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OverLord2011 [107]
3 years ago
11

Please help me gqvebqubgk yfawcyvgkbuh

Physics
2 answers:
Whitepunk [10]3 years ago
5 0

Answer:

d

Explanation:

because it is talking of the solid rock in the core of the earth

Sergeu [11.5K]3 years ago
4 0

Answer:CCCCCCCCCCCCCCCCC

Explanation:CCCCCCCCCCCCC

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A planet is discovered orbiting the star 51 Peg with a period of four days (0.01 years). 51 Peg has the same mass as the Sun. Me
artcher [175]

Answer:

Less than Mercury's

Explanation:

According to third Kepler's law, the square of the planet's orbital period is proportional to the cube of the average orbital radius of the planet's orbit. The constant of proportionality depends only on the mass of the star, recall that 51 Peg has the same mass as the Sun. Since the orbital period of this planet is less than Mercury's, its average orbital radius is less than Mercury's.

4 0
3 years ago
At what distance of separation, r, must two 3.20 x 10-9 Coulomb charges be positioned in order for the repulsive force between t
STatiana [176]
I'm not 100% sure but I think the answer is 60.4, because you multiply 3.20 by 10, to get 32 - 9 = 23, then subtract 23 from 83.4. Hope this helps you, and good luck!!!
4 0
3 years ago
Read 2 more answers
A car has uniformly accelerated from rest to a speed of 25m/s after traveling 75m. What is its acceleration in m/s^2
Roman55 [17]
<h2><em>So there is two truths given. After an amount of time Ttotal (lets call it ‘t’): </em></h2><h2><em> </em></h2><h2><em>The car’s speed is 25m/s </em></h2><h2><em>The distance travelled is 75m </em></h2><h2><em>Then we have the formulas for speed and distance: </em></h2><h2><em> </em></h2><h2><em>v = a x t -> 25 = a x t </em></h2><h2><em>s = 0.5 x a x t^2 -> 75 = 0.5 x a x t^2 </em></h2><h2><em>Now, we know that both acceleration and time equal for both truths. So we can say: </em></h2><h2><em> </em></h2><h2><em>t = 25 / a </em></h2><h2><em>t^2 = 75 / (0.5 x a) = 150 / a </em></h2><h2><em>Since we don’t want to use square root at 2) we go squared for 1): </em></h2><h2><em> </em></h2><h2><em>t^2 = (25 / a) ^2 = 625 / a^2 </em></h2><h2><em>t^2 = 150 / a </em></h2><h2><em>Since t has the same value for both truths we can say: </em></h2><h2><em> </em></h2><h2><em>625 / a^2 = 150 / a </em></h2><h2><em> </em></h2><h2><em>Thus multiply both sides with a^2: </em></h2><h2><em> </em></h2><h2><em>625 = 150 x a, so a = 625 / 150 = 4.17 </em></h2><h2><em> </em></h2><h2><em>We can now calculate t as well t = 25 * 150 / 625 = 6</em></h2>
4 0
3 years ago
What is the area of the bottom of a tank 30.0 cm long and 15.0 cm wide?​
Igoryamba

Answer:

perimeter of the bottom of the tank is 450 cm

Explanation:

If you want to get area you need the Length, width,and height.

5 0
2 years ago
Read 2 more answers
A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio
Nikitich [7]

Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft  × 4  sorings

K = 100 lb/ft

Amplitude of the microscope  \frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon  \frac{\omega}{W_n})^2}]

where;

\epsilon = 0

W_n = \sqrt { \frac{k}{m}}

= \sqrt { \frac{100*32.2}{200}}

= 4.0124

replacing them into the above equation and making X the subject of the formula:

X = Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}

X = 0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}

X = 5.676*10^{-3} \ mm

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

8 0
3 years ago
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