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2 years ago

What kind of light would be the best to use to look inside a cold dark cloud and see the warm stars forming inside?

Physics

1 answer:

lys-0071 [83]2 years ago

7 0

<h2>Answer: Infrared light</h2>

A dark nebula is a cloud of dust and cold gas, which does not emit visible light and hides the stars it contains.

These types of nebulae are composed mainly of the hydrogen they obtain from nearby stars, which is their fuel.

It is using infrared light that we can "observe" and analyze in detail what happens in the inner parts of these nebulae.

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If the magnitude of a positive charge is tripled, what is the ratio of the original value of the electric field at a point to th

ipn [44]

Answer:

b)1 :3

Explanation:

Lets that

The value of a positive charge = q

As we know that electric filed on a point charge given as

$E=\dfrac{Kq}{r^2}$

Where ,K=Constant

q=Charge ,r=Distance

If the value of the charge gets tripled ,q'= 3 q

Then electric filed E'

$E'=\dfrac{Kq'}{r^2}$

$E'=\dfrac{3Kq}{r^2}$

E' = 3 E

Therefore we can say that

$\dfrac{E}{E'}==\dfrac{1}{3}$

therefore the answer will be --

b)1 :3

3 0

2 years ago

A satellite is orbiting the earth. If a payload of material is added until it doubles the satellite's mass, the earth's pull of

dimulka [17.4K]

Answer:

True

Explanation:

When a satellite is orbiting the earth, the centripetal force is balanced by the gravitational force as :

$\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}$

$v=\sqrt{\dfrac{GM}{r}}$ ...........(1)

Where

M is the mass of the earth

m is the mass of the planet

From equation (1), the speed of the satellite depends only on the mass of the earth and the orbital radius.

So, If a payload of material is added until it doubles the satellite's mass, the earth's pull of gravity on this satellite will double but the satellite's orbit will not be affected. It is true.

6 0

3 years ago

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A small car with mass of 0.800 kg travels at a constant speed

Alexandra [31]

Answer:

The equation of equilibrium at the top of the vertical circle is:

\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}

The speed experimented by the car is:

\frac{N}{m}+g=\frac{v^{2}}{R}

v = \sqrt{R\cdot (\frac{N}{m}+g) }

v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}

v\approx 9.302\,\frac{m}{s}

The equation of equilibrium at the bottom of the vertical circle is:

\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}

The normal force on the car when it is at the bottom of the track is:

N=m\cdot (\frac{v^{2}}{R}+g )

N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)

N=21.690\,N

7 0

2 years ago

A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.

alisha [4.7K]

A)

It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

$S=S_{o}+v_{o}t+ \frac{at^2}{2} \\ 20= \frac{10t^2}{2} \\ t=2s$

Through Definition of Velocity, comes:

$\Delta v= \frac{\Delta S}{\Delta t} \\ v_x= \frac{36}{2} \\ \boxed {v_{x}=18m/s}$

B)

Using the Velocity Hourly Equation in vertical direction, we have:

$v_{y}=v_{y_{o}}+gt \\ v_{y}=10\times2 \\ \boxed {v_{y}=20m/s}$

The angle of impact is given by:

$cos(\theta) =\frac{v_{x}}{v_{y}} \\ cos(\theta) = \frac{18}{20} \\ cos(\theta) =0.9 \\ arccos(0.9)=\theta \\ \boxed {\theta \approx 25.84}$

If you notice any mistake in my english, please let me know, because i am not native.

7 0

2 years ago

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Using energy considerations, calculate the average force (in N) a 67.0 kg sprinter exerts backward on the track to accelerate fr

Illusion [34]

Answer:

$F_{sprinter}=110.4N$

Explanation:

Given data

Mass m=67.0 kg

Final Speed vf=8.00 m/s

Initial Speed vi=2.00 m/s

Distance d=25.0 m

Force F=30.0 N

From work-energy theorem we know that the work done equals the change in kinetic energy

W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²

And

$W=F_{total}.d$

$W=1/2mv_{f}^2-1/2mv_{i}^2\\F_{total}=\frac{1/2mv_{f}^2-1/2mv_{i}^2}{d} \\F_{total}=\frac{1/2(67.0kg)(8.00m/s)^2-1/2(67.0kg)(2.00m/s)^2}{25.0m} \\F_{total}=80.4N$

and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N

$F_{sprinter}=F_{total}+F_{wind}\\F_{sprinter}=80.4N+30N\\F_{sprinter}=110.4N$

7 0

2 years ago

Read 2 more answers