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sammy [17]
3 years ago
11

Where is the centre of mass of a uniform cube located?​

Physics
1 answer:
Stella [2.4K]3 years ago
8 0

Answer:

NICE QUESTION

Explanation:

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How do we know the sun is supported by nuclear fusion?
Leya [2.2K]
 we know that the sun is supported by nuclear fusion because the sun is a main-sequence star, meaning it has to use nuclear fusion to keep itself going, no nuclear fusion, no sun.

i hope this helps!
3 0
4 years ago
The laboratory exercise for this chapter addresses kinematic motion. You have experienced these motions in everyday life. Instea
ki77a [65]

Answer:

A) vectors: veloicty, force

scalar:  speed, work

B)  t = 1.75 s,  C)   v = - 17 2 m / s

Explanation:

We answer each part separately

A) A vector magnitude has magnitude and direction instead a scalar magnitude has only magnitude

vector quantities: the speed of a car number is the magnitude and direction is where it goes

Force, the number is the magnitude and above that applies gives direction

Scalar magnitude: how quickly the number of the speedometer of the car

Temperature, work

B) I = 15 m height to the soil and get to calculate time = 0

        y = y₀ + v₀ t - ½ g t²

as the ball is loose its initial velocity is zero

       0 = 0 +0  - ½ g t²

       t = \sqrt{2y_o/g}

       t = \sqrt{2 \ 15/ 9.8}

       t = 1.75 s

C) the velocity to the reach the floor

      v = vo - g t

      v = 0 - g t

      v = - 9.8 1.74

       v = - 17 2 m / s

The negative signt iindicates that the speed goes down

5 0
3 years ago
After touchdown a fighter jet passes a marker on the runway at an instantaneous speed of 100 m/s and constant negative accelerat
daser333 [38]

Given that,

Deacelectration = -15 m/s²

Negative sign shows the declaration

Slowly speed = 25 m/s

Acceleration = 35 m/s²

Speed = 170 m/s

We need to calculate the time

Using equation of motion

v=u+at

Where, v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the formula

25=0+15\times t

t=\dfrac{25}{15}

t=1.66\ sec

We need to calculate the distance from the original runway

Using equation of motion

s=ut+\dfrac{1}{2}at^2

s=0+\dfrac{1}{2}\times35\times(1.66)^2

s=48.2\ m

Hence, The distance from the original runway is 48.2 m.

3 0
4 years ago
A scooter is moving with a speed of 18 m/s. Its mass is 20 kg. What is the
kykrilka [37]
<h2><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>:-</h2>

A scooter is moving with a speed of 18 m/s. Its mass is 20 kg. What is the magnitude of its momentum?

A. 360 kg.m/s

B. 2 kg.m/s

C. 3600 kg.m/s

D. 1.1 kg.m/s

<h2><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>:-</h2>

<h3>Given:-</h3>

Speed (v) of a scooter = 18 m/s

Mass (m) = 20 kg

<h3>To Find:-</h3>

Magnitude of its momentum (p) = ?

<h3>Answer:-</h3>

Since, we know that,

p = mv

So,

p = 20 kg × 18 m/s

p = 360 kg.m/s

<h3>\therefore The correct option is (A) 360 kg.m/s </h3>

<h3>The magnitude of its momentum is <u>3</u><u>6</u><u>0</u><u> </u><u>k</u><u>g</u><u>.</u><u>m</u><u>/</u><u>s</u>. [Answer]</h3>
4 0
3 years ago
A small mass m on a string is rotating without friction in a circle. The string is shortened by pulling it through the axis of r
Komok [63]
I’m not sure, is there any other words you can explain it in ..?
4 0
3 years ago
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