All categories

2 years ago

A vehicle traveling at a steady speed takes 4 hours to travel 244 km. calculate the average speed of this vehicle.

Physics

1 answer:

worty [1.4K]2 years ago

3 0

The average speed is 61 mph.

You might be interested in

A car with a mass of 1500 kg is pulled by a rope that is horizontal to the ground. The tension in the rope is 2000 N and a frict

Tanzania [10]

Answer:

Explanation:

Assuming the ground is level as well.

F = ma

a = F/m

a = (2000 - 350) / 1500

a = 1.1 m/s²

7 0

2 years ago

Compare the current in the 8-ohm resistors to the current in the 4-ohm resistors.

Gemiola [76]

Answer:

a) i₈ = 0.5 i₄, b) i₁₀ = 0.3 i₃, i₁₀ = 0.8 i₈

Explanation:

For this exercise we use ohm's law

V = i R

i = V / R

we assume that the applied voltage is the same in all cases

let's find the current for each resistance

R = 4 Ω

i₄ = V / 4

R = 8 Ω

i₈ = V / 8

we look for the relationship between these two currents

i₈ /i₄ = 4/8 = ½

i₈ = 0.5 i₄

R = 3 Ω

i₃ = V3

R = 10 Ω

i₁₀ = V / 10

we look for relationships

i₁₀ / 1₃ = 3/10

i₁₀ = 0.3 i₃

i₁₀ / 1₈ = 8/10

i₁₀ = 0.8 i₈

7 0

2 years ago

10. Someone takes 11 minutes to walk up a hill 120m high. His weight is 550N.

belka [17]

Answer:

okay with you if you want to

8 0

2 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago

The internal energy of material is determined by

Mnenie [13.5K]

The combined amount of kinetic and potential energy of its molecules

6 0

3 years ago