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Nana76 [90]
3 years ago
12

6. Show that the weight of an object on the moon is 1/6 its weight on earth.​

Physics
1 answer:
mojhsa [17]3 years ago
5 0

Taking ratio of W & w. ≈ 6 . w = 1/6 W. Therefore , Weight of an object on the moon is 1/6 of its weight on the earth.

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Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. If their horizontal velocities at t
givi [52]
For t1:

t1 = square root of 2h1 / g = square root of 2 * 0.5 / 9.8 = 0.319 sec

For t2:

t2 = sqaure root of 2h2 / g = square root of 2 * 1.0 / 9.8 = 0.451 sec

Wherein:
t = time(s) for the vertical movement
h= height
g = gravity (using the standard 9.8 m/sec measurement)

d1 = 1*0.319 = 0.319 m
d2 = 0.5 * 0.451 = 0.225 m

Where:

d = hor. distance

ratio = d1:d2
= 0.319 : 0.225
=3.19 : 2.25

The answer is 3.19 : 2.25
8 0
3 years ago
Read 2 more answers
What is the equivalent resistance of the circuit?
Kaylis [27]

Answer:

80 Ω.

Explanation:

In this circuit the resistances are in series.The equivalent resistance of a series circuit is equal to the sum of the resistances. Req= 60 + 20 = 80 Ω.

8 0
3 years ago
An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Dete
dezoksy [38]

Answer:43.34 m

Explanation:

Given

acceleration(a)=2 m/s^2

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

v=0+2\times 6=12 m/s

After this object will start moving under gravity

height reached in first 6 s

s=ut+\frac{at^2}{2}

s=0+\frac{2\times 6^2}{2}

s=36 m

After fuel run out distance traveled in upward direction is

v^2-u^2=2as_0

here v=0

u=12 m/s

a=9.8 m/s^2

0-12^2=2(-9.8)(s)

s_0=\frac{144}{2\times 9.8}=7.34 m

s+s_0=36+7.34=43.34 m

7 0
3 years ago
Knowing the causes of respiratory system diseases, what are
Kamila [148]
Cover your cough
Wash your hands
Keep a clean space
Avoid smoking
Stay active

Hope this helps :)
8 0
4 years ago
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s 1045 rad/s ). If a particular disk is
Scilla [17]

Answer:

The magnitude of the average angular acceleration of the disk is 4139.74\ rad/s^2.

Explanation:

Given that,

Angular velocity, \omega_i=968.7\ rad/s

The disk comes to rest, \omega_i=0

Time, t = 0.234 s

We need to find the magnitude of the average angular acceleration of the disk. It is given by change in angular velocity per unit time. So,

\alpha =\dfrac{\omega f}{t}\\\\\alpha =\dfrac{968.7\ rad/s}{0.234\ s}\\\\\alpha =4139.74\ rad/s^2

So, the magnitude of the average angular acceleration of the disk is 4139.74\ rad/s^2.

5 0
3 years ago
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