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sammy [17]
3 years ago
12

HELP PLEASE!Can anyone help with any of these?

Physics
1 answer:
nadezda [96]3 years ago
6 0

1. The magnitude is 49.00 meters and his direction is 7.43 degrees to the right of his original position.

2. D = 7.7 km at 26.2° to ground ANS

3. 172 km

4.  172 km in a direction 34.2 degrees East of North

<em>I hope this helps you!!!</em>

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Do sound waves always travel in straight lines explain
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They travel like waves. Just throw rock at lake you will see wave. When it bumps to barrier barrier reflects some part of it . Not like a line lika a wave
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2 years ago
How fast would the International Space Station (ISS) have to travel to maintain a circular orbit a distance of 1400 km above the
professor190 [17]

Answer:

The International Space Station move at 7.22 km/s.

Explanation:

Orbital speed of satellite is given by  v=\sqrt{\frac{GM}{r}}, where G is gravitational constant, M is mass of Earth and r is the distance to satellite from centre of Earth.

r = R + h = 6350 + 1400 = 7750 km = 7.75 x 10⁶ m

G = 6.673 x 10⁻¹¹ Nm²/kg²

M = 5.98 x 10²⁴ kg

Substituting

              v=\sqrt{\frac{6.673\times 10^{-11}\times 5.98\times 10^{24}}{7.75\times 10^6}}=7223.86m/s=7.22km/s

  The International Space Station move at 7.22 km/s.      

4 0
2 years ago
Help, please. I am not sure what to do.
miss Akunina [59]

Answer:

option D) -3m

Explanation:

if 6m is diplaced by -3m then it would be -3+6=3m

feel free to ask if you are confused

3 0
2 years ago
(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

The charge chould be moved to infinity

7 0
3 years ago
Longitudinal waves transfer energy ___________ to the direction of the wave motion.
viva [34]
Longitudinal waves transfer energy parallel to the direction of the wave motion
4 0
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Read 2 more answers
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