Answer:
A. 456 seconds
Explanation:
We are given that two students walk in the same direction along a straight path at a constant speed.
One student walks with a speed=0.90 m/s
second student walks with speed=1.9 m/s
Total distance covered by each students=780 meter
We have to find who is faster and how much time extra taken by slower student than the faster student.
Time taken by one student who travel with speed 0.90 m/s=
Time=
Time taken by one student who travel with speed 0.90 m/s
=
Time taken by one student who travel with speed 0.90 m/s
=866.6 seconds
Time taken by second student who travel with speed 1.9 m/s=
=410.5 seconds
The second student who travels with speed 1.9 m/s is faster than the student travels with speed 0.90 m/s .
Extra time taken by the student travels with speed 0.90 m/s=866.6-410.5=456.1 seconds
Extra time taken by the student travels with speed 0.90 m/s=456 seconds
Hence, option A is true.
Because the temperature remains constant, we can apply Boyle's Law which states that
pV = constant
where
p = pressure
V = volume
Define the two states of the gas.
State 1
Pressure = p₁
Volume = 1000 ml
State 2
Pressure = p₂
Volume = 500 ml
Apply Boyle's law.
1000p₁ = 500p₂
2 = p₂/p₁
By halving the volume, the pressure doubles.
Answer:
The pressure increases by a factor of 2.
Answer:
Landed before it explodes
Explanation:
vf = vi + at,
0 = 145 - (9.8)t,
t = 14.79 s (Time to reach highest point)
14.79 x 2 = 29.59 s (Time to land on the ground)
It will have landed before it explodes because both the time to reach the highest point and the time to land on the ground are less than 32 seconds.
3780 km.
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A because of the resistors are four in this options first option is multiplied by 4