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earnstyle [38]
3 years ago
11

Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are

p1 and p2. After the collision, the magnitude of their combined momentum is p. Of what can one be certain?
Physics
1 answer:
slavikrds [6]3 years ago
7 0

Answer:

The correct answer is

p = p₁ + p₂

Explanation:

Newton's second law states that force = the change of momentum produced therefore since the collision is inelstic then the change of momentum of each car is p₁ and  p₂ and the force of the collition is proportional to p₁ + p₂ that is

F ∝ p₁ + p₂ and since force is directly proportional to p we have

p = p₁ + p₂

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9. A radioisotope has a half-life of 4.50 min and an initial decay rate of 8400 Bq. What will be
Akimi4 [234]

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

8 0
3 years ago
At the circus, a 100.-kilogram clown is fired at 15 meters per second from a 500.- kilogram cannon. What is the recoil speed of
photoshop1234 [79]

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.

6 0
3 years ago
Read 2 more answers
HOW do I find 25? *picture included* plz answerrr
pogonyaev
Subtract all numbers to your answer

3 0
3 years ago
A playground merry-go-round has radius 2.10m and moment of inertia 2500kg*m^2 about a vertical axle through its center, and it t
vladimir2022 [97]

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

                                    = \dfrac{18\times 2.1}{2500}

                                    = 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 = \dfrac{72}{16}

                                                                        = 4.5 watt        

8 0
3 years ago
When a metal bonds with a nonmetal, they form a(n)
MArishka [77]
I believe it would be B.
5 0
3 years ago
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