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Tju [1.3M]
3 years ago
6

Find the acceleration of a train whose speed increases from 9 m/s to 22 m/s in 160s

Physics
1 answer:
Rainbow [258]3 years ago
6 0
Acceleration = (v2-v1) / (t2-t1) = (22-9) / 160 = 13/160 m/s
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The Hubble Space Telescope has a mass of 1.16*10^ 4 kg and orbits the Earth at an altitude of 5.68 * 10 ^ 5 above Earth's surfac
andrezito [222]

Answer:

E=8.13\times 10^{12}\ J

Explanation:

Given that,

The mass of a Hubble Space Telescope, m_1=1.16\times 10^4\ kg

It orbits the Earth at an altitude of 5.68\times 10^5\ m

We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

E=\dfrac{Gm_1m_e}{r}

Where

m_e is the mass of Earth

Put all the values,

E=\dfrac{6.67\times 10^{-11}\times 1.16\times 10^4\times 5.97\times 10^{24}}{5.68\times 10^5}\\\\E=8.13\times 10^{12}\ J

So, the potential energy of the telescope is 8.13\times 10^{12}\ J.

5 0
3 years ago
A person carries a mass of 10 kg and walks along the +x-axis for a distance of 100m with a constant velocity of 2 m/s. What is t
gizmo_the_mogwai [7]
Since the direction of the force and the direction of the path is perpendicular, the person is not doing any physical work.
3 0
3 years ago
Read 2 more answers
Convert 8 light years to Astronomical Units
marusya05 [52]

Answer:

505929 AU

Explanation:

As you may know, one light-year is equivalent to approximately 63241.1 Astronomical Units. To get your answer, simply multiply 63241.1 * 8 to get ≈505929 AU

5 0
3 years ago
Which of the following are density labels? <br> a. Kg/L <br> b. g/m <br> c. g/mL <br> d. cm/g
bearhunter [10]
Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. It is a property that can be used to describe a substance.<span> </span><span>It has standard units of kg/m^3 or g/mL. So, the best answer is option C.</span>

4 0
3 years ago
Read 2 more answers
Determine the vector sum of the displacements Δd1 = 2.4 m [32° S of W]; Δd2 = 1.6 m [S]; and Δd3 = 4.9 m [27° S of E].
sergeinik [125]

Answer:

3.44 metres

Explanation:

To determine the vector sum of the displacements Δd1 = 2.4 m [32° S of W]; Δd2 = 1.6 m [S]; and Δd3 = 4.9 m [27° S of E], resolve the given parameters into x - component and y - component.

Resolving into x - component

- 2.4cos32 + 4.9cos27 = 2.3306

Resolving into y - component

- 2.4sin32 - 4.9sin27 - 1.6 = - 2.553

The vector sum of the displacement will be

Sqrt( 2.3^2 + 2.6^2) =

Sqrt ( 11.81)

3.44 m

Therefore, the vector sum of the displacements is 3.44 metres

6 0
3 years ago
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