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artcher [175]
3 years ago
14

Which sections of the heating curve illustrate this process?

Physics
1 answer:
kirill115 [55]3 years ago
5 0

Answer:

B followed by D

Explanation:

The heat energy absorbed at B goes into potential  energy that breaks the inter-molecular bonds and thus the constant temperature. Once the molecules have gained enough energy they escape the closely bonded  structure and thus are free to move in random directions due to high kinetic energy. At this point (part D)  an increase in heat energy leads to an increase in the kinetic energy leading to an increase in the temperature.

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Its going to be about -47.78°C

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What is the speed of a cheetah if it takes 20 sesonds to run 300 m
natta225 [31]

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A car stopped at a red light, not moving?
timofeeve [1]

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A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in t
HACTEHA [7]

Answer:

(a) 2.85 m

(b) 16.5 m

(c) 21.7 m

(d) 22.7 m

Explanation:

Given:

v₀ₓ = 19 cos 71° m/s

v₀ᵧ = 19 sin 71° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

(a) Find Δy when t = 3.5 s.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²

Δy = 2.85 m

(b) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2 aᵧ Δy

(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy

Δy = 16.5 m

(c) Find Δx when t = 3.5 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²

Δx = 21.7 m

(d) Find Δx when Δy = 0 m.

First, find t when Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧ t²

(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²

0 = t (18.0 − 4.9 t)

t = 3.67

Next, find Δx when t = 3.67 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²

Δx = 22.7 m

7 0
3 years ago
Two identical 3.0-kg cubes are placed on a horizontal surface in
DedPeter [7]

The magnitude of the force exerted by the left cube on the right cube is 17.64N.

<h3>What is frictional force?</h3>

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction.

Two identical 3.0-kg cubes are placed on a horizontal surface in contact with one another. The cubes are lined up from left to right and a force F₁ is applied to the left side of the left cube causing both cubes to move at a constant speed v. The coefficient of kinetic friction between the cubes and the surface is 0.3.

From the equilibrium of forces in vertical direction

Normal force N= 2m x g

friction force f = μN =μ(2m)g

From the equilibrium of forces in horizontal direction

F₁ =ma =0

using Newton's third law of motion, we get

F₁  - f =0

F₁  =f = μ(2m)g

Put the values, we get

F₁  = 17.64N

Thus, magnitude of the force exerted by the left cube on the right cube is 17.64N.

Learn more about friction force.

brainly.com/question/1714663

#SPJ1

5 0
1 year ago
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