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3 years ago

What best describes the dropping height of a ball that bounced back up to a height of 45 centimeters?

1 answer:

6 0

Answer:C:Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy

Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy.

Although the initial energy (potential energy is preserved), the energy of deformation as the ball strikes a surface creates energy dissipation in the form of frictional heat and audible sound energy.

Every time the ball bounces, its height will be less than its previous height.

Explanation:

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A battery charger can produce 3A at 12 Volt and charges a battery fer 2 hr. Calculate work in KJ.

Oksi-84 [34.3K]

Answer: 259.2 KJ

Explanation:

The formula calculate work don in a circuit is given by :-

$W=QV$ , where Q is charge and V is the potential difference.

The formula to calculate charge in circuit :-

$Q=It$ , where I is current and t is time.

Given : Current : $I=3A$

Potential difference : $V=12\ V$

Time : $t=2\ hr=2(3600)\text{ seconds}=7200\text{ seconds}$

Now, $Q=3(7200)=21,600\ C$

Then, $W=(21600)(12)=259,200\text{ Joules}=259.2\text{ KJ}$

Hence, the work done = 259.2 KJ

4 0

2 years ago

An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below

grin007 [14]

<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let $P$ be the pressure at a point.

Let $p$ be the density fluid at a point.

Let $v$ be the velocity of fluid at a point.

Bernoulli's equation states that $P+\frac{1}{2}pv^{2}+pgh=constant$ for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let $p_{up}$ be the pressure of a point just above the wing.

Let $p_{do}$ be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

$\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}$

So, $p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa$

Force is given by the product of pressure difference and area.

Given that area is $27ms^{2}$ .

So,lifting force is $16398.48\times 27=442758.96N$

6 0

3 years ago

A town is considering building a biodiesel power plant that would burn biomass to generate electricity. Which of the following c

zmey [24]

Answer: it reduces dependence on fossil fuels. Disadvantage: it would emit pollution into the air.

3 0

3 years ago

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

3 years ago

An object has a weight of 500 on earth what is the mass of this object

erastova [34]

550! OBVY! lol! ope this helps1

7 0

3 years ago