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MA_775_DIABLO [31]
2 years ago
12

A 53.0 kg sled is sliding on snow with μk=0.110. how much friction force does it feel?

Physics
2 answers:
S_A_V [24]2 years ago
7 0

Answer:

Force, F = 57.134 N  

Explanation:

It is given that,

Mass of the sled, m = 53 kg

Coefficient of kinetic friction, \mu_k=0.11

We need to find the force of friction felt by sled. The force of friction is given by :

F=\mu mg

F=0.11\times 53\ kg\times 9.8\ m/s^2

F = 57.134 N

So, the force of friction felt by the sled is 57.134 N. Hence, this is the required solution.

Anon25 [30]2 years ago
6 0

Answer:

57N

Explanation:

F_F = \mu F_N = \mu mg = 0.11 \times 53 kg \times 9,8 \frac{m}{s^{2} }

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Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

\frac{1}{2}mv^2 = mgh

where

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g=9.8 m/s^2 is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s

In the acceleration phase, we know that the initial velocity is

u=0

And the force exerted on the floor is 2.3 times the gravitational force, so

F=2.3 mg

This means the net force on you is

F_{net} = F-mg=2.3mg-mg=1.3 mg

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So the acceleration of the man is

a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g

Now we can use the  following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

v^2-u^2=2as

where s is the quantity we want to find. Solving for s,

s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m

2)

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

\Delta h = -0.77 m

This means that the lowest point reached by the center of mass above the floor during the crouch is

h'=h+\Delta h = 1.0 - 0.77 = 0.23 m

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

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3 years ago
List of energy changes when hydroelectricity city is made​
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Answer:

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8 0
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A hot iron horseshoe (mass = 0.35 kg ), just forged, is dropped into 1.40 L of water in a 0.45 kg iron pot initially at 22.0°C.
olga_2 [115]

Answer:

420 °C

Explanation:

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m_{w} = mass of the water = 1.40 L = 1.40 kg

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c_{w} = specific heat of water = 4186 J kg⁻¹ °C⁻¹

T_{hi} = initial temperature of the horseshoe = ?

T_{wi} = initial temperature of the water = 22 °C

T_{pi} = initial temperature of the iron pot = 22 °C

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(0.35)(450)(T_{hi} - 32) = (1.40)(4186)(32 - 22) + (0.45)(450)(32 - 22)

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Lunna [17]

If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.

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2 years ago
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