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uranmaximum [27]
3 years ago
9

I don’t get this question and i need help!!!

Physics
1 answer:
galina1969 [7]3 years ago
3 0

i would say number 2

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Write short letters.
Svetradugi [14.3K]

Answer:

No. 67

Peter Street

12th Road

Chennai

24th June 201_

Dear Amrish

I have come to know that since your school has closed for the Autumn Break you have plenty of free time at your disposal at the moment. I would like to tell you that even I am having holidays now.

It has been a long time since we have spent some time together. If you are free, I would welcome to have your company this weekend. Why don’t you come over to my house and spend a day or so with me?

I am anxiously waiting for your reply.

Yours affectionately

your name

4 0
3 years ago
Read 2 more answers
Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat
nikklg [1K]

Answers:

a) T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds V are given by:

V=\omega L=\frac{2\pi}{T} L (1)

Where:

\omega is the angular frequency

L is the radius of the orbit of each satellite

T is the period of the orbit of each satellite

Isolating T:

T=\frac{2 \pi L}{V} (2)

Applying this equation to each satellite:

T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s (3)

T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s (4)

T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s (5)

T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s (6)

T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s (7)

T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s (8)

Ordering this periods from largest to smallest:

T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) Acceleration a is defined as the variation of velocity in time:

a=\frac{V}{T} (9)

Applying this equation to each satellite:

a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2} (10)

a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2} (11)

a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2} (12)

a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2} (13)

a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2} (14)

a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2} (15)

Ordering this acceerations from largest to smallest:

a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

4 0
3 years ago
Which method will correctly determine whether the forces on an object are balanced or unbalanced?
MaRussiya [10]

For this case, the first thing you should do is define a reference system.

Once the system is defined, we must follow the following steps:

1) Do the sum of forces in a horizontal direction

2) Do the sum of forces in vertical direction

The forces will be balanced if for each direction the net force is equal to zero.

The forces will be unbalanced if for each direction the net force is nonzero.

Answer:

Add the forces in the horizontal and vertical directions separately.

6 0
3 years ago
Read 2 more answers
Which energy transfer occurs when a cube of ice is placed in a glass of<br> water?
Daniel [21]
Acetic acid is a weak acid and sodium hydroxide is strong base. Salts of the two will hydrolyse to give basic solution. So, at neutral point, pH of the solution will be greater than 8.
5 0
2 years ago
The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc
Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is 1.163 * 10^{-5} m

Explanation:

From the uncertainty Principle,

Δ_{y} Δ_{p} = \frac{h}{2\pi }

The momentum P_{y} = (mass of electron)(speed of electron)

                                = (9.109 * 10^{-31}kg)(995 * 10^{3} m/s)

                                = 9.0638 * 10^{-25}kgm/s

If the uncertainty is reduced to a 0.0010%, then momentum

                              = 9.068 * 10^{-30}kgm/s

Thus the uncertainty in the position would be:

                              Δ_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }

                              Δ_{y} \geq  1.163 * 10^{-5}m

5 0
3 years ago
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