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3 years ago

What is the distance from the crest to the equilibrium of a wave called? A. amplitude B. period C. frequency D. phase

Physics

2 answers:

Goryan [66]3 years ago

8 0

its the first one. A.

saw5 [17]3 years ago

7 0

Your answer is A. Amplitude

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The Hubble telescope’s orbit is 5.6 × 105 meters above Earth’s surface. The telescope has a mass of 1.1 × 104 kilograms. Earth e

djverab [1.8K]

The gravitational field is the Force divided by the mass

Call g the gravitational fiel, F the force exerted by the earth and m the mass of the telescope.

g = F / m

g=9.1x10^4 N / 1.1 x 10^4 kg = 8.27 N/kg

Note that the unit N/kg is equivalent to m/s^2

8 0

4 years ago

A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra

blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

$y=\sqrt{1+x^3}$

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Now at (2,3)

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s$

7 0

3 years ago

Help this Is easy for you

DochEvi [55]

Answer:

if its so easy why dont u do it .

Explanation:

6 0

3 years ago

A yet-to-be-built spacecraft starts from Earth moving at constant speed to the yet-tobe-discovered planet Retah, which is 20 lig

Gre4nikov [31]

Answer:

The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame

Explanation:

From the question we are told that

The distance between earth and Retah is $d = 20 \ light \ hours = 20 * 3600 * c = 72000c \ m$

Here c is the peed of light with value $c = 3.0*10^8 m/s$

The time taken to reach Retah from earth is $t = 25 \ hours = 25 * 3600 =90000 \ sec$

The velocity of the spacecraft is mathematically evaluated as

$v_s = \frac{d }{t}$

substituting values

$v_s = \frac{72000 * 3.0*10^{8} }{90000}$

$v_s = 2.40*10^{8} \ m/s$

The time elapsed in the spacecraft’s frame is mathematically evaluated as

$T = t * \sqrt{ 1 - \frac{v^2}{c^2} }$

substituting value

$T = 90000 * \sqrt{ 1 - \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }$

$T = 54000 \ s$

=> $T = 15 \ hours$

So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame

7 0

3 years ago

Refer to science 10 (mirror mirror on the wall)how do the height and width of the object compare with the height and width of th

Irina-Kira [14]

Depends on what type of mirror that is. I am going to assume this is a plain mirror (from the phrase), which means the height and width of the object and image is exactly the same.

6 0

3 years ago