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bogdanovich [222]
3 years ago
15

An object’s mass increases its

Physics
1 answer:
STatiana [176]3 years ago
4 0

Answer:  

It is commonly known that, if you accelerate an object, its mass will increase; however, to understand why this phenomenon occurs, we mustn’t think of the object’s mass increasing. Instead, we should think of its energy. In physics, mass is just simply locked up energy. We call this type of mass, ‘inertial mass.’

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A roller coaster car is going over the top of a 18-m-radius circular rise. at the top of the hill, the passengers "feel light,"
Mashutka [201]

The solution for this problem is:

If they feel 50% of their weight that means that the centripetal force is also 50% of their weight 1g - 0.5g = 0.5g 


Then 0.5* 9.8m/s² * 18m = 88.2 would be v² 

Then get the square root, the answer would be:
and v = 9.391 m/s is the answer.

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At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri
viva [34]

Answer:

v_{f2} =6.5%v_{i1}

Explanation:

Mass of the ball: m_{1} =0.12kg]

Initial velocity of the ball:   v_{i1}

final velocity of the ball: v_{f1} which is -30/100 of v_{i1} =-0.3v_{i1}

Mass of the bottle: m_{2} =2.4kg

Initial velocity of the bottle: v_{i2}=0m/s

final velocity of the bottle: v_{f2} is unknown (to find)

<em>by using conservation momentum, which stated that the initial momentum is equal to the final momentum.</em>

<em />m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}<em />

<em>so since the bottle is at rest firstly, therefore </em>v_{i2} =0<em />

<em />m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}<em />

<em />m_{1} v_{i1}  =m_{1} v_{f1} +m_{2} v_{f2}<em>         </em><em>equation 1</em>

so now substitute v_{f1} into equation 1

m_{1} v_{i1}  =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}

<em />m_{1} v_{i1}  = -0.3m_{1}v_{i1}  +m_{2} v_{f2}<em />

<em>collect the like terms</em>

m_{1} v_{i1}   +0.3m_{1}v_{i1}  =m_{2} v_{f2}

1.3m_{1} v_{i1}   =m_{2} v_{f2}

divide both  side by m_{2}

v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }

Now substitute

v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2}    =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}

v_{f2} =6.5%v_{i1}

<em />

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What is the largest planet in our galaxy
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If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in
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Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

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Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

The negative sign implies it's a negative acceleration.

Frictional torque that brought it to rest must be equal to the original torque.

Torqu = I x §

T = 0.33 x 0.33 = 0.11 Nm

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3 years ago
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