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padilas [110]
3 years ago
10

A 6 cm long spring extends to 9 cm when a 1 kg load is suspended from it. What would be its length if a 2 kg load were suspended

from it?
a. 12 cm
b. 15 cm
c. 18 cm
d. 21 cm
Physics
1 answer:
Ganezh [65]3 years ago
8 0
The answer is D that is the right question
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Svetlanka [38]

Answer:

Friction of the road on the motorcycle in the opposite direction

Explanation:

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On which factor potential energy depends?​
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The potential energy of the object depends on

  • the height of the object with respect to some reference points,
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When a guitar string plays the note "a," the string vibrates at 440 hz ?
Rama09 [41]
Yes, that's correct. The note "A" (which is used to tune the other strings of the guitar) corresponds to a frequency of 440 Hz.
8 0
3 years ago
An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi
Arturiano [62]

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

                  V_x'=0.8c\\V=0.6c\\m=4*10^5kg

  • Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

                  KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}

  • So, to V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
  • We have an expression from Lorents transformation for relativistic law of addition of velocities as,

                      V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V

  • Substituting values, we get,

          V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c

  • Thus, the KE will be,

              KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

3 0
2 years ago
A projectile has an initial horizontal velocity of 34.0 M/s at the edge of a roof top. Find the horizontal and vertical componen
Sveta_85 [38]

Answer:

v_x=34 m/s

v_y=53.9\ m/s

Explanation:

<u>Horizontal Launch</u>

When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

vx=v

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

v_y=g.t

The horizontal component of the velocity is always the same:

v_x=34 m/s

The vertical component at t=5.5 s is:

v_y=9.8*5.5=53.9

v_y=53.9\ m/s

8 0
3 years ago
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