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Eva8 [605]
3 years ago
6

What type of relationship exists between mass and distance?

Physics
2 answers:
sveticcg [70]3 years ago
5 0

Answer:

Gravitational force

Explanation:

Hope this helps :)

likoan [24]3 years ago
4 0
Answer: Gravitational force

Explanation: Gravitational Force is an attractive force that exists between all objects with mass; an object with mass attracts another object with mass; the magnitude of the force is directly proportional to the masses of the two objects and inversely proportional to the square of the distance between the two objects.
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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

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4 years ago
A skater slides across the ice with an initial velocity of 5.0 m/s. She slows 10 points
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Explanation:

Given that,

The initial velocity of a skater is, u = 5 m/s

She slows to a velocity of 2 m/s over a distance of 20 m.

We can find the acceleration of skater. It is equal to the rate of change of velocity. So, it can be calculated using third equation of motion as follows :

v^2-u^2=2as

a = acceleration

a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{(2)^2-(5)^2}{2\times 20}\\\\a=-0.525\ m/s^2

So, her acceleration is 0.525\ m/s^2 and she is deaccelerating. Also, her initial velocity is given i.e. 5 m/s.

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