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lapo4ka [179]
3 years ago
12

Two cannonballs are dropped from a second-floor physics lab at height h above the ground. Ball B has four times the mass of ball

A. When the balls pass the bottom of a first-floor window at height above the ground, the relation between their kinetic energies, KA and KB, is:_______.A) KA- 4KB B) KA 2KB C) KA KB D) KB 4KA.
Physics
1 answer:
frez [133]3 years ago
6 0

Formula of kinetic energy

e =   \frac{1}{2} m {v}^{2}

Therefore Kinetic energy of ball B is 4 times more than ball A.

ans is KB=4KA

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Many galaxies appear to have supermassive black holes in their centers powering active galactic nuclei (also called AGN). The Sc
Lena [83]

Answer:i dont remember

Explanation: ?>>> confused

5 0
4 years ago
The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increa
Nostrana [21]

Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s.  Then the mass m is 0.625kg.

Explanation: To find the answer, we need to know more about the simple harmonic motion.

<h3>What is simple harmonic motion?</h3>
  • A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
  • We have the equation of time period of a SHM as,

                                          T=2\pi \sqrt{\frac{m}{k} }

  • Where, m is the mass of the body and k is the spring constant.
<h3>How to solve the problem?</h3>
  • Given that,

               T_1=2s\\m_1=m\\m_2=m+2kg\\T_2=3s

  • We have to find the value of m,

              T_1=2\pi \sqrt{\frac{m}{k} } \\T_2=2\pi \sqrt{\frac{m+2}{k} } \\\frac{T_1}{T_2} =\sqrt{\frac{m}{m+2} }\\\frac{2}{3} =\sqrt{\frac{m}{m+2} }\\\\

               m=\frac{5}{8} =0.625kg

Thus, we can conclude that, the mass m will be 0.625kg.

Learn more about simple harmonic motion here:

brainly.com/question/28045110

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3 0
2 years ago
A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
MakcuM [25]

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

4 0
3 years ago
V=xf-xi/t solve for t
irina [24]

Answer:

t=\frac{x_f-x_i}{v}

Explanation:

Starting from the equation:

v=\frac{x_f-x_i}{t}

First of all, let's multiply by t on both sides:

v\cdot t = \frac{x_f-x_i}{t}\cdot t \\vt = x_f - x_i

And then, let's divide by v on both sides:

\frac{vt}{v}=\frac{x_f-x_i}{v}\\t=\frac{x_f-x_i}{v}

So, finally

t=\frac{x_f-x_i}{v}

7 0
3 years ago
An object falls from the top of a building that is 25 m high. Air resistance is negligible.
Vlada [557]

The velocity of the object s calculated as 22.1 m/s.

<h3>What is the speed of the object?</h3>

Given that we can write that;

v^2 = u^2 + 2gh

Now u = 0 m/s because the object was dropped from a height

v^2 = 2gh

v = √2 * 9.8 * 25

v = 22.1 m/s

Learn more about velocity:brainly.com/question/18084516

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5 0
2 years ago
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