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Margaret [11]
3 years ago
14

In a follow-up experiment, two identical gurneys are placed side-by-side on a ramp with their wheels locked to eliminate spinnin

g. Gurney 1 has a dummy placed on it to give it a total mass of 200 kg, while Gurney 2 is loaded with a dummy that makes it only 50 kg overall. If the ramp has a coefficient of friction of μs, which gurney is more likely to slide down the ramp?
Physics
1 answer:
tiny-mole [99]3 years ago
8 0

Answer:

Gurney 2 is more likely to slide.

Explanation:

We know static friction force is equal to the mass of the object times coefficient of friction.

Static friction force is the frictional force applied to the object which prevents the body to slip or slide when the body is at rest.

Now in the question, the gurney 1 which is placed in the ramp has a total mass of 200 kg and the coefficient of friction between the ramp and the gurney is μ.

And the gurney 2 place on the same ramp has a total mass of 50 kg and coefficient of friction μ between the ramp and the gurney 2.

Both the gurneys are at rest initially, means they are acted upon by static frictional force.

Mathematically, Static friction force = mass x coefficient of friction

So static friction depends on mass and since it is clear that the gurney 1 has a greater mass of 200 kg than the gurney 2 which has a total of 50 kg.

Therefore gurney 1 has greater static friction that is acted upon it which restricts it to slide or slip.

So, gurney 2 is more likely to slide down the ramp.

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Serggg [28]

A shadow forms on the side of an object that faces away from the sun. The length of shadows changes as Earth rotates. In the morning, the sun is low in the eastern sky and shadows are long. As time passes in the morning, the sun seems to move higher in the sky.

8 0
2 years ago
Desde una altura de 120 m se deja caer un cuerpo. Calcular a los 2,5 s i) la rapidez que lleva; ii) cuánto ha descendido; iii) c
stira [4]

Answer:

i) 24.5 m/s

ii) 30,656 m

iii) 89,344 m

Explanation:

Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer

i) Los parámetros dados son;

Altura inicial, s = 120 m

El tiempo en caída libre = 2.5 s

De la ecuación de caída libre, tenemos;

v = u + gt

Dónde:

u = Velocidad inicial = 0 m / s

g = Aceleración debida a la gravedad = 9.81 m / s²

t = Tiempo de caída libre = 2.5 s

Por lo tanto;

v = 0 + 9.8 × 2.5 = 24.5 m / s

ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación

s = u · t + 1/2 · g · t²

= 0 × 2.5 + 1/2 × 9.81 × 2.5² = 30.656 m

iii) La altura restante = 120 - 30.656 = 89.344 m.

6 0
3 years ago
A mercury thermometer is used to measure the temperature of boiling water.<br>Why?​
Rama09 [41]

Answer:

It has very high density, so a small bulb of a thermometer can contain much mercury. Mercury remains liquid state over a quite wide range of temperature because it freezes at 39°C and boils at 357°C.

Explanation:

6 0
3 years ago
(15 Points)
oksian1 [2.3K]

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

Read more about vertical weight here:

brainly.com/question/15244771

#SPJ1

5 0
2 years ago
Pls answer 50 points for an answer!!!!
lyudmila [28]

ummmn hi i dont know lol

5 0
2 years ago
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