A rocket initially at rest accelerates at a rate of 90 m/s 2 for 0.7 min. What is its speed at the end of this time? Answer in u
1 answer:
Answer:
Final speed of the rocket, v = 3780 m/s
Explanation:
It is given that,
Initial speed of the rocket, u = 0 (at rest)
Acceleration of the rocket,
Time taken, t = 0.7 min = 42 s
We need to find the speed at the end of this time. It can be calculated using first equation of motion as :
v = 3780 m/s
So, the speed of the rocket at the end at the end of 0.7 minutes is 3780 m/s. Hence, this is the required solution.
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Answer: a) 3.85 days
b) 10.54 days
Explanation:-
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = time taken for decomposition = 3 days
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process =
First we have to calculate the rate constant, we use the formula :
Now put all the given values in above equation, we get
a) Half-life of radon-222:
Thus half-life of radon-222 is 3.85 days.
b) Time taken for the sample to decay to 15% of its original amount:
where,
k = rate constant =
t = time taken for decomposition = ?
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process =
Thus it will take 10.54 days for the sample to decay to 15% of its original amount.
Answer:
Explanation:
is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:
A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:
is the centripetal force and is defined as:
Here is the proton's speed and
is the radius of the circular motion. Replacing this in (1) and solving for r:
Recall that 1 J is equal to , so:
We can calculate from the kinetic energy of the proton:
Finally, we calculate the radius of the proton path: