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morpeh [17]
3 years ago
5

Show with proof, the planet which an astronaut would have a greater weight if Planet A has twice the mass and twice the radius o

f Planet B.
Physics
1 answer:
meriva3 years ago
4 0

Answer:

First let's look at the gravitational formula by newton:

f = g \times  \frac{m1 \times m2}{ {r}^{2} }

Now the f and g should be capital but that's not possible with the equation system. But we can see that the force the astronaut is pulled at is dependent on the mass and the distance if we assume his mass stays the same (mass and weight aren't the same also g is a constant). If a planet has twice the radius the force will by four times as weak because of the ^2. This is not compensated by the twice as big mass. Therefore the astronaut will have a higher force and thus a higher weight on planet B

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A cyclist has a constant speed of 12 m/s. What is the magnitude of the displacement of the cyclist after 18 seconds?
Katyanochek1 [597]

Answer:

216 m

Explanation:

Assuming a straight line:

Δx = vt

Δx = (12 m/s) (18 s)

Δx = 216 m

7 0
3 years ago
Which statement about motor development is true?
BigorU [14]

Answer:

(a) Motor skills are interrelated.

Explanation:

As per the question,

<u>Motor development:</u>

It is defined as the development of movement abilities such as the development of a child's bones, muscles and ability to move around and manipulate his or her environment.

Importance of a Motor Development Model  are:

(1) Helps us make sense of developmental changes

(2) Provides a framework for observing the change

(3) Helps us include important factors of motor behavior in our observations.

Hence, Only option (a) is correct.

4 0
2 years ago
You are driving 12 meters per second. What is your speed in miles per hour? (1.6km=1 mike) EXPLAIN
Blizzard [7]

1.6km = 1 mike? Wow that guy is tall.

I think you meant 1.6 km = 1 mile? Okay then if we're going 12 meters per second how much would you travel in one hour? First we need to figure out how many seconds are in an hour.  There are 60 seconds in a minute and 60 minutes in an hour so:

60×60=3600 seconds in an hour

Now we will multiply that by 12 meters per second and we get:

12\frac{m}{s} \times 3600 s=43200m

And 43200 meters is 43.2 km (1000 meters in 1 kilometer) meaning 43.2 kilometers an hour.  Since there are 1.6 km in one mile we must divide 43.2km to 1.6.

\frac{43.2}{1.6}=27mph

And so your speed is 27 miles per hour.

8 0
3 years ago
Please help me with these questions
Marta_Voda [28]
Radio waves - radio waves have the longest wavelengths of all the electromagnetic waves. they range from around a foot long to several miles long. Radio waves are often used to transmit data and have been used for all sorts of applications including radio, satellites, radar, and computer networks. hope it helps
7 0
3 years ago
Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =
Sloan [31]

Answer:

The current is  I  = 8.9 *10^{-5} \  A

Explanation:

From the question we are told that

     The  radius is r =  3.17 \  mm  =  3.17 *10^{-3} \ m

      The current density is  J =  c\cdot r^2  =  9.00*10^{6}  \ A/m^4 \cdot r^2

      The distance we are considering is  r =  0.5 R  =  0.001585

Generally current density is mathematically represented as

          J  =  \frac{I}{A }

Where A is the cross-sectional area represented as

         A  =  \pi r^2

=>      J  =  \frac{I}{\pi r^2  }

=>    I  =  J  *  (\pi r^2 )

Now the change in current per unit length is mathematically evaluated as

        dI  =  2 J  *  \pi r  dr

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

         I  = 2\pi  \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr

         I  = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr

        I  = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ]  | \left    0.001585} \atop 0}} \right.

        I  = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]

substituting values

        I  = 2 *  3.142  *  9.00 *10^6 *   [ \frac{0.001585^4}{4} ]

        I  = 8.9 *10^{-5} \  A

5 0
3 years ago
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