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3 years ago

What force is required to cause a 5 kg bowling ball to accelerate at 4 m/s2

Physics

1 answer:

Scorpion4ik [409]3 years ago

4 0

Answer:

20 N

Explanation:

By Newton's 2nd law,

The rate of change of momentum is directly proportional to the unbalance force applied on the object,

By that you can get the equation,

F = ma

= 5 × 4 = 20 N

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Lesechka [4]

hi!

just use cross method

7 0

3 years ago

An object, with mass 83 kg and speed 15 m/s relative to an observer, explodes into two pieces, one 4 times as massive as the oth

Agata [3.3K]

Answer:

The amount of Kinetic energy added to the system is 2334.3J

Solution:

As per the question:

Mass of object, m = 83 kg

Relative velocity of the object, $v_{mb} = v_{m} - v_{o} = 15 m/s$

After the explosion,

mass of the fragment is M and the other fragment, M' = 4M

Velocity of the lighter fragment after collision, v = 0 m/s

Now,

Mass of heavier fragment, M' = $\frac{4}{5}m$

Mass of lighter fragment, M' = $\frac{1}{5}m$

Let the velocity of the heavier fragment be v'.

Therefore by the law of conservation of momentum, we have:

Momentum of the object before collision = Momentum of the object after collision

$mv_{mo} = Mv + M'v'$

$mv_{mo} = M.0 + \frac{4}{5}mv'$

$v' = \frac{5}{4}\times 15 = 18.75 m/s$

Now, the change in Kinetic Energy gives the amount of Kinetic energy added to the system:

$\Delta KE = KE_{final} - KE_{initial}$

$\Delta KE = \frac{1}{2}Mv'^{2} - \frac{1}{2}mv_{mo}^{2}$

Since, the lighter particle stops, it won't have any kinetic energy.

$\Delta KE = \frac{1}{2}\times \frac{4}{5}times 83\times {18.75}^{2} - \frac{1}{2}\times 83\times 15^{2}$

$\Delta KE = 11671.8 - 9337.5 = 2334.3 J$

7 0

2 years ago

By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d

77julia77 [94]

Answer:

The intensity of sound at rock concert is 10¹⁰ greater than that of a whisper.

Explanation:

The intensity of sound is given by;

$I(dB) = 10Log(\frac{I}{I_o} )$

where;

I is the intensity of the sound

I₀ is the threshold of sound intensity = 1 x 10⁻¹² W/m²

The intensity of sound at a rock concert

$120 = 10Log(\frac{I}{1*10^{-12}} )\\\\12 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{12}\\\\I = 1*10^{-12} *10^{12}\\\\I = 1*10^0\\\\I =1 \ W/m^2$

The intensity of sound of a whisper

$20 = 10Log(\frac{I}{1*10^{-12}} )\\\\2 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{2}\\\\I = 1*10^{-12} *10^{2}\\\\I = 1*10^{-10}\\\\I =10^{-10} \ W/m^2$