The crate is at rest on the incline. What can you say about the force of friction acting on the crate?A. the frictional force po
1 answer:
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<u>Answer:</u>
a) Speed of mailbag after 3 seconds = 32.4 m/s
b) Package is 44.1 meter below helicopter
c) If the helicopter was rising steadily at 3.00 m/s
Speed of mailbag after 3 seconds = 26.4 m/s
Package is 44.1 meter below helicopter
<u>Explanation:</u>
a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Initial velocity = 3 m/s, acceleration = 9.8 and time = 3 seconds.
v = 3+9.8*3 = 32.4 m/s
Speed of mailbag after 3 seconds = 32.4 m/s
b) We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 .
Distance traveled by helicopter = 9 meter.
Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 .
Distance traveled by package = 53.1 meter.
So package is (53.1-9)meter below helicopter = 44.1 m
c) Initial velocity = -3 m/s, acceleration = 9.8 and time = 3 seconds.
v = -3+9.8*3 = 26.4 m/s
Speed of mailbag after 3 seconds = 26.4 m/s
Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 .
Distance traveled by helicopter = 9 meter.
Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 .
Distance traveled by package = 35.1 meter.
So package is (35.1+9)meter below helicopter = 44.1 m
<u>Answer:</u> The final temperature of the mixture is 45.6°C
<u>Explanation:</u>
To calculate the mass of water, we use the equation:
Density of water = 1 g/mL
- <u>When volume is 240.0 mL</u>
Putting values in above equation, we get:
- <u>When volume is 100.0 mL</u>
Putting values in above equation, we get:
When two water solutions at different temperature are mixed, the amount of heat released by water present at higher temperature will be equal to the amount of heat absorbed by water present at lower temperature..
The equation used to calculate heat released or absorbed follows:
......(1)
where,
q = heat absorbed or released
= mass of water solution 1 = 240 g
= mass of water solution 2 = 100 g
= final temperature = ?°C
= initial temperature of water solution 1 = 25°C
= initial temperature of water solution 2 = 95°C
c = specific heat of water= 4.186 J/g°C
Putting values in equation 1, we get:
Hence, the final temperature of the mixture is 45.6°C