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dolphi86 [110]
1 year ago
10

Lexy used the formula shown to calculate the force of gravity on a space shuttle. Fg = G What does 3 × 105 kg represent? the dif

ference between Earth’s mass and the space shuttle’s mass the sum of Earth’s mass and the space shuttle’s mass the mass of Earth the mass of the space shuttle
Physics
1 answer:
STALIN [3.7K]1 year ago
5 0
<h2>Answer:</h2>

<u>This term shows the </u><u>mass of the space shuttle</u>

<h2>Explanation:</h2>

We know that the mass of the Earth is 5.972 × 10^24 kg. Similarly the sum of  mass of earth and the mass of shuttle must be a greater number as compared to the number given. It simply means that the mass of earth is itself 5.972 × 10^24 kg and the value given is 3 × 105 kg so it is obvious that if was the sum then it must be greater than the mass of earth. Therefore we can say that this not the mass of earth, neither the sum of mass of earth and shuttle, but this is only the mass of space shuttle which is the last multiple choice.

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A ping-pong ball weighs 0.025 N. The ball is placed inside a cup that sits on top of a vertical spring. If the spring is compres
kondor19780726 [428]

Answer:

Explanation:

The energy stored in the spring is used to throw the ball upwards . Let the height reached be h

stored energy of spring = 1/2 k y² , k is spring constant and y is compression created in the spring

stored energy of spring = potential energy of the ball

1/2 k y² = mgh , m is mass of the ball , h is height attained by ball

.5 k x .055² = .025  x 2.84

.0015125 k = .071

k = .071 / .0015125

= 46.9 N / m .

4 0
1 year ago
A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po
Umnica [9.8K]

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + 0.5mV1^{2} = mgh2 + 0.5mV2^{2}

where

  • speed at highest point = V2
  • speed at lowest point = V1
  • mass of the girl and swing = m
  • at the highest point, the  speed is minimum (V1 = 0)
  • at the lowest point the speed is maximum (V2 is the maximum speed)
  • therefore the equation becomes mgh1 + 0.5mV1^{2} = mgh2

      m(gh1 + 0.5V1^{2}) = m(gh2)

      gh1 + 0.5V1^{2} = gh2

      V1 = \sqrt{\frac{gh2 - gh1}{0.5}}

now we can substitute all required values into the equation above.

V1 = \sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}

V1 = \sqrt{\frac{32.34}{0.5}}

V1 =8.1 m/s

8 0
2 years ago
If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed?
sweet [91]

Answer:

Speed of the electron will be v=5.896\times 10^7m/sec

Explanation:

We have given that charge on electron e=1.6\times 10^{-19}C

Mass of electron m=9.11\times 10^{-31}kg

Potential difference = V=9.9KV=9.9\times 10^3volt

Now according to energy conservation eV=\frac{1}{2}mv^2

1.6\times 10^{-19}\times 9.9\times 10^3=\frac{1}{2}\times 9.11\times 10^{-31}v^2

v=5.896\times 10^7m/sec

4 0
1 year ago
Two speakers both emit sound of frequency 320 Hz, and are in phase. A receiver sits 2.3 m from one speaker, and 2.9 m from the o
satela [25.4K]

Answer:

Option B

Explanation:

The phase difference is found by subtracting the 2.3m for the receiver from the other speaker which is 2.9m hence

Phase difference= 2.9-2.3= 0.6

3 0
1 year ago
A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington
Roman55 [17]

Let \theta be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)

The current has velocity vector (relative to the Earth)

\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath

The swimmer's resultant velocity (her velocity relative to the Earth) is then

\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}

\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ

which is approximately 41º west of north.

6 0
1 year ago
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