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charle [14.2K]
2 years ago
6

a 1250 kg car accelerates from rest to 6.13m/s over a distance of 8.58m calculate the average force of traction

Physics
1 answer:
iogann1982 [59]2 years ago
3 0
Use formula, v^2= u^2 + 2as.
The "v" and the "s" of the formula are given.
Since u is 0, just use f=ma.
I hope this helped!
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Tyler stands at rest on a skateboard. He has a mass of 120 kg. His friend (m = 60 kg) jumps into his arms at a speed of 2 m/s. I
Andrews [41]
Momentum question. This is an inelastic collision, so 

m1v1+m2v2=Vf(m1+m2)
Vf=(m1v1+m2v2)/(m1+m2)=[(120kg)(0m/s)+(60kg)(2m/s)] / (120kg+60kg)
Vf=120kg m/s  /   180kg
Vf=0.67m/s

0.67m/s
5 0
2 years ago
The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2
sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed V_{0}

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}

Once we have the initial speed, we can isolate the final speed from following equation:

V_{f} =V_{0} +a*t  V_{f}= 4.165 \frac{m}{s}  

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }

3 0
2 years ago
Read 2 more answers
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
kap26 [50]

Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

3 0
2 years ago
Your town is installing a fountain in the main square. If the water is to rise 26.0 m (85.3 feet) above the fountain, how much p
Brums [2.3K]

Answer:

P = 3.55 \times 10^5 Pa

Explanation:

As we know that water from the fountain will raise to maximum height

H = 26.0 m

now by energy conservation we can say that initial speed of the water just after it moves out will be

\frac{1}{2}mv^2 = mgH

v = \sqrt{2gH}

v = \sqrt{2(9.81)(26)}

v = 22.6 m/s

Now we can use Bernuolli's theorem to find the initial pressure inside the pipe

P = P_0 + \frac{1}{2}\rho v^2

P = 10^5 + \frac{1}{2}(1000)(22.6^2)

P = 3.55 \times 10^5 Pa

6 0
2 years ago
At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt
Volgvan

Answer:

It takes you 32.27 seconds to travel 121 m using the speed ramp

Explanation:

<em>Lets explain how to solve the problem</em>

- The speed ramp has a length of 121 m and is moving at a speed of

 2.2 m/s relative to the ground

- That means the speed of the ramp is 2.2 m/s

- You can cover the same distance in 78 seconds when walking on

  the ground

<em>Lets find your speed on the ground</em>

Speed = Distance ÷ Time

The distance is 121 meters

The time is 78 seconds

Your speed on the ground = 121 ÷ 78 = 1.55 m/s

If you walk at the same rate with respect to the speed ramp that

you walk on the ground

That means you walk with speed 1.55 m/s and the ramp moves by

speed 2.2 m/s

So your speed using the ramp = 2.2 + 1.55 = 3.75 m/s

Now we want to find the time you will take to travel 121 meters using

the speed ramp

Time = Distance ÷ speed

Distance = 121 meters

Speed 3.75 m/s

Time = 121 ÷ 3.75 = 32.27 seconds

It takes you 32.27 seconds to travel 121 m using the speed ramp

8 0
2 years ago
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