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tensa zangetsu [6.8K]
2 years ago
3

A football coach sits on a sled while two of his players build their strength by dragging the sled across the field with ropes.

The friction force on the sled is 1500 N and the angle between the two ropes is 50∘. Assume both players pull with the same force. How hard must each player pull to drag the coach at a steady 2.10 m/s?
Physics
1 answer:
Pie2 years ago
7 0

Answer:

828.7N

Explanation:

When two players are pulling the sled by same force inclined at 50 degree Angle so the net force is given as

F_{net} = \sqrt{F^2 + F^2 + 2F^2cos\theta}

F_{net} = F\sqrt{2 + 2cos\theta}\\F_{net} = F\sqrt{2 + 2cos50}\\F_{net} = 1.81 F

now since the sled is moving with uniform speed so

F_{net} = F_f

now we have

1.81 F = 1500F = 828.7 N

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serious [3.7K]

Answer:

Q1: 3.2km

Q2: 4.8K

Explanation:

Q1:

So db is the distance of bird, and dr is the distance of runner

db = 2vr  and the distance of bird is going to be 2 times greater than the runner.

formulas: db = 2vr & db = 2dr

  1. db = 2dr
  2. L + (L - x) = 2x
  3. 2L - x = 2x
  4. 2L = 3x
  5. x = \frac{2}{3}L

Insert it in x = \frac{2}{3}L

\frac{2}{3}(2.4km) = 1.6km

Now we use formula db = 2dr

  1. db = 2L - x
  2. db = 2(2.4km) - 1.6km
  3. <u>db = 3.2km</u>

Q2:

Formulas: Vr = L /Δt & Vb = db/Δt

  1. Vr = L/ Δt ⇒ Δt = \frac{L}{Vr}
  2. \frac{2.4km}{6.8km/hr}
  3. \frac{6}{17}hr

(Km cancel each other)

  1. Vb = db/Δt ⇒ db = VbΔt
  2. 13.6km/hr(\frac{6}{17}hr )
  3. <u>4.8km</u>

(hr cancel each other)

Hope it helps you :)

6 0
2 years ago
Reginald slipped and broke his leg in his kitchen when he ran inside to grab a cookie. His mother had just mopped the floor. Wha
padilas [110]

Answer:

A. <u>The water decreases the friction between the floor and the feet </u>

Explanation: Think about it like this, when your in the shower and water is on your skin, you can scrub it fluidly, but when the water dries in your towel, there is more friction when your rub your skin, this is because the molecules in water aren't as compact as solids, so anything acting against it, is most likely to disperse from it.

3 0
2 years ago
A bullet blasts from the barrel of a gun upward in the vertical direction with an initial speed of 700 m/s . Find the maximum al
ryzh [129]

The maximum altitude that the bullet will reach is the point at which its velocity is zero. The equation that may be used in order to determine the altitude is,

<span>                                   D = ((Vi)2 – (Vf)2)/2g</span>

Where Vi and Vf are the initial and final velocities, respectively. g is the deceleration due to gravity.

Substituting the known values,

<span>                                    D = ((700 m/s)2 – (0 m/s)2) / (2(-9.8 m/s2))</span>

<span>                                    D = 25000 m</span>

 

Thus, the maximum height is 25000 m.

 

For the time needed to reach it, we use the equation,

<span>                                    Vf = Vi – (g)(t)</span>

Substituting,

<span>                                    0 = 700 m/s – (-9.8 m/s2)(t)</span>

The value of t is equal to 71.43 s.

<span> </span>

6 0
2 years ago
The speed of sound in air is 320 ms-1 and in water it is 1600 ms-1. It takes 2.5 s for sound to reach a certain distance from th
Nonamiya [84]

Answer:

Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.

Explanation:

Given:

Speed of sound in air = 320 m/s

Speed of sound in water = 1600 m/s

Time taken to reach certain distance in air = 2.5 sec

a.

We have to find the distance traveled by sound in air.

Distance = Product of speed and time.

⇒ Distance = Speed\times time\ taken

⇒ Distance = 320\times 2.5

⇒ Distance = 800 meters.

b.

Now we have to find how much time the sound will take to travel in water.

⇒ Time = Ratio of distance and speed

⇒ Time =\frac{distance}{speed}

⇒ Time =\frac{800}{1600}   <em>   ...distance = 800 m and speed = 1600 m/s</em>

⇒ Time =\frac{1}{2}

⇒ Time =0.5 seconds.

Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.

7 0
2 years ago
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