How long will it take for a car at rest to accelerate at 7m/s^2 to a speed of 45 m/s

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you.Your reaction time

shtirl [24]

Consider the motion of the car before brakes are applied:

v₀ = maximum initial velocity of the car before the brakes are applied

t = reaction time = 0.50 s

x₀ = distance traveled by the car before brakes are applied

since car moves at constant speed before brakes are applied

Using the equation

x₀ = v₀ t

x₀ = v₀ (0.50)

Consider the motion after brakes are applied :

v₀ = initial velocity of the car before the brakes are applied

a = acceleration = - 10 m/s²

v = final velocity of the car after it comes to stop = 0 m/s

x = stopping distance = initial distance - distance traveled before applying the brakes = 38 - x₀ = 38 - v₀ (0.50)

Using the equation

v² = v²₀ + 2 a x

inserting the values

0² = v²₀ + 2 (- 10) (38 - v₀ (0.50))

v²₀ = 20 (38 - v₀ (0.50))

v₀ = 23 m/s

3 0

3 years ago

The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and 589.6 nm. What are the freque

stiv31 [10]

Answer:

Explanation:

Given

wavelength of emissions are

$\lambda _1=589 nm$

$\lambda _2=589.6 nm$

Energy is given by

$E=\frac{hc}{\lambda }$

where h=Planck's constant

x=velocity of Light

$\lambda$ =wavelength of emission

$E_1=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589\times 10^{-9}}$

$E_1=3.374\times 10^{-19} J$

$E_1 in kJ/mol$

$E_1=203.2 kJ/mol$

frequency corresponding to this emission

$\nu =\frac{c}{\lambda }$

$\nu _1=\frac{3\times 10^8}{589\times 10^{-9}}$

$\nu _1=5.09\times 10^{14} Hz$

Energy corresponding to $\lambda _2$

$E_2=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589.6\times 10^{-9}}$

$E_2=3.371\times 10^{-19} J$

$E_2=203.02 kJ/mol$

frequency corresponding to this emission

$\nu =\frac{c}{\lambda }$

$\nu _1=\frac{3\times 10^8}{589.6\times 10^{-9}}$

$\nu _1=5.088\times 10^{14} Hz$

6 0

3 years ago

Two cars leave Calgary at the same time, travelling in opposite directions. Their average speeds differ by 5 km/h. After 2 h, th

dimulka [17.4K]

Let the first car's average speed be x

The second car's speed would be x+5

$2(x+(x+5))=210$

$2x+5=105$

$2x=100$

$x=50$

So the speed of the slower car is 50mph, and the speed of the faster car is 55mph

8 0

3 years ago

Read 2 more answers

According to Coulomb’s Law, the force between two charged objects is related to _____.

Natasha_Volkova [10]

Answer:

A.) the inverse of the square of the distance separating them

Explanation:

Coulombs law states that "the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them."

Mathematically, F = kq1q2/r²

Where q1 and q2 are the charges

r is the distance between the charges.

According to the law, the force between two charged objects is related to the inverse of the square of the distance separating them.

5 0

3 years ago

What conclusions can you draw about the connection between the Sun’s altitude and the latitude of the observer on seasonal chang

Cerrena [4.2K]

Answer:

Explanation:

Altitude of the Sun and the latitude position on the earth play an important role in the season change on the earth.

When the altitude of the sun is high then the average temperature of the earth is higher because the luminous intensity of the sun rays is higher due to the focusing of high energy sun rays over a small area.

But when the sun is at higher altitudes we receive less denser rays of the sun and hence we have less heat on the earth on an average.

But despite of the altitude some places on the earth have distinct temperature than the other place at the same time of the year. This is due to their latitudinal location. The places near the equator are warmer most of the times throughout the year because they receive the most direct rays while the poles receive slanting rays and hence are colder even in summer when the earth is at lower altitudes.

6 0

3 years ago