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3 years ago

Ow does the light-collecting area of an 8-meter telescope compare to that of a 2-meter telescope?

1 answer:

fomenos3 years ago

6 0

<span>The area will be the square of the quotient of the amounts for each telescope. Comparing the two microscopes, the 8-meter model is 4 times larger than the 2-meter model. Squaring this value gives a ratio of 16 times the amount of visual area for the larger telescope.</span>

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What influences the path that surface currents take?

Anastasy [175]

The answer is wind forces and Earth’s rotation

8 0

2 years ago

Read 2 more answers

An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 m/s, what is the frequency of the

bazaltina [42]

Answer:

the frequency of the second harmonic of the pipe is 425 Hz

Explanation:

Given;

length of the open pipe, L = 0.8 m

velocity of sound, v = 340 m/s

The wavelength of the second harmonic is calculated as follows;

L = A ---> N + N--->N + N--->A

where;

L is the length of the pipe in the second harmonic

A represents antinode of the wave

N represents the node of the wave

$L = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{4} \\\\L = \lambda$

The frequency is calculated as follows;

$F_1 = \frac{V}{\lambda} = \frac{340}{0.8} = 425 \ Hz$

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.

5 0

2 years ago

Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.

mr_godi [17]

Answer:

The magintude of the acceleration for both objects is $3.11m/s^2$

Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

$T_A-m_A*g=m_A*a$

and for mB (descending):

$T_B-m_B*g=-m_B*a$

$T_A=T_B$

because the two boxes has the same acceleration because they are attached together:

$m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2$

So the magintude of the acceleration for both objects is $3.11m/s^2$

5 0

2 years ago

PLS HELP ME OUT!!

MaRussiya [10]

In order to solve this problem, we must first find out the value of each line on the number line. However, we can make this problem more simple by ignoring every interval except for the ones between 0 and 6. There are three total intervals in between 0 and 6 (including 6 and excluding 0). Therefore, we can do 6/2, and get an interval value of 2. This means that each line adds a value of 2. Since the car is only one line past zero, we only have to add one value of 2. Since 0 + 2 = 2, our final answer is C. 2.

Hope this helps!

4 0

3 years ago

Read 2 more answers

A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?

Andru [333]

Answer:

F = 37.8 × 10^(6) N

Explanation:

The charges are 0.06 C and 0.07 C.

Thus;

Charge 1; q1 = 0.06 C

Charge 2; q2 = 0.07 C

Distance between them; r = 3 m

Formula for the force in between them is;

F = kq1•q2/r²

Where k is a constant = 9 × 10^(9) N.m²/C²

Thus;

F = (9 × 10^(9) × 0.06 × 0.07)/3²

F = 37.8 × 10^(6) N

3 0

2 years ago