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saw5 [17]
2 years ago
10

Idris has a mass of 65 kg. He is standing on the second floor of his apartment building, 5 meters above the ground. What is his

gravitational potential energy (GPE)? O A. 325J O B. 31,213 J O C. 3,185 J O D. 79.8J​
Physics
1 answer:
lapo4ka [179]2 years ago
8 0

Answer:A.325J

brainliest would be apreciated!

hope I helped!

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Cómo puedo saber si el agua destilada que nos vendieron es destilada de verdad?
Pepsi [2]

Answer:

Caliente el agua del grifo hasta que se convierta en vapor. Cuando el vapor se vuelve a condensar en agua

Explanation:

7 0
3 years ago
Please help! I will give brainlist.
ki77a [65]

Answer:

Either temperature or energies but I am pretty sure its temperature

Explanation:

4 0
3 years ago
Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass
kirza4 [7]

Answer:

a. 0.342 kg-m² b. 2.0728 kg-m²

Explanation:

a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m

I = 1/2MR²

= 1/2 × 56.5 kg × (0.11 m)²

= 0.342 kgm²

So the moment of inertia of the skater is

b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)

I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m

I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²

= 0.1802 kg-m² + 0.6852 kg-m²

= 0.8654 kg-m²

The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².

So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²

5 0
3 years ago
Two ropes support a load of 478 kg. The two ropes are perpendicular to each other, and the tension in the first rope is 2.2 time
sveta [45]

Answer:

T₂ = 1937.68 N

Explanation:

First, we will calculate the weight of the object:

W = mg = (478\ kg)(9.81\ m/s^2)\\W = 4689.18\ N

Now, we will calculate the resultant tension in the ropes. Since the ropes are perpendicular. Therefore,

T = \sqrt{T_1^2+T_2^2}\\

where,

T = Resultant Tension

T₁ = Tension in rope 1

T₂ = Tension in rope 2

According to the given condition tension in the first rope is 2.2 times the tension in the second rope:

T₁ = 2.2 T₂

Therefore

T = \sqrt{(2.2T_2)^2 + T_2^2}\\\\T =  2.42T_2

Now, the weight of the object must be equal to the resultant tension for equilibrium:

T = W\\2.42T_2 = 4689.18\ N\\\\

<u>T₂ = 1937.68 N</u>

5 0
3 years ago
Describe the conditions necessary for equilibrium on a balance.
tia_tia [17]

Answer:

Explanation:

A body under concurrent forces (concurrent forces are forces with their line of action acting at a point) is said to be in equilibrium if the sum total of the forces acting on the body is zero.

For forces acting on a balance, the sum of upward forces and the downward forced acting on the balance must be equal for the balance to be in equilibrium.

Also the sum of clockwise moment must be equal to the sum of anticlockwise moment acting on the balance.

Moment is the product of force and the perpendicular distance from the force to the pivot.

7 0
4 years ago
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