Answer:
The density will be 0.65 kg/m3
I think it won't float!! I am not sure though.
Answer:
For a velocity versus time graph how do you know what the velocity is at a certain time?
Ans: By drawing a line parallel to the y axis (Velocity axis) and perpendicular to the co-ordinate of the Time on the x axis (Time Axis). The point on the slope of the graph where this line intersects, will be the desired velocity at the certain time.
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How do you know the acceleration at a certain time?
Hence,
By dividing the difference of the Final and Initial Velocity by the Time Taken, we could find the acceleration.
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How do you know the Displacement at a certain time?
Ans: As Displacement equals to the area enclosed by the slope of the Velocity-Time Graph, By finding the area under the slope till the perpendicular at the desired time, we find the Displacement.
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If a car crashes into another car like this, the wreck should go nowhere. Besides this being an unrealistic question, the physics of it would look like this:
Momentum before and after the collision is conserved.
Momentum before the collision:
p = m * v = 50000kg * 24m/s + 55000kg * 0m/s = 50000kg * 24m/s
Momentum after the collision:
p = m * v = (50000kg + 55000kg) * v
Setting both momenta equal:
50000kg * 24m/s = (50000kg + 55000kg) * v
Solving for the velocity v:
v = 50000kg * 24m/s/(50000kg + 55000kg) = 11,43m/s
Answer: beta particles
Explanation:
Given mass = 14.0 g
Molar mass = 137 g/mol
According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number of particles.
1 mole of cesium contains atoms =
0.102 moles of cesium contains atoms =
The relation of atoms with time for radioactivbe decay is:
Where =atoms left undecayed
= initial atoms
t = time taken for decay = 3 minutes
= half life = 30.0 years = minutes
The fraction that decays :
Amount of particles that decay is =
Thus beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.
Answer:
Explanation:
Theorem of Binomial Distribution will apply here.
n = 29 , p = .67 , q = 0.33
mean = np = 29 x .67 = 19.43
Standard Deviation = √npq
= √29 x .67 x .33
= √6.4
= 2.53
=