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3 years ago

Why does the lens need to be thicker for viewing nearby objects?

1 answer:

5 0

Answer: To focus on a near object – the lens becomes thicker, this allows the light rays to refract (bend) more strongly. To focus on a distant object – the lens is pulled thin, this allows the light rays to refract slightly.

Explanation:

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At the distance of Jupiter (6 times further away from the Sun than Earth) the amount of sunlight received per square centimeter

givi [52]

The amount of sunlight received per square centimeter is different by 36 times less.

<h3>What is the speed of light?</h3>

A fundamental physical constant, typically abbreviated as c, the speed of light in a vacuum is significant in many branches of physics. 299792458 metres per second, or around 300000 kilometres per hour or 186,000 miles per hour, is the precise figure. The special theory of relativity states that the maximum speed at which ordinary matter, energy, or any signal containing information may move across space is given by the constant c.

Visible light is an example of electromagnetic radiation that moves at the speed of light. Light and other electromagnetic waves appear to move instantly for many practical uses, but their limited speed has dramatic implications over vast distances and particularly sensitive measurements. The starlight that can be seen from Earth has been out there for a very long time, allowing people to see far-off objects and learn about the evolution of the cosmos. It can take minutes to hours for signals to reach from Earth to a spacecraft when corresponding with far-off space missions. The absolute shortest communication latency between computers, to computer memory, and within a CPU is fixed in computing at the speed of light.

to learn more about speed of light go to - brainly.com/question/104425

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8 0

1 year ago

Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are

slavikrds [6]

Answer:

The correct answer is

p = p₁ + p₂

Explanation:

Newton's second law states that force = the change of momentum produced therefore since the collision is inelstic then the change of momentum of each car is p₁ and p₂ and the force of the collition is proportional to p₁ + p₂ that is

F ∝ p₁ + p₂ and since force is directly proportional to p we have

p = p₁ + p₂

7 0

4 years ago

Heart cells must contract simultaneously to move blood.

krek1111 [17]

Answer:

Explanation:

The heart cells must contract simultaneously to move blood.

This means that it needs to act fast and efficiently.

Therefore, the connections among heart cells are characterized by :

having many branches
having many communicating junctions

The correct option should be <u>A</u>

4 0

3 years ago

Read 2 more answers

A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and

cricket20 [7]

Answer:

The frequency $f$ = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = $\frac{d}{2}$ = $\frac{44}{2}$ = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = $\frac{d}{2}$ = $\frac{3.0}{2}$ = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = $\frac{d}{2}$ = $\frac{2.0}{2}$ = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

$P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2$

$pgy_1=\frac{1}{2}pv^2_2 +pgy_2$

$v_2=\sqrt{2g(y_1-y_2)}$

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = $\sqrt{2*9.8*(0.35-0)}$

v₂ = $\sqrt {6.86}$

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

$v_3 = \frac{v_2r_2^2}{v_3^2}$

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = $(2.62 m/s)*$ $(\frac{1.5mm^2}{1.0mm^2} )$

v₃ = 0.0589 m/s

∴ velocity of the fluid in the cylinder = 0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

$f=\frac{v_s}{4(h-v_3t)}$

where;

$v_s$ = velocity of sound

h = height of the fluid

v₃ = velocity of the fluid in the cylinder

$f=\frac{343}{4(0.40-(0.0589)(0.4)}$

$f= \frac{343}{0.6576}$

$f$ = 521.59 Hz

∴ The frequency $f$ = 521.59 Hz

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

$\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})$

$\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)$

$\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}$

where;

$v_s$ (velocity of sound) = 343 m/s

v₃ (velocity of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

$\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}$

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s when the cylinder has been filling for 4.0 s.

8 0

3 years ago

3 examples when friction is helpful?

Svetllana [295]

Some examples of when friction is helpful are: to help the movement of tires. When you walk, and also, when you erase. :)

4 0

3 years ago