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Brrunno [24]
3 years ago
12

How does newtons laws apply to the egg drop experiment

Physics
1 answer:
Anon25 [30]3 years ago
5 0
Because an object in rest stays in rest until an unequal force pushes it so gravity is pushing on the egg making it drop
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A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo
LuckyWell [14K]

Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

a) what is the work for one cycle

b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

W = (0.3 * 8.314 * 500) * ln(5000/1000)

W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But  ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

3 0
3 years ago
Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that ap
eduard

Answer:

The object can have zero velocity and, simultaneously, nonzero acceleration.

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

Explanation:

An object in simple harmonic motion has a total mechanical energy (sum of elastic potential energy and kinetic energy) that is constant:

E=U+K=1/2kx^2 + 1/2}mv^2

where,

k is equal to the spring constant

x is equal to the displacement

m is the mass

v is the speed

We can note that the force on the spring is given by Hook's law:

F=-kx

In Newton's law F = ma, this can be also be written as

ma=-kx

a=-k/mx

This implies that the acceleration is proportional to the displacement.

From the first equation, we can now states that:

When the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum

When the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum

In all the other intermediate situations, both velocity and acceleration are nonzero.

So the correct answers are

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

The object can have zero velocity and, simultaneously, nonzero acceleration.

4 0
3 years ago
Read 2 more answers
Which segment represent slowing down
lakkis [162]

Answer:

A downward sloping line

Explanation:

6 0
2 years ago
Read 2 more answers
PLS ANSWER ASAP
vivado [14]

Answer:

Cart A

Explanation:

Momentum can be computed by finding the product of mass and velocity. To solve this, you can use the formula below to find the greatest momentum:

p = mv

where:

p = momentum (kgm/s)       m = mass (kg)         v = velocity (m/s)

Because carts are moving along with the weights, we need to consider the whole system. This means that you need to add in the masses and the mass of the cart.

<u>Cart A:</u>

m = 200kg + 0 kg = 200 kg

v = 4.8 m/s

p = 200kg x 4.8 m/s = 960 kg-m/s

<u>Cart B:</u>

m = 200kg + 20 kg = 220 kg

v = 4.0 m/s

p = 220kg x 4.0 m/s = 880 kg-m/s

<u>Cart C:</u>

m = 200kg + 40 kg = 240 kg

v = 3.8 m/s

p = 240kg x 3.8 m/s = 912 kg-m/s

<u>Cart D:</u>

m = 200kg + 60 kg = 260 kg

v = 3.5 m/s

p = 260kg x 3.5 m/s = 910 kg-m/s

As you can see, Cart A has the greatest momentum.

5 0
2 years ago
Read 2 more answers
A race is held between a sports car and a motorcycle. The sports car can accelerate at 5.0 m/s^2 and the motorcycle can accelera
user100 [1]

Answer:

Vf₁ = 30 m/s

s₂ = 90 m

car wins the race.

Explanation:

To find the speed of car after 5 s, we use 1st equation of motion:

Vf₁ = Vi₁ + a₁t₁

where,

Vf₁ = Final Speed of Car = ?

Vi₁ = Initial Speed of Car = 0 m/s

a₁ = acceleration of car = 5 m/s²

t₁ = time = 5 s

Therefore,

Vf₁ = 0 m/s + (5 m/s²)(5 s)

<u>Vf₁ = 25 m/s</u>

<u></u>

To find the distance of motorcycle after 6 s, we use 2nd equation of motion:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₂ = distance covered by the motorcycle = ?

Vi₁ = Initial Speed of motorcycle = 0 m/s

a₂ = acceleration of motorcycle = 8 m/s²

t₂ = time = 6 s

Therefore,

s₂ = (0 m/s)(6 s) + (1/2)(5 m/s²)(6 s)²

<u>s₂ = 90 m</u>

<u></u>

We can use 2nd equation of motion to find time taken by each car and motorcycle to reach the finish point:

For Car:

s₁ = Vi₁ t₁ + (1/2)a₁t₁²

where,

s₁ = 200 m - 50 m = 150 m (since, car starts 50 m ahead)

Therefore.

150 m = (0 m/s)(t₁) + (1/2)(5 m/s²)t₁²

t₁ = √60 s²

t₁ = 7.74 s

For Motorcycle:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₁ = 200 m

Therefore.

200 m = (0 m/s)(t₂) + (1/2)(5 m/s²)t₂²

t₂ = √80 s²

t₂ = 8.94 s

Since, the car takes less time to reach finish line.

Therefore, car wins the race.

5 0
2 years ago
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