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Airida [17]
3 years ago
6

It's five miles from Tim's house to Rita's house. Roughly how long is this distance in feet?

Physics
2 answers:
Ber [7]3 years ago
5 0
The answer is 26400
sergey [27]3 years ago
3 0

Answer:

26400 ft

Explanation:

that would be my answer

hope this helps!!! :)

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A car slows down from speed of 72
Nastasia [14]

Explanation:

Given parameters:

Initial velocity = 72km/hr

Final velocity  = 0km/hr

Time taken  = 25s

Unknown:

Acceleration = ?

Solution:

To solve this problem, convert km/hr to m/s;

           1000m = 1km

           3600s = 1hr

  72km/hr;

          1km/hr  = 0.278m/s

         72km/hr = 0.278 x 72  = 20.02m/s

Acceleration is the change in velocity divided by the time taken;

       Acceleration = \frac{final velocity - initial velocity }{time}  

       Acceleration  = \frac{0 - 20.02}{25}   = -0.8m/s

The car  is actually decelerating at a rate of 0.8m/s

5 0
2 years ago
Does the force of kinetic friction depend on the weight of the block? explain.
drek231 [11]
Yes for an object moving on a horizontal plane, R = mg (where mg = weight). therefore, for an object moving on a horizontal plane: F = μmg



7 0
3 years ago
When you ask you friend for hoemwork answers and he answers like a year later hahahah
Hitman42 [59]
G that’s a nice friend
5 0
3 years ago
Read 2 more answers
You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0
3 years ago
The distance between a speaker and a listener is 8.0 m. Determine the frequency of the sound wave if exactly 20 waves are formed
dmitriy555 [2]

The frequency of the wave is 6800 Hz

<u>Explanation:</u>

Given:

Wave number, n = 20

Speed of light, v = 340 m/s

Frequency, f = ?

we know:

wave number = \frac{frequency}{speed of light}

20 = \frac{f}{340 m/s} \\\\f = 20 X 340 s^-^1\\\\f = 6800 Hz

Therefore, the frequency of the wave is 6800 Hz

4 0
3 years ago
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