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Eddi Din [679]
3 years ago
13

2. A brick is sitting on a building 22 m high. It has a mass of 7.9 kg. What amount of potential

Physics
1 answer:
Citrus2011 [14]3 years ago
6 0

Answer:

1703.24J

Explanation:

Given parameters:

Mass of brick  = 7.9kg

Height of building  = 22m

Unknown:

Potential energy of the brick  = ?

Solution:

The potential energy of a body is the energy at rest of the body. Mathematically;

 

         P.E = mgh

m is the mass of the brick

g is the acceleration due to gravity

h is the height of the building

  Insert the given parameters and solve;

        P.E  = 7.9 x 9.8 x 22  = 1703.24J

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Hari planted to go abroad. change into negative​
marishachu [46]

Answer:

Hari didn't plan to go abroad

Explanation:

Abroad planned to go to hari.

7 0
3 years ago
A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦
blagie [28]

Answer:

W = 0.63 KJ

Explanation:

Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.  

W= F*d *cosα Formula (1)  

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)  

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force,  inclined at  θ=24.7° above the horizontal

Calculated of the W

W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the  tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7°  (N)

Calculated of the FN  

We apply the formula (2)  

∑Fy = m*ay ay = 0  

FN +Ty- W = 0  

FN = W-Ty  

FN =  89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*(  89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax  ,  ax= 0 ,because the speed of the cart  is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

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I believe that the answer is C. Hope this Helps:)))
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