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masya89 [10]
3 years ago
15

an object on a planet has a mass of 243 Kg. what is the acceleration of the object, if the radius of the planet is 2.32 x 10^7m

and the mass of the planet is 6.35 x 10^30 Kg estimate g as 6.67 x 10^-11 N (m/kg)
Physics
1 answer:
Ksenya-84 [330]3 years ago
3 0

Answer:

7.87x10^5m/s^2

Explanation:

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The secondary coil of a neon sign transformer provides 7500 V at 0.01 A. The primary coil operates on 120 V. What is the input c
nikitadnepr [17]

Answer:

0.625 A

Explanation:

Vs = 7500 V, Is = 0.01 A

Vp = 120 V

Let the primary current be Ip.

As the transformer is ideal, so input power is equal to the output power

Vp x Ip = Vs x Is

120 x Ip = 7500 x 0.01

Ip = 0.625 A

8 0
3 years ago
Devise an experiment using a Landsat satellite and hypothesize<br> results.
guapka [62]

An experiment that involves using a Landsat satellite is given below;

Paddy lands detection through the use of Landsat-8 satellite images and object-based classification in  cape town, South Africa.

<h3>What is the experiment about?</h3>

Rice is known to be one of the most vital  food staples in a lot of countries, especially South Africa . Due to the irrigated rice production that tend to  differs from other kinds of  agricultural fields, this study was said to have created a paddy field mapping model via the use of  phenological aspects, a lot of  satellite sensor data, and also the use of object-based approach.

This study  uses the phonological features of rice plants and also the use of an annual data regarding surface temperature (LST) to make the paddy map.

The core remote sensing data is made up of the yearly LST that is obtained from MODIS and multi-temporal Landsat-8 satellite imagery.

Based on the study, the total  accuracy and kappa coefficient for the pixel-based classification method is seen to be  92% and 0.89.

Hence, An experiment that involves using a Landsat satellite is Paddy lands detection through the use of Landsat-8 satellite images and object-based classification in  cape town, South Africa.

Learn more about Landsat satellite from

brainly.com/question/25656875

#SPJ1

8 0
2 years ago
If you were traveling away from Earth at speed 0.5c, would you notice a change in your heartbeat? Would your mass, height, or wa
FrozenT [24]

Answer:

If you were traveling away from earth at speed 0.5c, you wouldn't notice any change in your heartbeat, you won't notice your mass, height and waistline change. This is because you are on the same frame of reference as the ship in spacetime and any measurement done from the ship will give normal readings from an observer on the ship.

For an observer on earth, your heartbeat will be seen to slowdown (because your time on the ship will be perceived to slow down to an

observer on earth). Also, your mass will be seen to increase, you height will also be seen to increase, and your waistline will be seen to decrease when viewed from earth.

8 0
3 years ago
Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect a
rjkz [21]

Answer:

The speed will be "18km/s". A further explanation is given below.

Explanation:

According to the question, the values are:

Wavelength,

\lambda = 656.46 \ nm

\Delta \lambda = 0.04

c=3\times 10^8

As we know,

⇒  \frac{\Delta \lambda}{\lambda} =\frac{v}{c}

On substituting the values, we get

⇒  \frac{656.46}{0.04} =\frac{v}{3\times 10^8}

⇒         v=\frac{656.46}{0.04} (3\times 10^8)

⇒            =16411.5\times 3\times 10^8

⇒            =18280 \ m/s

or,

⇒            =18 \ km/s

8 0
3 years ago
A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea
viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}                           (1)

time taken = 5 years

We know that :

time = \frac{distance}{speed}

5 = \frac{L''}{v}                                                      (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}

Solving the above eqn:

t'' = 20.56 years

4 0
3 years ago
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