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eimsori [14]
2 years ago
5

A ball at rest rolls across a frictionless floor at 12.0 m/s/s. How far will it travel in

Physics
1 answer:
professor190 [17]2 years ago
3 0

Answer:

The distance, d travelled by the ball is 768 metres.

Explanation:

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of final speed from the initial speed all over time.

Hence, if we subtract the final speed from the initial speed and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{initial \; speed  -  final \; speed}{time}

a = \frac{v  -  u}{t}

Where,

a is acceleration measured in ms^{-2}

v and u is initial and final speed respectively, measured in ms^{-1}

t is time measured in seconds.

Given the following data;

Acceleration = 12.0m/s²

Time, t = 8secs

Velocity =?

First, we would calculate its velocity;

a = \frac{v  -  u}{t}

Since the ball rolls at rest, initial velocity is zero (0).

V = a * t

V = 12 * 8

Velocity = 96ms^{-1}

We can now solve for the distance;

Velocity = \frac{distance}{time}

Therefore,

Distance = velocity * time

Distance = 96 * 8

Distance = 768m.

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Sammy squirrel is steering his boat at a heading of 327 degree at 18mph. The current is flowing at 4mph at a heading of 60 degre
Tanya [424]

Answer:

  • 59.97 º at 18.23 mph

Explanation:

To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.

To add the two vectors analytically you decompose each vector into their vertical and horizontal components.

<u>1. 18 mph 327º</u>

  • Horizontal component: 18 mph × cos (327º) = 15.10 mph

  • Vertical component: 18 mph × sin (327º) = - 9.80 mph

  • Vector notation:

       15.10\hat i-9.80\hat j

<u>2. 4 mph 60º</u>

  • Horizontal component: 4 mph × cos (60º) = 2.00 mph

  • Vertical component: 4 mph × sin (60º) = 3.46 mph

  • Vector notation:

       2.00\hat i+3.46\hat j

<u>3. Addition:</u>

You add the corresponding components:

15.10\hat i-9.80\hat j+2.00\hat i+3.46\hat j\\ \\ 17.10\hat i-6.34\hat j

To find the magnitude use Pythagorean theorem:

  • \sqrt{17.1^2+6.34^2}= 18.23

<u>4. Direction:</u>

Use the tangent ratio:

  • tan(\alpha )=opposite/adjacent=3.46/2.00=1.73

Find the inverse:

  • arctan (1.73) ≈ 59.97º
5 0
3 years ago
Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and
lawyer [7]

1 kilometre=1000 metre

      1 hour = 3600 second

       1\ km/hr=\frac{1000}{3600} m/s

       1\ km/hr=\frac{5}{18} m/s

The initial velocity of car A is 35.0 km/hr i.e

                                         35.0\ km/hr=35*\frac{5}{18} m/s

                                                                   = 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as  25\ km/hr^2

                                            =\ 25*\frac{1000}{3600*3600} m/s^2

                                            =0.00192901234 m/s^2

The time taken by car A = 15 min.

From equation of kinematics we know that-

                                 v= u+at      [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A,  v = 9.72 m/s +[0.00192901234×15×60]m/s

                                   =11.456111106 m/s

The acceleration of B is given as    15\ km/hr^2

                                    =0.00115740740740 m/s^2

The time taken by car B =20 min

The final velocity of B is -

                             v= u+at

                               = u-at    [Here a is negative due to deceleration]

                               =12.5 m/s +[0.0011574074074×20×60]

                               =13.8888888.....

                               =13.9

The acceleration of C is given as    40\ km/hr^2          

                                                            =\ 0.003086419753 m/s^2

The time taken by car C =30 min

The final velocity of C is-

                                v = u+at

                                   =8.89 m/s+[0.003086419753×30×60] m/s

                                   =14.4455555555..m/s

                                   =14.45 m/s

The car C is decelerating.The deceleration is given as-  60\ km/hr^2

                                                                      =0.0046296296296m/s^2

The time taken by car D= 45 min.

The final velocity of the car D is-

                     v =u+at

                        =30.56 -[0.00462962962962×45×60]m/s

                        =18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

                 


3 0
3 years ago
If the 250 kg bumper car that you are riding in hits another bumper car that is sitting still while driving 3.5 m/s, how much fo
Vilka [71]

Answer:

875 N

Explanation:

From this question, you didn't state the time taken for the bumper car to move or to hit the other bumper car. In calculations of force, time is often needed, because

Force = mass * acceleration, while

Acceleration = velocity / time, basically

Force = mass * velocity / time.

We have our mass, we have our velocity, but we haven't time. So, for this calculation, I'd assume our time to be 1s.

Going by the formula I stated, we can then say that

Force = 250 * 3.5 / 1

Force = 875 N

This means the force my bumper car have while moving at 3.5 m/s for an estimated time of 1s is 875 N

3 0
2 years ago
3) A 30 kg child slides freely across a "Slip and Slide" on LEVEL GROUND. While the child slides, the force applied to keep them
zavuch27 [327]

Answer:

a = 0.1962 m/s^2

Explanation:

The magnitude of kinetic friction exerted is given by

F_k=\mu_kN

Where, μ_k= coefficient of kinetic friction= 0.02 and N = reaction force = mg

Where m= mass = 30 Kg and, g is acceleration due to gravity =9.81 m/s^2

F_k=0.02×30×9.81 =5.886 N

Now, since, there is no applied force this kinetic friction force will cause acceleration of the child

⇒ ma = F_k

here, a is the acceleration

⇒30a = 5.886

⇒ a = 0.1962 m/s^2

4 0
2 years ago
How much force is needed to lift a 25-kg mass at a constant verlocity?
Troyanec [42]

-- In order to achieve constant verlocity, the net force on the mass must be zero.  So if there ARE any forces acting on it, they must be balanced.

-- There is already a force on the mass that can't be eliminated . . . the force of gravity.

-- That force due to gravity is (mass x gravity) = (25 kg)(9.8 m/s²) = <em><u>245N</u></em> in the <u><em>downward</em></u> direction.

-- In order to 'balance' the forces and make them add up to zero, we have to provide another force of <em>245N</em>, all in the <em>upward</em> direction.

-- Then the forces on the object will be balanced, the NET force on it will be zero, and whichever way you start it moving, it will continue to move at a cornstant verlocity.

5 0
2 years ago
Read 2 more answers
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