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elena-s [515]
3 years ago
7

A car is traveling at a 20.0 m/s for 7.00 s and then suddenly comes to a stop over a 3 s period.

Physics
1 answer:
Delicious77 [7]3 years ago
7 0

Answer: A) Deceleration of the car is -6.6667 m/s² while it came to stop.

B) The total distance the car travels is 200 meter during the 10 s period.

Explanation:

Given Data

Initial velocity of the car ($$v_{i}$$) =   20.0 m/s

Final velocity of the car (v_{f}) = 0 m/s

Time (in motion) =7.00 s

Time (in rest) =3 s

To find - A) car's deceleration while it came to a stop

              B) the total distance the car travels in 10 s

A) The formula to find the deceleration is

Deceleration = (( final velocity - initial velocity ) ÷ Time)        (m/s²)

Deceleration = ((v_{f}) - ($$v_{i}$$)) ÷ time     (m/s²)

Deceleration =  ( 0.0 - 20 ) ÷ 3      (m/s²)

Deceleration =   (- 20) ÷ 3  (m/s²)

Deceleration   =  - 6.6667 m/s²

(NOTE : Deceleration is the opposite of acceleration so the final result must have the negative sign)

The car's deceleration is  - 6.6667 m/s² while it came to a stop

B) The formula to find the distance traveled by the car is  

Distance traveled by the car is equals to the product of the speed and time

Distance = Speed × Time  (meter)

Distance = 20.0 × 10

Distance = 200 meters

The total distance the car travels during the period of 10 s is 200 meters

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A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B, which has a mass of 3 kg. After th
Zolol [24]

Answer:

(a) The velocity of the bullet and B after the first impact is 4.4554 m/s.

(b) The velocity of the carrier is 0.40872 m/s.

Explanation:

(a) To solve the question, we  apply the principle of conservation of linear momentum as follows.

we note that the distance between B and C is 0.5 m

Then we  have

Sum of initial momentum = Sum of final momentum

0.03 kg × 450 m/s = (0.03 kg + 3 kg) × v₂

Therefore v₂ = (13.5 kg·m/s)÷(3.03 kg) = 4.4554 m/s

The velocity of the bullet and B after the first impact = 4.4554 m/s

(b) The velocity of the carrier is given as follows

Therefore from the conservation of linear momentum we also have

(m₁ + m₂)×v₂  = (m₁ + m₂ + m₃)×v₃

Where:

m₃ = Mass of the carrier = 30 kg

Therefore

(3.03 kg)×(4.4554 m/s) = (3.03 kg+30 kg) × v₃

v₃ = (13.5 kg·m/s)÷ (33.03 kg) = 0.40872 m/s

The velocity of the carrier = 0.40872 m/s.

3 0
3 years ago
A runner of mass 56.0kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its cent
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Answer: -0.84 rad/sec (clockwise)

Explanation:

Assuming no external torques act on the system (man + turntable), total angular momentum must be conserved:

L1 = L2

L1 = It ω + mm. v . r = 81.0 kg . m2 .21 rad/s – 56.0 kg. 3.1m/s . 3.1 m  

L1 = -521.15 kg.m2/sec (1)

(Considering to the man as a particle that is moving opposite to the rotation of  the turntable, so the sign is negative).

Once at rest, the runner is only a point mass with a given rotational inertia respect from the axis of rotation, that can be expressed as follows:

Im = m. r2 = 56.0 kg. (3.1m)2 = 538.16 kg.m2

The total angular momentum, once the runner has come to an stop, can be written as follows:

L2= (It + Im) ωf = -521.15 kg.m2/sec  

L2= (81.0 kg.m2 + 538.16 kg.m2) ωf = -521.15 kg.m2/sec  

Solving for ωf, we get:

ωf = -0.84 rad/sec  (clockwise)

5 0
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