center of mass of the rectangular plate is given as:
where:
length of the plate
breadth of the plate
We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .
Now we find the distance between the center of mass and the corner:
Now using parallel axis theorem:
Answer:
The tension in each half of the rope, is approximately 4,908.8 N
Explanation:
The mass of the acrobat, m = 60 kg
The length of the rope, l = 10 m
The extent by which the center dropped = 30 cm = 0.3 m
Let, 'T' represent the tension in each half of the rope
Weight, W = Mass, m × The acceleration due to gravity, g
∴ W = m × g
The acceleration due to gravity, g ≈ 9.8 m/s²
∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N
The angle the dropped rope makes with the horizontal, θ is given as follows;
θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°
At equilibrium, the sum of vertical forces, = 0
The vertical component of the tension, , in each half of the rope is given as follows;
= T × sin(θ)
∴ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)
Plugging in the values, with θ = arctan(0.06) for accuracy, we get;
588 N + 2 × T × sin(arctan(0.06) = 0
∴ 2 × -T × sin(arctan(0.06) = 588 N
-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N
The tension in each half of the rope, T ≈ 4,908.8 N.
