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Y_Kistochka [10]

3 years ago

13

A car travels 300km in six hour. What is the average speed

1 answer:

KiRa [710]3 years ago

5 0

Answer:

<h2>50 km/hr</h2>

Explanation:

The average speed of the car can be found by using the formula

$s = \frac{d}{t} \\$

d is the distance

t is the time taken

From the question we have

$s = \frac{300}{6} = 50 \\$

We have the final answer as

<h3>50 km/hr</h3>

Hope this helps you

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When two waves meet they combine according to the ____________________ principle?

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3 years ago

Linear Velocity What is the linear velocity in cm>min for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10

daser333 [38]

3,89,988 cm/min is the linear velocity

Given,

Diameter of CD = 12 cm

So, Radius of CD = 6 cm

CD is spinning at 10350 rev/min

Firstly , convert rev/min into rad/min

1 rev = 2π radians

10350 rev/min = 10350 × 2π

= 64998 rad/min

Formula used,

$v=rw$ where,

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$w$ is the angular velocity

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= 3,89,988 cm/min

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Learn more about Angular speed here brainly.com/question/540174

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2 years ago

What term best describes the geologic event taking place in the above illustration?

grigory [225]

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3 years ago

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How are cactus adapted to survive in deserts?

andreev551 [17]

1. They have evolved their leaves into spikes for minimum water loss through transpiration.

2. They have a waxy layer for minimum water loss.

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2 years ago

A stone dropped in a pond sends out a circular ripple whose radius increases at a constant rate of 4 ft/sec. After 12 seconds, h

Natali5045456 [20]

Answer:

After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

Explanation:

Area of a circle = πr²

where;

r is the circle radius

Differentiate the area with respect to time.

$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$

dr/dt = 4 ft/sec

after 12 seconds, the radius becomes = $\frac{dr}{dt} X 12 = 4 \frac{ft}{sec} X 12 sec = 48 ft$

To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt

$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$

$\frac{dA}{dt} = 2\pi (48)(4)$

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Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

7 0

3 years ago