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Free_Kalibri [48]
2 years ago
6

If mass = 7.89 kg, weight on Earth =

Physics
1 answer:
marysya [2.9K]2 years ago
8 0

Answer:

77.322

Explanation:

you have to multiply the mass with gravity (which is 9.8).. which will give u that result

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A pallet of bricks is to be suspended by attaching a rope to it and connecting the other end to a couple of heavy crates on the
SashulF [63]

Answer:

<h3><u>C</u><u>.</u><u>2</u><u>5</u><u>0</u><u> </u><u>l</u><u>b</u><u> </u><u>i</u><u>s</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>.</u><u>.</u><u>.</u></h3>
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3 years ago
What are 3 wether instuerments i know anemometer and barometer but the third one and what there job!!
BaLLatris [955]

Well, those are good ones. Now how about a <u><em>thermometer</em></u> to <em>measure the temperature</em> ?

7 0
3 years ago
A cyclist travels at 15 m/s during a sprint finish. What is this speed in km/h
g100num [7]

Answer:

54 km/hr

Explanation:

m/s to km/hr => 18/5

15 m/s to km/hr => 15 x 18/5 =>3 x 18 => 54km/hr

8 0
3 years ago
Five hundred joules of heat are added to a closed system. The initial internal energy of the system is 87 J, and the final inter
Aleonysh [2.5K]

We can solve the problem by using the first law of thermodynamics:

\Delta U= Q-W

where

\Delta U is the variation of internal energy of the system

Q is the heat added to the system

W is the work done by the system

In this problem, the variation of internal energy of the system is

\Delta U=U_f-U_i=134 J-87 J=47 J

While the heat added to the system is

Q=500 J

therefore, the work done by the system is

W=Q-\Delta U=500 J-47 J=453 J

5 0
3 years ago
Read 2 more answers
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
3 years ago
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