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givi [52]
2 years ago
10

4. A disobedient student dropped his Physics textbook (mass 0.1kg) from the window (15m above the ground). How fast was it going

when it hit the ground?
Physics
1 answer:
vaieri [72.5K]2 years ago
8 0

Answer:

v= 17.15 m/s

Explanation:

mass of the book=0.1 Kg

height above ground, h= 15 m

Using conservation of energy

Potential energy is converted into kinetic energy

mgh = \frac{1}{2}mv^2

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 15}

v=\sqrt{294}

v= 17.15 m/s

Hence, the book will hit the ground at the speed of 17.15 m/s.

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Answer:

21741 s i believe

Explanation:

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Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?
Alexandra [31]

11) 1.04\cdot 10^7 J

12) 1.04\cdot 10^7 J

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

PE=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

g=9.8 m/s^2 (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

PE=(8325)(9.8)(127)=1.04\cdot 10^7 J

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

KE_t + PE_t = KE_b + PE_b

where

KE_t is the kinetic energy at the top

PE_t is the potential energy at the top

KE_b is the kinetic energy at the bottom

PE_b is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

KE_t=0

Also, at the bottom the height is zero, so the potential energy is zero

PE_b=0

Therefore, we find:

KE_b=PE_t=1.04\cdot 10^7 J

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

KE=1.04\cdot 10^7 J

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

KE_1 + PE_1 = KE_2 + PE_2

where

KE_1 is the kinetic energy at the top of the 1st hill

PE_1 is the potential energy at the top of the 1st hill

KE_2 is the kinetic energy at the top of the 2nd hill

PE_2 is the potential energy at the top of the 2nd hill

From the previous questions, we know that

KE_1=0

and

PE_1=1.04\cdot 10^7 J

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J

So, the kinetic energy at the second hill is

KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J

And so, the speed is

v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s

4 0
2 years ago
How does the mass of an object affect the outcome when an unbalanced force acts on it?
Verdich [7]

Answer:

The magnitude of acceleration is reduced.

Explanation:

Force is defined as push or pull

The force is said to be<em> balance force  </em>if the force are equal in size but opposite in direction. ie the object does not move or move with constant speed.

The force are to be<em> unbalanced force </em>if the force cause change in motion. ie the object has force greater than zero and has acceleration.

According to <em>Newton second law of motion </em>, acceleration depends on force acting on the object and mass of object.

     F=ma

     a=\frac{F}{m}

When unbalanced force act on the mass of object it reduces magnitude of acceleration without changing the direction.

6 0
3 years ago
Kon'nichiwa~<br>please help me with this question!!​
andrezito [222]

Let's check the relationship

\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}

\\ \rm\hookrightarrow g\propto G

So

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  • Also walking on ground would become more difficult as g increases.

Option C is wrong by now .Let's check D once

\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}

  • So time period of simple pendulum would decrease.

4 0
2 years ago
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