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irina [24]
3 years ago
13

Pls help ASAP now! This is physical science

Physics
2 answers:
zalisa [80]3 years ago
7 0

Answer:

A

Explanation:

D and C are both eliminated because it being shaped like a horseshoe has nothing yo do with it and c is wrong cause the core is really hot there for it can not be similar so A because it has north and south poles meaning its magnetic ability

const2013 [10]3 years ago
6 0

Option A is correct answer

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The interaction of electric currents or fields and magnetic fields.
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the engine won't start or it sputters when it should be running perfectly. if the gasoline is old and stale, it will have lost a portion of its volatility. the lighter components of the gasoline (remember, gasoline is a mixture of different hydrocarbons) have probably evaporated off or disappeared.
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2 years ago
Which tool would you use to measure the amount of rainfall? a. graduated cylinder b. stopwatch c. scale d. thermometer
Fittoniya [83]

A graduated cylinder measures the volume of a liquid.

A stopwatch measures the amount of time that elapses.

A scale measures the mass of objects.

A thermometer measures the temperature of any object.

Because we are measuring rain, a liquid, we would want to use a tool that would allow us to collect the rain for measuring. Therefore, the tool e would use to measure the amount of rainfall would be A. a graduated cylinder.

5 0
3 years ago
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A 95.0 kg satellite moves on a circular orbit around the Earth at the altitude h=1.20*103 km. Find: a) the gravitational force e
Natali5045456 [20]

Answer:

a) F = 660.576\,N, b) a_{c} = 6.953\,\frac{m}{s^{2}}, c) v \approx 7255.423\,\frac{m}{s}, \omega = 9.583\times 10^{-4}\,\frac{rad}{s}, d) T \approx 1.821\,h

Explanation:

a) The gravitational force exerted by the Earth on the satellite is:

F = G\cdot \frac{m\cdot M}{r^{2}}

F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}

F = 660.576\,N

b) The centripetal acceleration of the satellite is:

a_{c} = \frac{660.576\,N}{95\,kg}

a_{c} = 6.953\,\frac{m}{s^{2}}

c) The speed of the satellite is:

v = \sqrt{a_{c}\cdot R}

v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}

v \approx 7255.423\,\frac{m}{s}

Likewise, the angular speed is:

\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}

\omega = 9.583\times 10^{-4}\,\frac{rad}{s}

d) The period of the satellite's rotation around the Earth is:

T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)

T \approx 1.821\,h

6 0
3 years ago
The objective lens and the eyepiece of a telescope are spaced 85 cm apart. If the eyepiece is123 D what is the total magnificati
dedylja [7]

Answer:

The magnification would be "103.55". A further explanation is given below.

Explanation:

The given values are:

Distance between lens and eyepiece,

L = 85 cm

Eyepiece is,

= 123 D

Now,

The refractive power of eye piece will be:

⇒ \frac{1}{f_e}=123D

   f_e=\frac{1}{123D}

   f_e=0.813 \ cm

The length of the telescope will be:

⇒ L=f_0+f_e

⇒ f_0=L-f_e

On substituting the values, we get

⇒     =85-0.813

⇒     =84.187 \ cm

Now,

The magnification of the telescope will be:

⇒ M=\frac{f_0}{f_e}

⇒      =\frac{84.187}{0.813}

⇒      =103.55

5 0
2 years ago
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