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uysha [10]
3 years ago
12

An object is measured 6.0cm in length, 3.0cm in width and 5.0cm in height. What is it’s volume?

Physics
1 answer:
sergey [27]3 years ago
5 0

Answer:

90

Explanation:

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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
1 year ago
The tires of a car make 75 revolutions as the car reduces its speed uniformly from 95 km/hr to 55 km.hr. the tires have a diamet
sveta [45]
<span>95 km/h = 26.39 m/s (95000m/3600 secs) 55 km/h = 15.28 m/s (55000m/3600 secs) 75 revolutions = 75 x 2pi = 471.23 radians radius = 0.80/2 = 0.40m v/r = omega (rad/s) 26.39/0.40 = 65.97 rad/s 15.28/0.40 = 38.20 rad/s s/((vi + vf)/2) = t 471.23 /((65.97 + 38.20)/2) = 9.04 secs (vf - vi)/t = a (38.20 - 65.97)/9.04 = -3.0719 The angular acceleration of the tires = -3.0719 rad/s^2 Time is required for it to stop (0 - 38.20)/ -3.0719 = 12.43 secs How far does it go? 65.97 - 38.20 = 27.77 M</span>
7 0
3 years ago
PLEASE HELP Scientists have observed that the average sunspots have increased in the last 50 years. Which of these statements be
Serga [27]

Answer:

option c is correct, earth temp increase

6 0
3 years ago
Which of these are part of our solar system? select all that apply
Len [333]
A)planets
b)the sun
c)moons
e)comets
f)asteroids
6 0
3 years ago
Read 2 more answers
A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to os
Bingel [31]

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}

\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\

<u>k = 3.5 N/m</u>

4 0
3 years ago
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