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kherson [118]
2 years ago
7

Snow has been lying on a mountainside. Suddenly, it starts to move down the mountain. Which types of friction are observed in th

is avalanche? Where does each type occur?
Physics
1 answer:
Alexus [3.1K]2 years ago
8 0
<h2>Snow starts to move down the mountain </h2>

The energy which is due to position is potential energy. So when the snow is lying on the mountain. It possess potential energy but when suddenly, it starts to move down the mountain, the potential energy is converted into the kinetic energy. Yet some force is exerting on the snow to stop the smooth flow of snow through mountains.

This example of frictional force may be due to presence of rough surface or stones. Generally, there are four types of friction as static, rolling, sliding and fluid friction. Though in this case when snow is lying it possess static friction, when flows then it possesses sliding and fluid friction both.

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A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve
Troyanec [42]

The velocity of tennis racket after collision is 14.96m/s

<u>Explanation:</u>

Given-

Mass, m = 0.311kg

u1 = 30.3m/s

m2 = 0.057kg

u2 = 19.2m/s

Since m2 is moving in opposite direction, u2 = -19.2m/s

Velocity of m1 after collision  = ?

Let the velocity of m1 after collision be v

After collision the momentum is conserved.

Therefore,

m1u1 - m2u2 = m1v1 + m2v2

v1 = (\frac{m1-m2}{m1+m2})u1 + (\frac{2m2}{m1+m2})u2

v1 = (\frac{0.311-0.057}{0.311+0.057})30.3 + (\frac{2 X 0.057}{0.311 + 0.057}) X-19.2\\\\v1 = (\frac{0.254}{0.368} )30.3 + (\frac{0.114}{0.368}) X -19.2\\ \\v1 = 20.91 - 5.95\\\\v1 = 14.96

Therefore, the velocity of tennis racket after collision is 14.96m/s

7 0
2 years ago
A car is travelling to the right with a speed of 42\,\dfrac{\text m}{\text s}42 s m ​ 42, space, start fraction, m, divided by,
Effectus [21]

Answer:

d = 84 m

Explanation:

As we know that when an object moves with uniform acceleration or deceleration then we can use equation of kinematics to find the distance moved by the object

here we know that

initial speed v_i = 42 m/s

final speed v_f = 0

time taken by the car to stop

t = 4s

now the distance moved by the car before it stop is given as

d = \frac{v_f + v_i}{2} \times t

now we have

d = \frac{42 + 0}{2} \times 4

d = 84 m

7 0
2 years ago
Read 2 more answers
A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average
frez [133]
 (u) = 20 m/s 
(v) = 0 m/s 
<span> (t) = 4 s 
</span>
<span>0 = 20 + a(4) 

</span><span>4 x a = -20 
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>
8 0
2 years ago
Read 2 more answers
A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d
marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

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

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 35.4}

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0
2 years ago
A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

    T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

 T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0
2 years ago
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