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klemol [59]
2 years ago
9

On a ring road, 12 trams are spaced at regular intervals and travel at a constant speed. How many trams need to be added to the

circuit so that, maintaining the same speed, the intervals between them will decrease by 1 5
Physics
1 answer:
Sphinxa [80]2 years ago
4 0

3 trams must be added

Explanation:

In this problem, there are 12 trams along the ring road, spaced at regular intervals.

Calling L the length of the ring road, this means that the space between two consecutive trams is

d=\frac{L}{12} (1)

In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become

d'=(1-\frac{1}{5})d=\frac{4}{5}d

And the number of trams will become

12+n

So eq.(1) will become

\frac{4}{5}d=\frac{L}{n+12} (2)

And substituting eq.(1) into eq.(2), we find:

\frac{4}{5}(\frac{L}{12})=\frac{L}{n+12}\\\rightarrow n+12=15\\\rightarrow n = 3

Learn more about distance and speed:

brainly.com/question/8893949

#LearnwithBrainly

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Two identical carts travel at the same speed toward each other, and then a collision occurs. The graphs show the momentum of eac
madam [21]

Explanation :

The interaction between two objects is termed as the collision. The collision can be of two types i.e. elastic collision and inelastic collision.

In this case, two identical carts travel at the same speed toward each other, and then a collision occurs. In an inelastic collision, the momentum before and after the collision remains the same but its kinetic energy gets lost.

After the collision, both the object sticks over each other and moves with one velocity.

Out of the given graph, the graph that shows a perfectly inelastic collision is attached. It shows that after the collision both the carts move with the same velocity.

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2 years ago
Consider a father pushing a child on a playground merry-go-round. the system has a moment of inertia of 84.4 kg · m2. the father
-Dominant- [34]
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6 0
2 years ago
Read 2 more answers
You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped a
bekas [8.4K]

Answer:

A) 12.08 m/s

B) 19.39 m/s

Explanation:

A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:

mg(sinθ) – F_k = ma

Where; F_k is frictional force due to kinetic friction given by the formula;

F_k = (μ_k) × F_n

F_n is normal force given by mgcosθ

Thus;

F_k = μ_k(mg cosθ)

We now have;

mg(sinθ) – μ_k(mg cosθ) = ma

Dividing through by m to get;

g(sinθ) – μ_k(g cosθ) = a

a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)

a = -3.71 m/s²

We are told that distance d = 24.0 m and v_o = 18 m/s

Using newton's 3rd equation of motion, we have;

v = √(v_o² + 2ad)

v = √(18² + (2 × -3.71 × 24))

v = 12.08 m/s

B) Now, μ_k = 0.10

Thus;

a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)

a = 1.08 m/s²

Using newton's 3rd equation of motion, we have;

v = √(v_o + 2ad)

v = √(18² + (2 × 1.08 × 24))

v = 19.39 m/s

6 0
2 years ago
A sock with a mass of 0.03 kg is stuck to the inside of a clothes dryer spins
ValentinkaMS [17]

Answer:

15.71 m/s

Explanation:

We are given;

Time; t = 0.2 s

Radius; r = 0.5 m

The circumference will give us the distance covered.

Formula for circumference is 2πr

Thus; Distance = 2πr = 2 × π × 0.5 = π

Linear speed = distance/time = π/0.2 = 15.71 m/s

5 0
2 years ago
The number of gallons of water in a storage tank at time t, in minutes, is modeled by w(t) = 25 − t2 for 0≤t≤5. At what rate, in
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Answer:

Rate of change of water will be -6 gallon/minute

Explanation:

We have given water in the tank as the function of time as

w(t)=25-t^2

We have to find the rate of change of water in the tank at t = 3 minute

For rate of change we have to differentiate both side

So \frac{dw}{dt}=0-2t

At t = 3 minute

\frac{dw}{dt}=0-2\times 3=-6gallon/minute

8 0
2 years ago
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