According to the law of reflection-

The rate of change of velocity is called:.

Pavel [41]

Answer: Acceleration

Detailed Explanation:

Acceleration is defined as the rate of change of velocity.

6 0

2 years ago

What is the mass of a truck if it is accelerating at a rate of 5 m/s2 and hits a parked car with a

sineoko [7]

Answer:

<h3>The answer is 2800 kg</h3>

Explanation:

The mass of the truck can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{14000}{5} \\$

We have the final answer as

Hope this helps you

5 0

3 years ago

The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?

BigorU [14]

1. Frequency: $7.06\cdot 10^{14} Hz$

The frequency of a light wave is given by:

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

where

$c=3\cdot 10^{-8} m/s$ is the speed of light

$\lambda$ is the wavelength of the wave

In this problem, we have light with wavelength

$\lambda=425 nm=425\cdot 10^{-9} m$

Substituting into the equation, we find the frequency:

$f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz$

2. Period: $1.42 \cdot 10^{-15}s$

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

The frequency of this light wave is $7.06\cdot 10^{14} Hz$ (found in the previous exercise), so the period is:

$T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s$

4 0

3 years ago

Read 2 more answers

How high does a rocket have to go above Earth's surface before its weight is half of what it is on Earth?

Contact [7]

Answer:

$h=1.6\times 10^6\ m$

Explanation:

As we know that the acceleration due to gravity decreases with height.

At certain height it will get to the half of its value on the surface of the earth.

As we know that the weight on the surface of the earth is given as:

$w=m.g$

where:

m = mass of the object

g = acceleration due to gravity of the substance

Since mass of the substance is constant so the variation is weight is possible only due to change in the acceleration due to gravity.

<u>We know that the variation of the acceleration due gravity with height is given as:</u>

$g_{_h}=g\times (1-\frac{2h}{R} )$

where:

$g_{_h}=$ value to acceleration due to gravity at height h

g = acceleration due to gravity at the earth's surface

h = height of the object

R = radius of the earth = $6400\ km$

according to question the weight becomes half, so,:

$4.9=9.8\times (1-\frac{2h}{6400\times10^3} )$

$h=1.6\times 10^6\ m$ is the height a rocket has to go above Earth's surface before its weight is half of what it is on Earth.

4 0

3 years ago

A race card drives one lap around a race track that is 500 meters in length. How is this displacement different from the distanc

gulaghasi [49]

Answer:

Distance is 500 m, displacement is 0

Explanation:

Distance and displacement are defined in two different ways:

- Distance is the total length of the path covered by an object in motion - so it depends on the path taken. In this problem, the distance travelled by the car corresponds to the length of one lap, which is the length of the track, so 500 m

- Displacement is the distance in a straight line between the final point and the initial point of the motion. This means that displacement does not depend on the path taken, but only on the starting and ending point of the motion. In this problem, the car completes one lap, so the final position of the car is equal to its starting position - therefore the displacement is zero, since the distance between these two points is zero.

5 0

3 years ago