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1 year ago

The ability of a telescope to separate two closely spaced stars is called _____.

Physics

1 answer:

larisa [96]1 year ago

3 0

The ability of a telescope to separate two closely spaced stars is called (angular) resolution.

What is a telescope ?

Astronomers use a telescope to observe distant things. Curved mirrors are used by the majority of telescopes, including all large telescopes, to collect and concentrate light from the night sky. The original telescopes employed lenses, which are simply curved pieces of clear glass, to focus light.

The capacity of a telescope to distinguish two point sources into distinct pictures is known as resolution. The resolving power is constrained by diffraction effects in ideal circumstances, such as above the atmosphere where there is no turbulence (seeing).

to learn more about telescope click on the link below:

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Very large accelerations can injure the body, especially if they last for a considerable length of time. The severity index (SI)

Ludmilka [50]

Answer:

a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.

Explanation:

a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:

$a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$a' = 3.569$

The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:

$v_{f} = v_{o} + a \cdot t$

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v_{f}$ - Final speed, measured in meter per second.

$a$ - Acceleration, measured in $\frac{m}{s^{2}}$ .

$t$ - Time, measured in seconds.

$t = \frac{v_{f}-v_{o}}{a}$

$t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)}{35\,\frac{m}{s^{2}} }$

$t = 0.095\,s$

Lastly, the severity index is now determined:

$SI = \frac{a'^{5}}{2\cdot t}$

$SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}$

$SI = 3047.749$

b) The initial and final speed of the injured are $1.944\,\frac{m}{s}$ and $5.278\,\frac{m}{s}$ , respectively. The travelled distance can be determined from this equation of motion:

$v_{f}^{2} = v_{o}^{2} + 2\cdot a \cdot \Delta s$

Where $\Delta s$ is the travelled distance, measured in meters.

$\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}$

$\Delta s = \frac{\left(5.278\,\frac{m}{s} \right)^{2}-\left(1.944\,\frac{m}{s} \right)^{2}}{2\cdot \left(35\,\frac{m}{s^{2}} \right)}$

$\Delta s = 0.345\,m$ .

8 0

3 years ago

Jack has two boxes: one is 148g and one is 78g. if jack pushes both boxes with the same amount of force which will accelerate fa

kifflom [539]

1) A The 78g
2) C Push on the wagon in the opposite direction as Jack with a force that is the same as Jack is applying.

5 0

3 years ago

A 15g bullet is fired horizontally into a 3kg block of wood suspended by a long cord. Assume that the bullet remains in the bloc

Varvara68 [4.7K]

Answer:

261.3 m/s

Explanation:

Mass of bullet=m=15 g= $\frac{15}{1000}=0.015 kg$

1 kg=1000g

Mass of block=M=3 kg

d=0.086 m

Total mass =M+m=3+0.015=3.015 kg

K.E at the time strike=Gravitational potential energy at the end of swing

$\frac{1}{2}(m+M)^2V^2=(m+M)gh$

Using g= $9.8m/s^2$

Substitute the values

$\frac{1}{2}(3.015)V^2=3.015\times 9.8\times 0.086$

$V^2=\frac{2\times 3.015\times 9.8\times 0.086}{3.015}$

$V=\sqrt{2\times 3.015\times 9.8\times 0.086}}$

$V=1.3m/s$

Velocity after collision=V=1.3 m/s

Velocity of block=v'=0

Using conservation law of momentum

$mv+Mv'=(m+M)V$

Using the formula

$0.015v+3(0)=3.015(1.3)$

$0.015v=3.015(1.3)$

$v^2=\frac{3.015(1.3)}{0.015}=261.3$

$v=261.3 m/s$

5 0

3 years ago

WILL GIVE BRAINLIEST

NemiM [27]

Yes! you are :) bc you are FORCING the page to turn, and the broom ti sweep

6 0

3 years ago

Read 2 more answers

A pot on the stove contains 200 g of water at 20°C. An unknown mass of ice that is originally at −10°C is placed in an identical

Mumz [18]

Answer:

a) The mass of the ice is smaller than the mass of the water

b) The ice reaches first 80°C ,

Explanation:

Since the heat Q that should be provided to ice

Q = sensible heat to equilibrium temperature (as ice) + latent heat + sensible heat until final temperature ( as water)

m ice * c ice * ( T equil -T initial ) + m ice* L + m ice* c water * ( T final - T equil)

and the heat Q that should be provided to water is

Q= m water * c water * ( T final - T equil )

since the rate of heat addition q = constant and the time t taken to reach the final temperature is the same , then the heat absorbed Q=q*t is the same for both, therefore

m water * c water * ( T final - T equil ) = m ice* [c ice *( T equil -T initial ) + L + c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial ) + L + c water * ( T final - T equil)]/ [ c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial ) + L ]/[c water * ( T final - T equil) ] + 1

since [c ice * ( T equil -T initial ) + L ]/[c water * ( T final - T equil) ] >0 , then

m water/ m ice > 1

m water > m ice

so the mass of ice is smaller that the mass of water

b) Since the heat Q that should be provided to the ice, starting from 55°C mass would be

Q ice= m ice * c water * ( T final2 - T final1 )

and for the water mass

Q water = m water * c water * ( T final2 - T final1 )

dividing both equations

Q water / Q ice = m water / m ice >1

thus

Q water > Q ice

since the heat addition rate is constant

Q water = q* t water and Q ice=q* t ice

therefore

q* t water > q* t ice

t water > t ice

so the time that takes to reach 80°C is higher for water , thus the ice mass reaches it first.

5 0

3 years ago