All categories

3 years ago

When a car is traveling 120 km/h how many times more kinetic energy would it have than if it were

Physics

1 answer:

LekaFEV [45]3 years ago

4 0

Answer: 9

Explanation:

Kinetic energy is the energy possessed by a moving object. It is measured in joules, and depends on the mass (m) of the object and the speed (v) by which it moves i.e K.E = 1/2 x mass x velocity^2

So, car is traveling 120 km/h

KE = 1/2x m x (120 km/h)^2

KE = 0.5 x m x 14400

KE1 = 7200m

So, car is traveling 40 km/h

KE = 1/2x m x (40 km/h)^2

KE = 0.5 x m x 1600

KE2 = 800m

Now, divide the value of kinetic energy obtained during 120 km/h by that obtained during 40 km/h

i.e 7200m / 800m

= 9

Thus, the kinetic energy will be 9 times more at a speed of 120 km/h than at 40 km/h

You might be interested in

Why does a comet's tail point away from the Sun? 1Choice 1 The solar wind blows the tail away from the Sun. 2 It is being pulled

AnnyKZ [126]

Choice 1

The Sun's radiation and solar wind cause the dust and gas around the comet (coma) to stretch the coma. The solar wind electromagnetically blows the ions in the coma away.

4 0

3 years ago

How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

$T_{f}$ = (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

$T_{f}$ = 466.8 K

so change in temperature ΔT

ΔT = 466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x $_{i\\}$ ² - x $_{f\\}$ ² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q = 16 J

Therefore, the required heat energy is 16 J

5 0

3 years ago

two electronics students are discussing electrical units. student A says that electrical power is measured in units coulombs, st

Savatey [412]

The statements of both students are incorrect.

-- Electrical power, just like mechanical power, is expressed in units of watts.
-- 'Coulomb' is the unit of electrical charge.
-- '400 k ohms' means 400,000 ohms of resistance.
-- 'Volt' is the unit of electromotive force (or potential difference).

There are no 'following statements'.

All in all, a very disappointing question.

8 0

3 years ago

A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotat

AleksandrR [38]

Answer:

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

Explanation:

As we know that there is no external torque on the system of two disc

then the angular momentum of the system will remains conserved

So we will have

$L_i = L_f$

now we have

$L_i = (\frac{1}{2}MR^2)\omega_o$

also we have

$L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now from above equation we have

$(\frac{1}{2}MR^2)\omega_o = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now we have

$\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}$

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

6 0

3 years ago

Small cubes that are 10 cm on a side and larger ones that are 12 cm on a side are submerged in water. Cubes A and B are made of

Digiron [165]

Answer

given,

small cube side = 10 cm

larger cube side = 12 cm

density of steel = 7 g/cm³

density of aluminium = 2.7 g/cm³

density of the water (ρ₁)= 1 g/cm³

Cube A and B made of steel

buoyant force of Cube A

B₁ = ρ₁ V g = 1 x 10 x 10 x 10 x g= 1000 g

for cube B

B₂ = ρ₁ V g = 1 x 12 x 12 x 12 x g= 1728 g

buoyant force of Cube C

B₃ = ρ₁ V g = 1 x 10 x 10 x 10 x g= 1000 g

for cube D

B₄ = ρ₁ V g = 1 x 12 x 12 x 12 x g= 1728 g

buoyant force acting on the cube depends on the density of the fluid

hence,

B₂ = B₄ > B₁ = B₃

8 0

4 years ago