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____ [38]
3 years ago
10

A 9 kg box is placed on a 6 m tall shelf. What will be the kinetic energy of the box right before it hits the ground? (use g = 1

0 m/s/s)
Physics
1 answer:
m_a_m_a [10]3 years ago
5 0

Answer:

540 J

Explanation:

U at top equals to K at bottom if it's an isolated system therefore U at top is equals to 540 J so we can assume that at 0m U=0 (mgh) therefore the box has gained some velocity due to the acceleration due to g and we can calculate it using 1/2mv²

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natulia [17]
A because the dot nearst to a
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4 years ago
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A parachutist of mass 20.1 kg jumps out of an airplane at a height of 662 m and lands on the ground with a speed of 7.12 m/s. Th
Klio2033 [76]

Answer:

1.30\cdot 10^5 J

Explanation:

The energy lost due to air friction is equal to the mechanical energy lost by the parachutist during the fall.

The initial mechanical energy of the parachutist (at the top) is equal to his gravitational potential energy:

E_i=mgh

where

m = 20.1 kg is his mass

g=9.8 m/s^2 is the acceleration due to gravity

h = 662 m is the initial heigth

The final mechanical energy (at the bottom) is equal to his kinetic energy:

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

where

v = 7.12 m/s is the final speed of the parachutist

Therefore, the energy lost due to air friction is:

\Delta E=E_i-E_f=mgh-\frac{1}{2}mv^2=(20.1)(9.8)(662)-\frac{1}{2}(20.1)(7.12)^2=1.30\cdot 10^5 J

4 0
4 years ago
A merry-go-round a.k.a "the spinny thing" is rotating at 15 RPM, and has a radius of 1.75 m
arlik [135]

Answer:

A.) 4 revolution

B.) 0.2 revolution

C.) 4 seconds

D.) 2.75 m/s

Explanation:

Given that a merry-go-round a.k.a "the spinny thing" is rotating at 15 RPM, and has a radius of 1.75 m

Solution

1 revolution = 2πr

Where r = 1.75m

A. How many revolutions will it make in 3 minutes?

(2π × 1.75) / 3

10.9955 / 3

3.665 RPM

Number of revolution = 15 / 3.665

Number of revolution = 4 revolution

B. How many revolutions will it make in 10.0 seconds?

First convert 10 seconds to minutes

10/60 = 0.167 minute

(2π × 1.75) / 0.167

10.9955 / 0.167

65.973

Number of revolution = 15 / 65.973

Number of revolution = 0.2 revolution

C. How long does it take for a person to make 1 complete revolution?

15 = 1 / t

Make t the subject of formula

t = 1/15

t = 0.0667 minute

t = 4 seconds

D. What is the velocity in m/s of person standing on its edge?

Velocity in m/ s will be:

Velocity = (15 × 2pi × r) / 60

Velocity = 164.9334 / 60

Velocity = 2.75 m/s

8 0
3 years ago
The ball will oscillate along the z axis between z=dz=d and z=−dz=−d in simple harmonic motion. What will be the angular frequen
Eddi Din [679]

Answer:

\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }

Explanation:

Additional information:

<em>The ball has charge </em>-q_0<em>, and the ring has  positive charge </em>+Q<em> distributed uniformly along its circumference. </em>

The electric field at distance z along the z-axis due to the charged ring is

E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.

Therefore, the force on the ball with charge -q_0 is

F=-q_oE_z

F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}

and according to Newton's second law

F=ma=m\dfrac{d^2z}{dz^2}

substituting F we get:

- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}

rearranging we get:

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0

Now we use the approximation that

z^2+a^2\approx a^2 <em>(we use this approximation instead of the original </em>d^2+a^2\approx a^2<em> since </em>z<em>, our assumption still holds )</em>

and get

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0

Now the last equation looks like a Simple Harmonic Equation

m\dfrac{d^2z}{dz^2}+kz=0

where

\omega=\sqrt{ \dfrac{k}{m} }

is the frequency of oscillation. Applying this to our equation we get:

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}

\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}

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4 years ago
What is thermodynamics?
rusak2 [61]

Answer:

The branch of physical science that deals with the relations between heat and other forms of energy (such as mechanical, electrical, or chemical energy), and, by extension, of the relationships between all forms of energy.

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4 years ago
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