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vladimir2022 [97]
2 years ago
11

Suppose we have a space ship orbiting the earth. What must happen in order for the spaceship to leave orbit and fall back toward

earth?
The spaceship will fall back toward earth on its own but you have to just wait for it to fall.

The spaceship's forward motion must be slowed down so the earth's gravitational pull on it will be stronger than the ship's forward motion.

The spaceship's forward motion must be sped up so the earth's gravitational pull on it will be weaker than the ship's forward motion.

It is not possible for a spaceship orbiting the earth to leave orbit and fall back to earth once it is in orbit.
Physics
2 answers:
professor190 [17]2 years ago
8 0
It’s option 2
Hope it will help you:)
sdas [7]2 years ago
3 0

Answer:

#2) The spaceship's forward motion must be slowed down so the earth's gravitational pull on it will be stronger than the ship's forward motion.

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Let the resistance of an electrical component remain constant while the potential difference across the two ends of the componen
slamgirl [31]

Answer:

The current through it will also decrease to half of its former value because according to Ohm's law the current flowing through a resistor is directly proportional to the potential difference applied across its ends provided that the temperature and some other necessary conditions remain constant.

This is mathematically represented as follows;

V=IR.........(1)

The current is thus given as

I=\frac{V}{R}..............(2)

if R is constant and V is reduced to half, then we have the following;

I=\frac{V/2}{R}

Simplifying further we obtain

I=\frac{V}{2R}...........(3)

Equation (3) shows that the current I is also reduced to half.

8 0
2 years ago
Read 2 more answers
A 120-V rms voltage at 60.0 Hz is applied across an inductor, a capacitor, and a resistor in series. If the rms value of the cur
klemol [59]

Answer:

The impedance of this circuit is 200 ohm.

Explanation:

Given that,

rms voltage = 120 v

Frequency = 60.0 Hz

rms current = 0.600 A

We need to calculate the impedance

Using formula of impedance

Z=\dfrac{V_{rms}}{I_{rms}}

Where, V_{rms} = rms voltage

I_{rms} = rms current

Z= impedance

Put the value into the formula

Z=\dfrac{120}{0.600}

Z=200\ Omega

Hence, The impedance of this circuit is 200 ohm.

8 0
3 years ago
Joe is measuring the time it takes for a ball to roll down a ramp. In this experiment Joe takes the measurement 5 times and gets
PolarNik [594]

 Step by step solution :

standard deviation is given by :

\sigma = \sqrt\dfrac{{\sum (x-\bar{x})^2}}{n}

where, \sigma is standard deviation

\bar{x} is mean of given data

n is number of observations

From the above data, \bar{x}=24.88

Now, if x=24.8, then (x-\bar{x})^2=0.0064

If  x=23.9, then (x-\bar{x})^2=0.9604

if x=26.1, then (x-\bar{x})^2=1.4884

If x=25.1, then (x-\bar{x})^2=0.0484

If x=24.5, then (x-\bar{x})^2=0.1444

so, \sum (x-\bar{x})^2 =\frac{0.0064+0.9604+1.4884+0.0484+0.1444}{5}

\sum (x-\bar{x})^2 =2.648

\sqrt{\sum \frac{(x-\bar{x})^2}{n}}

\sigma =0.7277

No, Joe's value does not agree with the accepted value of 25.9 seconds. This shows a lots of errors.

6 0
2 years ago
You have a 35X objective lens in place, and the numerical aperture of the objective lens is 0.75. The numerical aperture of the
olasank [31]

Answer:

350x

Explanation:

In a microscope the objective has higher magnification than the eyepiece so, this is a microscope

The magnification of a microscope is given by the product of the magnifications of the eyepiece and and the objective.

Objective lens magnification = 35x =m_o

Eyepiece magnification = 10x =m_e

Total magnification

M=m_o\times m_e\\\Rightarrow M=35\times 10\\\Rightarrow M=350

Total magnification is 350x

8 0
3 years ago
As the wavelength increases, the frequency (2 points) decreases and energy decreases. increases and energy increases. decreases
frozen [14]
Bohr's equation for the change in energy is
\Delta E= \frac{hc}{\lambda}
where
h = Planck's constant
c == the velocity of light
λ = wavelength.

The velocity is related to wavelength and frequency, f, by
c = fλ

Let us examine the given answers on the basis of the given equations.

a. As λ increases, f decreases and ΔE decreases.
     TRUE

b. As λ increases, f increases and ΔE increases.
    FALSE

c. As λ increases, f increases and ΔE decreases.
    FALSE

Answer: 
As the wavelength increases, the frequency decreases and energy decreases.

3 0
3 years ago
Read 2 more answers
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