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2 years ago

What is the weight of an object with a mass of 6.0 kg on Earth?

Physics

2 answers:

Rom4ik [11]2 years ago

6 0

Weight = Mass × The gravitational acceleration of the planet

Earth's gravitational acceleration ≅ 9.81 m/s²

Weight = 6 × 9.81 ≅ 58.86 N

gregori [183]2 years ago

3 0

<h2>since weight is measured in newtons, convert the 6 kg to newtons</h2><h3>the formula to convert is kg x 9.807 = N</h3>

6 x 9.807
= 58.842 N

hope that helps :))

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Suppose you monitor a large number (many thousands) of stars over a period of 3 years, searching for planets through the transit

nasty-shy [4]

Answer:

2. You must be able to precisely measure variations in the star's brightness with time.

5. As seen from Earth, the planet's orbit must be seen nearly edge–on (in the plane of our line-of-sight).

6. You must repeatedly obtain spectra of the star that the planet orbits.

Explanation:

The transit method is a very important and effective tool for discovering new exoplanets (the planets orbiting other stars out of the solar system). In this method the stars are observed for a long duration. When the exoplanet will cross in front of theses stars as seen from Earth, the brightness of the star will dip. To observe this dip following conditions must be met:

1. The orbit of the planet should be co-planar with the plane of our line of sight. Then only its transition can be observed.

2. The brightness of the star must be observed precisely as the period of transit can be less than a second as seen from Earth. Also the dip in brightness depends on the size of the planet. If the planet is not that big the intensity dip will be very less.

3. The spectrum of the star needs to be studied and observe during the transit and normally to find out the details about the planets.

4. Also, the orbital period should be less than the period of observation for the transit to occur at least once.

4 0

3 years ago

After a laser bean passes through two thin parallel slits, thefirst completely dark fringes occur at ± 15.00with the original di

PSYCHO15rus [73]

Answer:

143 °

Explanation:

a ) If d be the distance between slits , λ be wavelength of light used and at angle θ nth dark fringe is formed then

d sinθ = ( 2n+1) λ/2

for first dark fringe

d sinθ = λ/2

d /λ = 1/ 2 sinθ

1 / 2 sin15

= 1.93

b )

For intensity of fringe at angle θ, the relation is

I = I₀ cos²θ

I / I₀ = cos²θ/2

Given I / I₀ =0. 1

0.1 = cos²θ/2

θ/2 = 71.5

θ = 143 °

4 0

2 years ago

b. What would the momentum be if the mass of the bowling ball were doubled and its velocity still was 3 m/s?

umka2103 [35]

Answer:

Twice.

Explanation:

The momentum of an object is given by :

p = mv

Where

m is mass and v is the velocity

If the mass of the ball were doubled, m'=2m and v'=v=3 m/s

New momentum,

p'=m'v'

p'=2m × v

p'=2mv

p'=2p

So, the new momentum becomes twice the initial momentum.

6 0

2 years ago

which of the following is true? a)speed is velocity w a vector b)speed is a scalar w direction c)speed is velocity w direction d

Aleksandr [31]

The answer is D velocity is speed w a direction

8 0

3 years ago

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

shutvik [7]

Answer:

Answer is explained in the explanation section.

Explanation:

Solution:

For this to find, we need to calculate the centripetal acceleration on the equator.

The centripetal acceleration of the equator:

a = 4 $\pi ^{2}$ RcosФ/ $T^{2}$ ,

where,

R is the radius of the earth

R = 6378 KM = 6.3 x $10^{6}$ and

T is the time period

T = 24 h = 86164.1 s

At Equator, Ф = 0°

So, CosФ = 1

Hence,

a = 4 $\pi ^{2}$ R/ $T^{2}$

By plugging in the values, we get:

a = 4 x ( $3.14^{2}$ ) x (6.3 x $10^{6}$ ) / $86164.1^{2}$

a = 0.03 m/ $s^{2}$

Hence, this is the centripetal acceleration on the equator. And we also know that, acceleration due to gravity is 9.8 m/ $s^{2}$ which is very higher than the centripetal acceleration on the equator.

B) Normal force exerted by chair will always be equal and opposite to the mass times gravitational acceleration (F = mg). Otherwise, I would be thrown away from chair in case the normal force is not equal and opposite or I would be drag down to the earth due to greater mass times gravitational acceleration. Hence, both are equal and opposite.

C) Of course, this is not a lie, it is true because the acceleration due to gravity is 9.8 m/ $s^{2}$ and as we calculated the acceleration on the equator is 0.03 m/ $s^{2}$ which way too low to experience.

For percentage difference,

9.8 - 0.03 = 9.77

So, % diff = (9.8 - 9.77)/9.8 x 100

% diff = 0.00306 x 100

% diff = 0.306%

Obviously, this is way too low to experience.

D) With the help of same formula as discussed above, we have:

a = 4 $\pi ^{2}$ RcosФ/ $T^{2}$ ,

Here, Ф = 44.4°

Just putting the values. we get

a = 0.0241 m/ $s^{2}$

Acceleration while sitting in my chair.

6 0

2 years ago