A. Is very attractive. If it's sublimation directly from water vapor in the air to ice on the glass, then yes. But from liquid water mist to water ice, no. Ice is less dense than water. That's why cubes float in your soda. Better leave 'A' alone. . . . D. Ice pellets turn to liquid. That one's good.
Answer:
The appropriate solution is "61.37 s".
Explanation:
The given values are:
Boat moves,
= 10 m/s
Water flowing,
= 1.50 m/s
Displacement,
d = 300 m
Now,
The boat is travelling,
= 
= 
Travelling such distance for 300 m will be:
⇒ 
On putting the values, we get
Throughout the opposite direction, when the boat seems to be travelling then,
= 
= 
Travelling such distance for 300 m will be:
⇒ 
On putting the values, we get
hence,
The time taken by the boat will be:
= 
= 
Answer:
Explanation:
The voltage of a disconnected charged capacitor increases when the plate area is decreased.
When plate area decreases , capacitance C decreases , but charge Q remains constant .
Q = C V where C is capacitance and V is voltage .
when C decreases , V increases for keeping Q constant .
So the statement is true.
The electric field is dependent on the charge density on the plates.
This statement is true .
The voltage of a connected charged capacitor remains the same when the plate area is decreased .
For a connected capacitor , V or voltage is constant which is equal to voltage of charging battery .
So the statement is true .
John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.
The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds) = 4,000 foot-pounds of work.
If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds) = 645.2 foot-pounds per second.
The rate of doing work is called "power".
(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)
So back to our problem.
John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.
Well, 550 foot-pounds per second is called 1 "horsepower".
So as John runs up the steps to the balcony, he's doing the work
at the rate of
(645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)
= 1.173 Horsepower. GO JOHN !
(I'll betcha he needs a shower after he does THAT 3 times.)
_______________________________________________
Oh my gosh ! Look at #26 ! There are the metric units I was talking about.
Do you need #26 ?
I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.
a). 5
b). 750 Joules
c). 800 Joules
d). 93.75%
You're welcome.
And #27 is 0.667 m/s .
