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3 years ago

If a football player has more mass they will also have more ______ ? Fill in the blank

1 answer:

3 0

If a football player has more mass, they will also have more <u>momentum</u>. This is because mass is directly proportional to momentum.

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A wheel has a radius of 9 centimeters and an axle radius of 3 centimeters. The IMA of the system is ___?

otez555 [7]

Answer:

Answer is C.

Explanation:

To calculate the IMA for a wheel and axle, find the ratio of the larger radius to the smaller radius:

IMA = R/r

IMA = 9/3

IMA = 3

4 0

3 years ago

a block with a mass of 3 kg is swung on a cord in a horizontal circle with a radius of 2m. the speed of the block is 6m/s^. the

Ivenika [448]

Given Information:

Mass = m = 3 kg

Speed = v = 6 m/s

Radius = r = 2 m

Required Information:

Magnitude of the acceleration = a = ?

Answer:

Magnitude of the acceleration = 18 m/s²

Explanation:

The acceleration of the block traveling along a circular path with some velocity is given by

a = v²/r

a = 6²/2

a = 36/2

a = 18 m/s²

Therefore, the magnitude of the acceleration of the block is most nearly equal to 18 m/s².

Bonus:

The corresponding force acting on the block can be found using

F = ma (a = v²/r)

F = mv²/r

7 0

3 years ago

Read 2 more answers

Which science will

PtichkaEL [24]

4) Chemistry .........

4 0

3 years ago

Why is the efficiency of a machine always less than 100%?

never [62]

Answer:The percentage of the work input that becomes work output is the efficiency of a machine. Because there is always some friction, the efficiency of any machine is always less than 100 percent.

Explanation:

Sub to rockingcoolstyle15 on yt

8 0

3 years ago

A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

elena-14-01-66 [18.8K]

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ( $\dot Q$ ), measured in watts, that is, joules per second, by the following formula:

$\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}$ (1)

Where:

$m$ - Mass of the sphere, measured in kilograms.

$c$ - Specific heat of the material, measured in joules per kilogram-degree Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures of the sphere, measured in degrees Celsius.

$\Delta t$ - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

$\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}$ (2)

Where:

$m_{I}$ , $m_{X}$ - Masses of the iron and unknown spheres, measured in kilograms.

$\Delta t_{I}$ , $\Delta t_{X}$ - Times of the iron and unknown spheres, measured in seconds.

$c_{I}$ , $c_{X}$ - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

$c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}$

If we know that $\Delta t_{I} = 6.35\,s$ , $\Delta t_{X} = 4.59\,s$ , $m_{I} = 0.515\,kg$ , $m_{X} = 1.263\,kg$ and $c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}$ , then the specific heat of the unknown material is:

$c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)$

$c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}$

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

3 0

3 years ago