Answer:
525 Bq
Explanation:
The decay rate is directly proportional to the amount of radioisotope, so we can use the half-life equation:
A = A₀ (½)^(t / T)
A is the final amount
A₀ is the initial amount,
t is the time,
T is the half life
A = (8400 Bq) (½)^(18.0 min / 4.50 min)
A = (8400 Bq) (½)^4
A = (8400 Bq) (1/16)
A = 525 Bq
The recoil velocity of cannon is (4) 5.0 m/s
Explanation:
We can find the recoil velocity from the law of conservation of momentum.
The recoil velocity is velocity of body 2 after release of body 1, i.e. velocity of cannon after release of clown.
Let v2 be cannon's velocity, v1 be clown's velocity given = 15 m/sec
m1 be clown's mass = 100kg and m2 be cannon's mass given = 500kg.
So recoil velocity of cannon v2 is given by,
v2 = -(m1÷m2)v1
v2 = -(100÷500)15
v2 = -5 m/s
where the minus sign refers to the direction of cannon's recoil velocity being opposite to that of clown.
Hence, option (4)5.0 m/s is the correct answer.
Subtract all numbers to your answer
Answer
Radius of the wheel r = 2.1 m
Moment of inertia I = 2500 Kg m²
Tangential force applied F = 18 N
Time interval t = 16 s
Initial angular speed ω1 = 0
Final angular speed ω2 = ?
Let α be the angular acceleration.
Torque applied τ = Iα
F r = Iα
Angular acceleration α = F r/I
= 
= 0.015 rad/s²
(a)From rotational kinematic relation
Final angular speed ω₂ = ω₁ + αt
= 0 + (0.015 rad/s^2 * 16 s)
= 0.24 rad/s
(b) Work done W = 0.5 Iω₂² - (1/2)Iω₁²
= 0.5*( 2500 Kg m²)(0.24 rad/s)^2 - 0
= 72 J
(c) Average power supplied by the child P = W/t = 
= 4.5 watt