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mixas84 [53]
3 years ago
15

As a rock sinks deeper and deeper into water of constant density, what happens to the buoyant force on it? As a rock sinks deepe

r and deeper into water of constant density, what happens to the buoyant force on it? It increases. It may increase or decrease, depending on the shape of the rock. It may increase or decrease, depending on the weight of the rock. It remains constant. It decreases. g
Physics
1 answer:
Sidana [21]3 years ago
3 0

Answer:

It remains constant

Explanation:

As we know that buoyant force on an object given as

Fb = ρ Vd g

ρ= Density of fluid

Vd=Volume displace by body

g=10 m/s²

Fb =buoyant force

So from above we can say that buoyant force does not depends on the depth. It only depends on the fluid density and volume displace by body.

So when rock gets deeper and deeper the buoyant force will remain constant.

It remains constant

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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

6 0
4 years ago
Mer and count Assessment items Assessment started:
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Answer:

Asteroids

Explanation:

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These are believed to be the remnants particles that are present in space right after the formation of the solar system.

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