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algol13
2 years ago
10

BRAINLIEST WILL BE MARKED FOR THE FIRST ANSWER

Physics
1 answer:
Andreyy892 years ago
4 0

Answer:

98.4 N

Explanation:

Given that the body weighs 800 N on earth.

Thus,

Weight = mass x acceleration due gravity

i.e W = mg

800 = m x 10

m = \frac{800}{10}

   = 80 kg

The mass of the body is 80 kg.

To be able to determine its weight on the planet, we have to first calculate the gravitational pull of the planet.

But,

g = \frac{GM}{r^{2} }

Where: G is the Newton's universal gravitation, M is the mass of the planet and r is the radius of the planet.

g = \frac{6.674*10^{-11} *2.986*10^{24} }{(12.742*10^{6})^{2} }

  = \frac{1.9929*10^{14} }{1.6236*10^{14} }

  = 1.2275

g = 1.23 m/s^{2}

Thus,

Weight of the body on the planet = mg

                          = 80 x 1.23

                          = 98.4 N

The weight of the body on the planet is 98.4 N

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Serggg [28]

Answer:

Explanation:

We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for

λ

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λ

=

v

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Let's plug in our given values and see what we get!

λ

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3 0
2 years ago
Four blocks of weights are required using which any body whose weight is between 1kg and 40 kg can be weighed. Find the four wei
Vikentia [17]

Answer:

The weights are 1 kg, 3kg, 9kg and 27kg.

Explanation:

The weights are 1 kg, 3kg, 9kg and 27kg.

1+3+9+27= 40

27+9+3= 39

27+9+3-1=38

27+9+1=37

27+9=36

27+9-1=35

27+9+1-3=34

27+9-3=33

27+9-3-1=32

27+3+1=31

27+3=30

27+3-1=29

27+1=28

27

27-1=26

27+1-3=25

27-3=24

27-3-1=23

27+3+1-9=22

27+3-9=21

27+3-9-1=20

Like this all the weights from 1 to 40 kg can be made using 1,3,9 and 27 kg.

6 0
3 years ago
Why does the spectroscopic parallax method only work for main sequence stars?.
Brums [2.3K]

Answer:

Only main sequence stars have a well-defined relationship between spectral type and luminosity.

Explanation:

Low-mass stars have much longer lifetimes than high-mass stars.

5 0
1 year ago
What quantity of heat must be removed from 20g<br>of water at 0°C to change it to ice at 0°C?​
seraphim [82]

The quantity of heat must be removed is 1600 cal or 1,6 kcal.

<h3>Explanation : </h3>

From the question we will know if the condition of ice is at the latent point. So, the heat level not affect the temperature, but it can change the object existence. So, for the formula we can use.

\boxed {\bold {Q = m \times L}}

If :

  • Q = heat of latent (cal or J )
  • m = mass of the thing (g or kg)
  • L = latent coefficient (cal/g or J/kg)
<h3>Steps : </h3>

If :

  • m = mass of water = 20 g => its easier if we use kal/g°C
  • L = latent coefficient = 80 cal/g

Q = ... ?

Answer :

Q = m \times L \\ Q = 20 \times 80 = 1600 \: cal

So, the quantity of heat must be removed is 1600 cal or 1,6 kcal.

<u>Subject : Physics </u>

<u>Subject : Physics Keyword : Heat of latent</u>

4 0
3 years ago
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes
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Answer:

The value of F= - 830 N

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Final velocity, v2 = 0 (since the car came to rest after brake was applied)

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F= - 830 N

The car was driving toward the north, and since the force is negative, it implies direction of the force applied was due south.

3 0
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