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professor190 [17]
4 years ago
7

A car moving 20 km/h skids 30 m with locked brakes. how far will the car skid with locked breaks if it were traveling at 100 km/

hr
Physics
1 answer:
Inessa05 [86]4 years ago
5 0

Kinetic energy = (1/2) (mass) (speed)²

Since the 'speed' in the KE formula is squared, if the car's speed
increases by 5 times, its kinetic energy increases by (5²) = 25 times.

The loss of kinetic energy in a skid is just the wo0rk done by friction
between the tires and pavement.  So the skid distance is proportional
to the initial kinetic energy, and the car must skid 25 times as far when
it enters the skid at the higher speed.

         25 x 30m  =  750 meters
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Describe the results of Ernest Rutherford's gold-foil experiment and explain how his results changed ideas about the distributio
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In Rutherford's gold foil experiment, some of the positive particles would pass through the foil and some would bounce off. This led to a new theory that all of the positive subatomic particles were in the center of the atom instead of evenly spread throughout.
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3 years ago
You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a
Zolol [24]

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

d = v*t \rightarrow v = \frac{d}{t}    

<u>Where</u>:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

t = \sqrt{\frac{2d}{g}}

<u>Where:</u>

g: is gravity = 9.8 m/s²

t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s

Now, the horizontal velocity of the rock is:

v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s      

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.          

2. To calculate the distance at which the projectile will land, first, we need to find the time:

t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s

So, the distance is:

d = v*t = 6.67 m/s*1.43 s = 9.54 m    

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!        

7 0
4 years ago
Help meh in this question plzzz <br>​
iragen [17]

The Moment of Inertia of the Disc is represented by I = \frac{15}{32}\cdot M\cdot R^{2}. (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

I = I_{D} - I_{H} (1)

Where:

  • I_{D} - Moment of inertia of the Disk.
  • I_{H} - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right) (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole (m):

\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}

m = \frac{1}{2}\cdot M

And the resulting equation is:

I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)

I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}

I = \frac{15}{32}\cdot M\cdot R^{2}

The moment of inertia of the Disc is represented by I = \frac{15}{32}\cdot M\cdot R^{2}. (Correct answer: A)

Please see this question related to Moments of Inertia: brainly.com/question/15246709

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<span>Porous material has many spaces that can hold(store) groundwater.</span>
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What is someone who leaves there country to settle in another?
Eddi Din [679]

Answer:

an immigrant

Explanation:

8 0
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