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4 years ago

what is change in internal energy if 30j of heat are released from a system that does 50j of work on its surroundings

Physics

1 answer:

slamgirl [31]4 years ago

3 0

Answer:

-80 J

Explanation:

The first law of thermodynamics states that:

$\Delta U = Q - W$

where

$\Delta U$ is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system

In this problem, we have:

Q = -30 J is the heat released by the system (negative because the system releases it)

W = +50 J is the work done by the system on the surrounding (positive since it is done by the system)

Therefore, the change in internal energy is

$\Delta U = -30 J - (+50 J) = -80 J$

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At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m

eimsori [14]

Answer:

1) The time it takes the diver to move off the end of the diving board to the pool surface, $t_w$ , is approximately 1.392 seconds

2) The horizontal distance from the edge of the pool to where the diver enters the water, $d_w$ , is approximately 5.76 meters

Explanation:

1) The given parameters are;

The height of the diving board above the pool's surface, h = 9.5 m

The length by which the diving board over hangs the pool L = 2 m

The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s

Taking $t_w$ = The time it takes the diver to move off the end of the diving board to the pool surface

Therefore, we have from the equation of free fall;

h = 1/2 × g × $t_w$ ²

Where;

g = The acceleration due to gravity = 9.81 m/s²

Substituting the values, gives;

9.5 = 1/2 × 9.81 × $t_w$ ²

$t_w$ = √(9.5/(1/2 × 9.81)) ≈ 1.392 s

The time it takes the diver to move off the end of the diving board to the pool surface = $t_w$ ≈ 1.392 s

2) The horizontal distance, $d_w$ , in meters from the edge of the pool to where the diver enters the water is given as follows;

$d_w$ = L + v₀ × $t_w$ = 2 + 2.7× 1.392 ≈ 5.76 m

∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.

7 0

3 years ago

The diagram below shows the stars that are nearest to our solar system.

abruzzese [7]

Answer:

no u tried of the same dam

thing

Explanation:

7 0

3 years ago

Based on the evidence for the impact theory, what was the most probable order of events for the collision that led to the format

anastassius [24]

Answer:

Answered

Explanation:

According to the impact theory, a glancing collision between a Mars-sized object and Earth led to the formation of the Moon. The iron core of Earth had formed before this collision, leading to the similarity between the composition of the Moon and Earth's mantle. After the collision, any iron core of the Mars-sized object would have been left behind on Earth and eventually merged with Earth's core. The Moon then formed out of the debris thrown into space by the collision.

5 0

3 years ago

A car of mass 2400kg accelerates from 10m/s to 30m/s if the force is 12000N what is the distance

leonid [27]

Explanation:

Initial velocity (u) = 10 m/s
Final velocity (v) = 30 m/s
Mass (m) = 2400 kg
Force (F) = 12000 N

Let us find the time taken first.

→ F = ma

Acceleration (a) = (v – u)/t

→ 12000 = 2400 × (30 – 10)/t

→ 12000 ÷ 2400 = (20)/t

→ 5 = 20/t

→ 5t = 20

→ t = 20 ÷ 5

→ t = 4 seconds

Now, find the acceleration.

→ a = (v – u)/t

→ a = (30 – 10)/4

→ a = 20/4

→ a = 5 m/s²

Now, by using the third equation of motion,

→ v² – u² = 2as

s = distance

→ (30)² – (10)² = 2 × 5 × s

→ 900 – 100 = 10s

→ 800 = 10s

→ 800 ÷ 10 = s

→ 80 m = s

Therefore, distance travelled is 80 m.

4 0

3 years ago

A 1.2 kg ball drops vertically onto a floor from a height of 32 m, and rebounds with an initial speed of 10 m/s.

iren [92.7K]

Explanation:

Given that,

Mass of the ball, m = 1.2 kg

Initial speed of the ball, u = 10 m/s

Height of the floor from ground, h = 32 m

(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :

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

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 32}$

v = -25.04 m/s (negative as it rebounds)

The impulse acting on the ball is equal to the change in momentum. It can be calculated as :

$J=m(v-u)$

$J=1.2\times (-25.04-10)$

J = -42.048 kg-m/s

(b) Time of contact, t = 0.02 s

Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :

$J=\dfrac{F}{t}$

$F=J\times t$

$F=42.048\times 0.02$

F = 0.8409 N

Hence, this is the required solution.

5 0

3 years ago