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svetoff [14.1K]
3 years ago
10

Which constants and directions always apply when using gravity?​

Physics
2 answers:
Lostsunrise [7]3 years ago
6 0

Answer:

.

Explanation:

Romashka-Z-Leto [24]3 years ago
3 0

Answer:

Explanation::

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I need help to figure out how to solve this problem and solve it!!!
wariber [46]

well it looks like the walk at a constant increasing pace then at a constant pace then increaseing pace then constant pace then they slow down then walk at a constant pace then walk at a constantly increasing pace

plz rate me  brainliest

4 0
4 years ago
Read 2 more answers
In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin
alexandr402 [8]

Answer:

a) v=0.999124c

b) E=7.566*10^{22}

c) E_a=760 times\ larger

Explanation:

From the question we are told that

Distance to Betelgeuse d_b=430ly

Mass of Rocket M_r=20000

Total Time in years traveled T_d=36years

Total energy used by the United States in the year 2000 E_{2000}=1.0*10^20

Generally the equation of speed of rocket v mathematically given by

v=\frac{2d}{\triangle t}

v=860ly/ \triangle t

where

\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}

\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}

\triangle t=\sqrt{(860)^2+(36)^2}

\triangle t=860.7532

Therefore

v=\frac{860ly}{ 860.7532}

v=0.999124c

b)

Generally the equation of the energy E required to attain prior speed mathematically given by

E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2

E=7.566*10^{22}

c)Generally the equation of the energy E_a required to accelerate the rocket mathematically given by

E_a=\frac{E}{E_{2000}}

E_a=\frac{7.566*10^{22}}{1.0*10^{20}}

E_a=760 times\ larger

8 0
3 years ago
How do you do this? Plz help answer
Contact [7]

Answer: Really

Explanation:

Just look it up for this page and maybe you will find an anwser sheet.

7 0
3 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.83 times a second. A tack is stuck in the tire a
timama [110]

Answer:

The tangential speed of the tack is 6.988 meters per second.

Explanation:

The tangential speed experimented by the tack (v), measured in meters per second, is equal to the product of the angular speed of the wheel (\omega), measured in radians per second, and the distance of the tack respect to the rotation axis (R), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second:

\omega = 2.83\,\frac{rev}{s} \times \frac{2\pi\,rad}{1\,rev}

\omega \approx 17.781\,\frac{rad}{s}

Then, the tangential speed of the tack is: (\omega \approx 17.781\,\frac{rad}{s}, R = 0.393\,m)

v = \left(17.781\,\frac{rad}{s} \right)\cdot (0.393\,m)

v = 6.988\,\frac{m}{s}

The tangential speed of the tack is 6.988 meters per second.

7 0
3 years ago
If an X-ray beam of wavelength 1.4 × 10-10 m makes an angle of 20° with a set of planes in a crystal causing first order constru
Flura [38]

Answer:

43.16°

Explanation:

λ = Wavelength = 1.4×10⁻¹⁰ m

θ₁ = 20°

n can be any integer

d = distance between the two slits

Since for the first bright fringe, n₁ = 1

n₂ = 2 for second order line

The relation between the distance of the slits and the angle through which it is passed is:

dsinθ=nλ

As d and λ are constant

\frac{n_1\lambda}{sin \theta_1}=\frac{n_2\lambda}{sin \theta_2}\\\Rightarrow \frac{1}{sin20}=\frac{2}{sin\theta_2}\\\Rightarrow sin\theta_2=\frac{2}{\frac{1}{sin20}}\\\Rightarrow \theta_2=sin^{-1}{\frac{2}{\frac{1}{sin20}}}\\\Rightarrow \theta_2=43.16^{\circ}

∴ Angle by which the second order line appear is 43.16°

5 0
3 years ago
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