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3 years ago

When is the car exceeding a 10 km/hr speed limit?

1 answer:

6 0

If the car travels a distance greater than 10 km in one hour or less than one hour, it has exceeded the speed limit.

The given parameter:

speed limit = 10 km/hour

To find:

when the car will exceed the speed limit

Speed is defined as the total distance traveled divided by the total time taken for the journey.

$speed = \frac{distance}{time}$

If the car travels any distance greater than 10 km in 1 hour or less than 1 hour, it has exceeded the speed limit of 10 km/hr.

For example, let's say the distance traveled by the car in one hour is 11 km. The speed of the car is 11 km/hr which is greater than the speed limit.

Thus, we can conclude that if the car travels a distance greater than 10 km in one hour or less than one hour, it has exceeded the speed limit.

Learn more here:brainly.com/question/17337430

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Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3

VARVARA [1.3K]

Answer:

Part a)

$f = 4.76 \times 10^{14} Hz$

Part b)

$d = 3.48 \times 10^{-4} m$

Part c)

$\theta = 0.311 degree$

Explanation:

Part a)

As we know that the speed of light is given as

$c = 3 \times 10^8 m/s$

$\lambda = 630 nm$

now the frequency of the light is given as

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

so we have

$f = \frac{3 \times 10^8}{630 \times 10^{-9}}$

$f = 4.76 \times 10^{14} Hz$

Part b)

Position of Nth maximum intensity on the screen is given as

$y_n = \frac{n\lambda L}{d}$

so here we know for 3rd order maximum intensity

$y_3 = 0.76 cm$

n = 3

L = 1.4 m

$0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}$

$d = 3.48 \times 10^{-4} m$

Part c)

angle of third order maximum is given as

$d sin\theta = 3 \lambda$

$3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})$

$\theta = 0.311 degree$

8 0

3 years ago

A car drives with a constant speed of 22 miles per hour. How far can it travel in 1 hour?

horsena [70]

Answer:
22 miles

Explanation:

8 0

3 years ago

Which option below correctly compares the average annual dose of background radiation to the dose liked to an increased cancer r

julia-pushkina [17]

The answer for this question would be choice "<span>B. The average annual dose of background radiation is 250 times smaller than the dose linked to increased cancer risk."

You only have to compare 4.0 x 10^-4 and 1.0 x 10^-1. And if you can observe carefully, when you try to multiply the average annual dose of background radiation by 250, you would get 0.1 which is equivalent to the amount of annual dose linked to increased cancer risk. Therefore, the answer is B.</span>

6 0

3 years ago

A 5.00 kilogram block slides along a horizontal,frictionless surface at 10.0 meters per second. for 4.00 seconds. The magnitude

finlep [7]

In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
= (5 * 10) Kg m/s
= 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".

5 0

3 years ago

A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0s, what is the magnitude

kondaur [170]

Answer:

81.8 m/s

Explanation:

The initial velocity of the plane is:

$v_0=115 m/s$ (toward east)