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mr Goodwill [35]

2 years ago

8

How potential and kinetic energy changed during the spacecraft launches.

2 answers:

Drupady [299]2 years ago

7 0

Answer:

Moving the spacecraft against the magnetic force provided the energy needed to launch it. The energy expended to move a magnet against a magnetic force is stored in the magnetic field as potential energy.

Explanation:

LenaWriter [7]2 years ago

4 0

Answer:

Have a blessed day!

Explanation:

The energy to launch the spacecraft came from moving the spacecraft against the magnetic force. The energy used to move a magnet against a magnetic force is stored as potential energy in the magnetic field. This magnetic force can convert potential energy stored in the magnetic field to kinetic energy.

Please give brainliest!

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aleksklad [387]

Answer:

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Explanation:

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3 years ago

the electrical resistance of copper wire varies directly with its length and inversely with the square of the diameter of the wi

leonid [27]

Answer:24.47

Explanation:

Given

$L_1=30 m$

$d_1=3 mm$

$R_1=25 \Omega$

$L_2=40 m$

$d_2=3.5 mm$

we know Resistance $R=\frac{\rho L}{A}$

Where R=resistance

$\rho =resistivity$

L=Length

A=area of cross-section

$A=\frac{\pi d^2}{4}$

Thus $R\propto \frac{L}{d^2}$

therefore

$R_1\propto \frac{L_1}{d_1^2}$ --------1

$R_2\propto \frac{L_2}{d_2^2}$ --------2

divide 1 and 2 we get

$\frac{R_1}{R_2}=\frac{L_1}{L_2}\times \frac{d_2^2}{d_1^2}$

$\frac{R_1}{R_2}=\frac{30}{40}\times \frac{3.5^2}{3^2}$

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8 0

3 years ago

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of

Pachacha [2.7K]

Answer:

<em>The frequency changes by a factor of 0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

$f$ = $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ ÷ $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ x $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = <em>0.27</em>

7 0

3 years ago

A 29.0-kg child on a 3.00-m-long swing is released from rest when the ropes of the swing make an angle of 25.0° with the vertica

KiRa [710]

Answer:

v = 2.348 m/s

Explanation:

Let:

h be the height of the child above the swing's low point,

v be the speed at the low point.

When the ropes make angle 3

25 deg. with the vertical:

h = 3.00(1 - cos(25)).

Equating potential energy at the high point to kinetic energy at the low point:

29.0 * 9.81 * 3.00(1 - cos(25)) = 29.0v^2 / 2

79.96 =29.0v^2 / 2

v = 2.348 m/s

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3 years ago