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Virty [35]
3 years ago
10

Can someone help pls !

Physics
1 answer:
Gemiola [76]3 years ago
6 0
A sort of electricity is a light bulb or a phone / computer charger. plants food water. the sun and rain . that’s what i’m guessing!
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Air is a good medium for sound waves because it is
Paul [167]

Answer:

air does not have a modulus of rigidity.

Explanation:

Since air is completely elastic medium, that is, it does not have a modulus of rigidity, therefore sound waves in air are longitudinal.

4 0
2 years ago
a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the
ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: F=m*a with this we can get both accelerations; solving for acceleration a=\frac{F}{m}. Now a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2}) anda_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2}). Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, x=\frac{1}{2}*a_{girl}*t^{2} and15.0-x=\frac{1}{2}*a_{sled}*t^{2}, solving for the time we get:t=\sqrt{\frac{2x}{a_{girl} } } and t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } } with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get:\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }. Finally we get:\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} } and replacing the values we have got:\frac{x}{0.14} =\frac{(15.0-x)}{0.73} so 5.33*x=15-x so x=2.37 (meters).

5 0
3 years ago
Assume that a satellite orbits mars 150km above its surface. Given that the mass of mars is 6.485 X 10^23kg, and the radius of m
Kisachek [45]
<span>3598 seconds The orbital period of a satellite is u=GM p = sqrt((4*pi/u)*a^3) Where p = period u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits. a = semi-major axis of orbit. Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2 The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So 150000 m + 3.396x10^6 m = 3.546x10^6 m Substitute the known values into the equation for the period. So p = sqrt((4 * pi / u) * a^3) p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3) p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3) p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3) p = sqrt(1.2945785x10^7 s^2) p = 3598.025212 s Rounding to 4 significant figures, gives us 3598 seconds.</span>
8 0
3 years ago
Two particles A and B start simultaneously from a Point P with velocities 20 m/s and 30 m/s respectively. A and B move with acce
zysi [14]

Answer:

<u>20 m/s</u>

Explanation:

<u>Given</u>

  • u(A) = 20 m/s
  • u(B) = 30 m/s
  • acceleration equal in magnitude but opposite in direction

<u>Solving</u>

  • Velocity of A at Q = 30 m/s
  • From, P to Q, <u>Δv(A) = 30 - 20 = +10 m/s</u>
  • Therefore, velocity of B at Q will be decreased by 10 as it is equal in magnitude but opposite in direction to A
  • Δv(B) = v(B at Q) - u(B at P)
  • -10 m/s = v(B at Q) - 30 m/s
  • v(B at Q) = 30 - 10 = <u>20 m/s</u>
6 0
2 years ago
By what percent must one increase the tension in a guitar string to change the speed of waves on the string from 301 m/s to 343
garri49 [273]

Answer:

29.8 %

Explanation:

7 0
3 years ago
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