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DENIUS [597]
3 years ago
5

_ is where there are no particles or very few particles that are spaced out very far apart. (example: outer space)

Physics
1 answer:
irina1246 [14]3 years ago
4 0
The answer is d - Vacuum
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You want to make a photo album for your friend. In three to five sentences, describe at least one advantage and one disadvantage
DanielleElmas [232]

Answer:

if it were to be digital then you would be able to carry it around where ever and look at it to remind you of the memories, however it probably wouldnt last as long as the physical one. eg if you phone breaks there is a chance to restore the photo but you would had to have backed it up beforehand.

4 0
3 years ago
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Time period of a simple pendulum is measured at Karachi. What change will occur in the time period, if it is measured on mount e
Reil [10]
The period of a pendulum is given by T=2 \pi  \sqrt{ \frac{l}{g} }
since Karachi is near sea level, g is larger than it is on Mt. Everest.  That means the pendulum will have a larger period on Mt. Everest than it would in Karachi.
5 0
3 years ago
Which of the following types of technologies has best helped scientists to study very high-energy objects in outer space, such a
erica [24]

Very high-energy objects and events spit out very high-energy photons, so the instrument you need in order to detect them is the       X-ray telescope. <em>(C)  </em>

Inconveniently, X-ray telescopes only work when they're up in orbit, because X-rays get seriously soaked up in Earth's atmosphere, and most of them never make it down to the surface ... (lucky for us !) .

3 0
3 years ago
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An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction  between the object and th
Gnom [1K]

Answer:

8m * (μg/v)^2

Explanation:

k, the spring constant = ?

(k in terms of μ, m, g, and v.)

Frictional force = μmg

Note: lost KE is converted to work done against the friction + PE of the spring

1/2mv2 = μmgx + 1/2kx^2....equation i

Cancel the 1/2 on both sides

mv^2 = μmgx + kx^2

Lets recall that:

Due to frictional effect, further enegy will be lost when the spring recoils backward

Therefore

1/2kx^2 = μmgx..... equation ii

Let's substitute 1/2kx^2 in equation I for ii

So we can say that:

1/2mv^2 = (μmgx)+ μmgx

1/2mv^2 = 2 (μmgx)

1/4mv^2 = μmgx

Cancel out m on both sides

1/4v^2 = μgx

Make x subject of the formula

x = (1/4v^2) / (μg)...... equation iii

substitute x to equation ii

But first make k in equation ii subject of the formula

1/2kx^2 = μmgx

k = 2μmg/x

Now substitute x

k = 2μmg / ((1/4v^2) / (μg))

k = 2μmg * ((μg) / (1/4v^2))

k = 8m * (μg/v)^2

8m * (μg/v)^2

7 0
3 years ago
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
Anarel [89]

Answer:

R = 24.3 m

Explanation:

As we know that the orbital speed is given as

v = \sqrt{\frac{GM}{R + h}}

here we know that

v = 5500 m/s

R = 4.48 \times 10^6 m

h = 630 km

now we have

5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}

M = 2.34 \times 10^24 kg

now acceleration due to gravity of planet is given as

a = \frac{GM}{R^2}

a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}

a = 7.7 m/s^2

now range of the projectile on the surface of planet is given as

R = \frac{v^2 sin2\theta}{g}

R = \frac{14.6^2 sin(2\times 30.8)}{7.7}

R = 24.3 m

3 0
4 years ago
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