How can I get the temperature? Sound Speed = 331 m/s + (0,6xTemperature)

car was moving in a straight road of length 320 km it covered 240 km with an average velocity 75 km/hr then it ran out of fuel a

Stella [2.4K]

The average velocity of the car for the whole journey is 69.57 km/h.

The given parameters:

Length of the road, L = 320 km
Distance covered = 240 km at 75 km/h
time spent refueling, t₂ = 0.6 hr
Final velocity, = 100 km/hr

The time spent by the before refueling is calculated as follows;

$t = \frac{d}{v} \\\\t_1 = \frac{240}{75} \\\\t_1 = 3.2 \ hours$

The time spent by the car for the remaining journey;

$t_3 = \frac{320 - 240}{100} \\\\t_3 = 0.8 \ hr$

The total time of the journey is calculated as follows;

$t = t_1 + t_2 + t_3\\\\t = 3.2 \ hr \ + \ 0.6 \ hr \ + \ 0.8 \ hr\\\\t = 4.6 \ hours$

The average velocity of the car for the whole journey is calculated as follows;

$v = \frac{total \ distance }{total \ time} \\\\v = \frac{320}{4.6} \\\\v = 69.57 \ km/h$

Learn more about average velocity here: brainly.com/question/6504879

6 0

3 years ago

People cannot see certain types of light waves because:

enot [183]

Answer:

Light comes in different colors like radio, ultra violet, gamma-ray, etc, and they are invisible to the bare eye

Explanation:

5 0

2 years ago

¿Cuál es el parámetro que indica la cantidad de energía liberada en un movimiento sísmico?

artcher [175]

Answer:

Explanation:

en un movimiento sísmico?

3 0

2 years ago

Since the Wagon is being pulled down hill is it increasing (C)

Ket [755]

Since the wagon is being pulled down hill with a constant velocity, all the forces of the wagon would be (C) increasing.
You are correct! **

4 0

3 years ago

Situation: two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a

Ahat [919]

Answer:

$r_{min}=\frac{kq^2}{5m_ev^2}$

Explanation:

The total kinetic energy of both electrons will be electrostatic potential energy, when the electrons reach the minima distance due to electrostatic repulsion. Then, you have:

$E_{k}=U_E\\\\\frac{1}{2}m_ev_1^2+\frac{1}{2}m_ev_2^2=k\frac{q^2}{r_{min}}$

me: mass of the electron

q: charge of the electron

k: Coulomb's constant

you take into account that v2=3v1=3v and do rmin the subject of the formula:

$\frac{1}{2}m_e[v^2+9v^2]=5m_ev^2=k\frac{q^2}{r_{min}}\\\\r_{min}=\frac{kq^2}{5m_ev^2}$

4 0

4 years ago