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3 years ago

The following diagram represents a cart with an initial velocity of 1.0 m/s sliding along a frictionless track from point A:

Physics

2 answers:

FrozenT [24]3 years ago

6 0

the answer is point c

vichka [17]3 years ago

5 0

Answer:

The kinetic energy will be maximum at point C.

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What mechanical layer lies below the lithosphere

joja [24]

The asthenosphere is under the lithosphere.

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3 years ago

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crowbar of 5 metre is used to lift an object of 800 metre if the effort arm is 200cm calculate the force applied

Klio2033 [76]

Answer:

F = 5226.6 N

Explanation:

To solve a lever, the rotational equilibrium relation must be used.

We place the reference system on the fulcrum (pivot point) and assume that the positive direction is counterclockwise

F d₁ = W d₂

where F is the applied force, W is the weight to be lifted, d₁ and d₂ are the distances from the fulcrum.

In this case the length of the lever is L = 5m, t the distance desired by the fulcrum from the weight to be lifted is

d₂ = 200 cm = 2 m

therefore the distance to the applied force is

d₁ = L -d₂

d₁ = 5 -2

d₁= 3m

we clear from the equation

F = W d₂ / d₁

W = m g

F = m g d₂ / d₁

we calculate

F = 800 9.8 2/3

F = 5226.6 N

4 0

3 years ago

An echo is not heard in a small room.why ? give reason

JulsSmile [24]

Answer:

They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the room should be 17.2 m.

hopw it helps

7 0

3 years ago

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A flatbed truck is carrying a 20.0 kg crate along a level road. the coefficient of static friction between the crate and the bed

Dahasolnce [82]

3.92 m/s^2 Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so 0.4 * 9.8 m/s^2 = 3.92 m/s^2 If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so 20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N Multiply by the coefficient of static friction, giving 196 N * 0.4 = 78.4 N So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass 78.4 N / 20.0 kg = 78.4 kg*m/s^2 / 20.0 kg = 3.92 m/s^2 And you get the same result.

6 0

3 years ago

Calculate the critical angle for light going from Glycerine to air.

Georgia [21]

The refractive index for glycerine is $n_g=1.473$ , while for air it is $n_a = 1.00$ .

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
$\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}$

6 0

3 years ago