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maksim [4K]
2 years ago
5

Suppose that the angular separation of two stars is 0.1 arcseconds, and you photograph them with a telescope that has an angular

resolution of 1 arcsecond. How will the stars appear in the photograph?
Physics
1 answer:
Crank2 years ago
4 0

Answer: the photograph will likely show only one star.

Explanation:

Since their angular separation is smaller than the telescope's angular resolution, the picture will apparently show only one star rather than two.

You might be interested in
Zoning laws establish _______.
Aleonysh [2.5K]

Answer:

Zoning ordinances detail whether specific geographic zones are acceptable for residential or commercial purposes. Zoning ordinances may also regulate: - size

- placement

- density

- height of structures

HOPE IT HELPS :)

PLEASE MARK IT THE BRAINLIEST!

5 0
3 years ago
A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm
Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g

The mass of the meter stick is 126.99115 g

6 0
2 years ago
Read 2 more answers
A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (F
Marrrta [24]

Answer:

Part a)

x = 0.4 m

Part b)

v_i = 11.7 m/s

Part c)

Speed is more than the required speed so it will reach the top

Explanation:

Part a)

As we know that there is no frictional force while block is moving on horizontal plane

so we can use energy conservation on the block

\frac{1}{2}mv^2 = \frac{1}{2}kx^2

\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2

x = 0.4 m

Part b)

If the track has average frictional force of 7 N then work done by friction while block slides up is given as

W_f = -7( \pi R)

W_f = -7(\pi \times 1.00)

W_f = -22 J

work done against gravity is given as

W_g = - mg(2R)

W_g = -(0.500)(9.8)(2\times 1)

W_g = -9.8 J

Now by work energy equation we have

\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2

\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2

v_f = 4.1 m/s

Part c)

now minimum speed required at the top is such that the normal force must be zero

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = 3.13 m/s

so here we got speed more than the required speed so it will reach the top

5 0
3 years ago
A. What are the three longest wavelengths for standing waves on a 240-cm-long string that is fixed at both ends?
vovangra [49]

Answer:

a) the three longest wavelengths = 4.8m, 2.4m, 1.6m

b) what is the frequency of the third-longest wavelength = 75Hz

Explanation:

The steps and appropriate formula and substitution is as shown in the attached file.

5 0
3 years ago
A ball of moist clay falls 15.0 m to the ground. It is in contact with the ground for 20.0 ms before stopping. (a) What is the m
rosijanka [135]

Answer:

a= -0.86 m/s²

The negative sign shows that ball down the ground or moving down

Explanation:

Vf² - Vo² = 2gS

where

Vf = velocity of clay as it hits the ground

Vo = initial velocity of clay = 0

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

S = distance travelled by clay = 15 m

Substituting appropriate values,

Vf² - 0 = 2(9.8)(15)  

Vf = 17.15 m/sec.

Formula to use is,  

V - Vf = aT

where

V = velocity of clay when it stops = 0

Vf = 17.15 m/sec (as determined above)

a = acceleration

T = 20 ms  

Put the values to find acceleration

a=(V-Vf)/T

a=(0-17.15)/20

a= -0.86 m/s²

The negative sign shows that ball down the ground

3 0
3 years ago
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