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Alina [70]
3 years ago
14

A heavy sled and a light sled, both moving horizontally with the same speed, suddenly slide onto a rough patch of snow and event

ually come to a stop. The coefficient of kinetic friction between the sleds and the rough snow is the same for both of them.
Which of the following statements about these sleds are correct?

A) Both sleds will slide the same distance on the rough snow before stopping.

B) The heavy sled will slide farther on the rough snow than the light sled.

C) The light sled will slide farther on the rough snow than the heavy sled.

D) The friction from the snow will do the same amount of work on both sleds.
Physics
2 answers:
Alex777 [14]3 years ago
5 0

Answer:

A) Both sleds will slide the same distance on the rough snow before stopping.

D) The friction from the snow will do more negative work on the heavy sled than on the light sled.

Explanation:The coefficient of kinetic friction is the ratio of the force of friction of a given object or material to the normal force acting on it,it depends mainly on the nature of the surface in consideration. The higher the roughness of a surface is, the higher the coefficient of kinetic friction. For an object at rest on a plane surface with no other force acting on it,the normal force will be considered as the force of gravity.

maria [59]3 years ago
4 0

Answer:

The correct answers is option A) "Both sleds will slide the same distance on the rough snow before stopping".

Explanation:

For objects moving horizontally at the same speed, an equal coefficient of kinetic friction would mean that both objects will move the same distance before stopping, independently of their mass. The coefficient of kinetic friction is obtain from dividing the force needed to pull an object between the force that holds or prevents the object from moving. The coefficient of kinetic friction is a dimensionless scalar, which means that is independent of the size of the object because it depends only in the forces that act upon it.

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A 61.0-kg person jumps from rest off a 10.0-m-high tower straight down into the water. Neglect air resistance. She comes to rest
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Answer:

Explanation:

In this case, law of conservation of energy will be implemented. It states that "the energy of the system remains conserved until or unless some external force act on it. Energy of the system may went through the conversion process like kinetic energy into potential and potential into kinetic energy.But their total always remain the same in conserved systems."

Given data:

Height of tower = 10.0 m

Depth of the pool = 3.00 cm

Mass of person = 61.0 kg

Solution:

Initial energy = Final energy

U_{i} =  (K.E) + U_{f}

As the person was at height initially so it has the potential energy only.

mg(h_{1} +h_{2}) = K.E + mgh_{2}

K.E = mgh_{1}

K.E = (61.0)(9.8)(10)\\K.E = 5978 J

Lets find out the magnitude of the force that the water is exerting on the diver.

W =ΔK.E

F.h_{2} = 5978\\

F = \frac{5978}{3}

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7 0
3 years ago
vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses
dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}

(This is correct because the horizontal motion has acceleration zero). Then:

v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}

Then, plugging in the given values, we obtain:

k=\frac{(61.2kg)(15.0m/s)^2}{(2.76m)^2}\\\\k=1808N/m

Finally, the effective spring constant of the firing mechanism is 1808N/m.

3 0
3 years ago
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