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4 years ago

6

In scientific notation, (6.2x10^4)x(3.3x10^2) equals_____? answer should be written 1.0x10^2 exmple

1 answer:

atroni [7]4 years ago

8 0

(6.2 × 10⁴) × (3.3 × 10²)
(6.2 × 3.3) × (10⁴ × 10²)
20.46 × 10⁽⁴⁺²⁾
20.46 × 10⁶ ⇒ The value 20.46 is not in standard form, so we need to rewrite this in standard form
2.046 × 10 × 10⁶
2.046 × 10⁽⁶⁺¹⁾
2.046 × 10⁷

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An airplane accelerates from a velocity of 22 m/s to 40 m/s with an acceleration of 2 m/s2. How long does it

andrey2020 [161]

The time it takes the plane to change its velocity is 9s.

<h3>What is time?</h3>

Time can be defined the measured or measurable period during which an action, process, or condition exists or continues.

To calculate the time it takes the airplane to change its velocity, we use the formula below.

Formula:

t = (v-u)/a.......... Equation 1

Where:

a = Acceleration
v = Final velocity
u = Initial velocity
t = time

From the question,

v = 40 m/s
u = 22 m/s
a = 2 m/s²

Substitute these values into equation 1

t = (40-22)/2
t = 18/2
t = 9s

Hence, the time it takes the plane to change its velocity is 9s.

Learn more about time here: brainly.com/question/2854969

4 0

2 years ago

A dog is 60m away while moving at constant velocity of 10m/s towards you. Where is the dog after 4 seconds?

evablogger [386]

20m away

the dog was 60m away from. you subtract 40m since it is 10m/s x 4 seconds

4 0

3 years ago

Read 2 more answers

A person carries a mass of 10 kg and walks along the +x-axis for a distance of 100m with a constant velocity of 2 m/s. What is t

gizmo_the_mogwai [7]

Since the direction of the force and the direction of the path is perpendicular, the person is not doing any physical work.

3 0

3 years ago

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A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c

stealth61 [152]

Answer :

(a) The acceleration of the car is, $-2m/s^2$

(b) The distance covered by the car is, 150 m

Explanation :

By the 1st equation of motion,

$v=u+at$ ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

$5m/s=25m/s+a\times (10s)$

$a=-2m/s^2$

The acceleration of the car is, $-2m/s^2$

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = $-2m/s^2$

Now put all the given values in the above equation 2, we get:

$s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2$

By solving the term, we get:

$s=150m$

The distance covered by the car is, 150 m

8 0

3 years ago

Jason walks 20 m East, turns around and 20 m West, Finally, he walks 10 rn North. This takes 20 s. what is Jason's velocity

serious [3.7K]

Answer:

0.5 m/s north

Explanation:

Take east to be +x, west to be -x, north to be +y, and south to be -y.

His displacement in the x direction is:

x = 20 m − 20 m = 0 m

His displacement in the y direction is:

y = 10 m

His total displacement is therefore 10 m north.

His velocity is equal to displacement divided by time.

v = 10 m north / 20 s

v = 0.5 m/s north

3 0

3 years ago