Answer:
Some planets have seasons some don't bc of the distance from the sun some of them are too cold or too hot to have seasons
Answer:
Explanation:
Let the volume of the unknown bulb = X L
The volume of the system , after opening valve = (X + 0.72 L )
Use Boyles law gas equation,
P1V1 = P2V2 ( at temperature is constant )
Given:
P1 = 1.2 atm
P2 = 683 torr
Converting mmHg to atm,
1 atm = 760 mmHg(torr)
683 mmHg = 683/760
= 0.8987 atm
1.2X = 0.8987*(X + 0.720)
1.2X = 0.8987X + 0.6471
0.3013X = 0.6471
X = 2.15 L
Answer:
Explanation:
During the first .8 s , the elevator is under acceleration . It starts from initial velocity u = 0 , final velocity v = 1.2 m /s , time = .8 s
v = u + at
1.2 = 0 + .8 a
a = 1.2 / .8
= 1.5 m /s²
During the acceleration in upward direction , let reaction force of ground on man be R .
Net force on man = R - mg
Applying Newton's 2 nd law
R - mg = ma
R = m ( g + a )
= 72 ( 9.8 + 1.5 )
= 813.6 N .
This reaction force will be measured by spring scale , so reading of spring scale will be 813.6 N .
First put your turn signal on, next check for any ongoing traffic and wait until it is clear lastly start to drift into the lane you need to clear away from traffic
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4. For this problem, we have to write and solve a proportion. We would set this proportion up as 12/15 = 8/x. This is because we're looking for the length of the shadow and we know the height of the items, so we line them up horizontally and x goes with 8, because we're looking for the shadow length. Let's cross multiply the values. 15 * 8 = 120. 12 * x = 12. You get 120 = 12x. Now, we must divide each side by 12 to isolate the "x". 120/12 is 10. x = 10. There. The cardboard box casts a shadow that is 10 ft long.
5. For this question, you do the same thing. This time, you're finding the height of the tower, so you would do 1.2/0.6 = x/7. Cross multiply the values in order to get 8.4 = 0.6x. Now, divide each side by 0.6x to isolate the "x". 8.4/0.6 is 14. x = 14. There. The tower is 14 m tall.
If you need more help on proportions and using proportions in real life situations, feel free to search on the internet to find more information about how you solve them.
