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IrinaK [193]
3 years ago
9

A 1500kg car traveling at 25m/s skids to a stop. The force of friction between the tires and the road is 10500N. How far does th

e car skid?
Physics
1 answer:
ale4655 [162]3 years ago
7 0

44.64m

Explanation:

Given parameters:

Mass of the car = 1500kg

Initial velocity = 25m/s

Frictional force = 10500N

Unknown:

Distance moved by the car after brake is applied = ?

Solution:

The frictional force is a force that  opposes motion of a body.

To solve this problem, we need to find the acceleration of the car. After this, we apply the appropriate motion equation to solve the problem.

     -Frictional force = m x a

the negative sign is because the frictional force is in the opposite direction

m is the mass of the car

 a is the acceleration of the car

    a = \frac{frictional force}{mass} = \frac{10500}{1500} = -7m/s²

Now using;

   V² = U² + 2as

   V is the final velocity

   U is the initial velocity

   a is the acceleration

   s is the distance moved

  0² = 25² + 2 x 7 x s

  0 = 625 - 14s

  -625 = -14s

      s = 44.64m

   

learn more:

Velocity problems brainly.com/question/10932946

#learnwithBrainly

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How much work is done if a 50N weight is lifted to a height of 10m? ​
borishaifa [10]

500J  

Explanation:

Given parameters:

Weight of the body = 50N

Height = 10m

Unknown:

Work done = ?

Solution;

Work done is the force that moves a body through a particular distance in the direction of the force.

In this problem, we can solve the problem by relating work done to the potential energy used in lifting the mass.

  The weight on a body, a force in the presence of gravity

         weight(weight) = mg

  where m = mass of body

              g = acceleration due to gravity

  Work done = Fxd

   Where F = force on the body

               d = distance moved

Potential energy = mgh

    where h is the height

     P.E = work done =  Weight x height = 50 x 10 = 500J  

Learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

6 0
3 years ago
A college student likes to take her books to class by placing them in a box and pulling the box around behind her. She pulls on
just olya [345]

Answer:

It would result an a negatice answer.

Explanation:

The accelarion should be pulled as a kite not a box :) columbus said that musical stuff no just no

Hope this helps!!

       - Katty queen

             

8 0
3 years ago
What is the velocity (in m/s) of a 550 kg roller coaster cart at the bottom of the track if it started with 990,000 J of gravita
sesenic [268]

Answer:

60m/s

Explanation:

initial energy = final energy

g.p.e = k.e

k.e = 0.5 × mass × velocity²

g.p.e = 990000J as per Question

990000Nm = 0.5 × 550 × V²

V² = 3600

V = 60m/s

4 0
3 years ago
A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete
NeX [460]

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}

From Rayleigh criteria

sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm

The focal length of the camera's lens is 300cm

3 0
3 years ago
HURRY!!!!
VMariaS [17]

Answer:

Ted is correct

Explanation:

The equation for gravitational potential energy is PE = m·g·h

The equation for gravitational kinetic energy is KE = 1/2·m·v²

Where:

m = Mass of the object (The racing car)

g = Acceleration due to gravity

h = The height to which the object is raised

v = Velocity of motion of the object

From the principle of conservation of energy, energy can neither be created nor destroyed but changes from one form to another, we have;

Potential energy gained from location at height h = Kinetic energy gained as the object moves down the level ground

m·g·h = 1/2·m·v² canceling like terms gives

g·h = 1/2·v²

v = (√2·g·h)

If the speed is doubled, we have

2·v = 2× (√2·g·h) =  (√2·g·4·h)

Therefore, if 2·v = v₂ then v₂ =  (√2·g·4·h)

Since g, the acceleration due to gravity, is constant, it means that the initial  height must be multiplied or increased 4 times to get the new height, that is we have;

v₂ =  (√2·g·4·h) = (√2·g·h₂)

Where:

4·h = h₂

Which gives;

v₂² = 2·g·h₂

1/2·v₂² = g·h₂

1/2·m·v₂² = m·g·h₂ Just like in the first relation

Therefore, Ted is correct s they need to go up four times the initial height to double the speed.

5 0
3 years ago
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