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gizmo_the_mogwai [7]
3 years ago
5

The peak luminosity of a white dwarf supernova is around 1010 Lsun, and it remains brighter than 108 Lsun for about 150 days. In

comparison, the luminosity of a bright Cepheid variable star is about 10,000 Lsun. The Hubble Space Telescope is sensitive enough to make accurate measurements of apparent brightness for Cepheid variable stars up to a distance of about 100 million light-years. Estimate the distance of a fading white dwarf supernova of luminosity 108 Lsun whose apparent brightness is comparable to that of a bright Cepheid variable star 100 million light-years from Earth. How does the distance of that supernova compare to the size of the observable universe?

Physics
1 answer:
Airida [17]3 years ago
8 0

Answer:

Explanation: find the attached solution below

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Manuel is holding a 5kg box. How much force is the box exerting on him?In what direction
soldi70 [24.7K]
The force the box is exerting on Manuel is the weight of the box, downward:
W=mg=(5 kg)(9.81 m/s^2)=49.05 N
and this force is perfectly balanced by the constraint reaction applied by Manuel's hand, pushing upward.
3 0
3 years ago
The illustration shows a rollercoaster and indicates four different positions the car might be at as it moves along the track. A
Vanyuwa [196]

Answer:

Point a

Explanation:

The potential energy of an object is given by :

P = mgh

m is mass, g is acceleration due to gravity, h is height above ground level.

Potential energy is directly proportional to the position of an object.

In the attached figure, the maximum height is shown at point (a). It means it will have maximum potential energy at a as compared to b,c and d.

5 0
3 years ago
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
While Newton's Second Law often deals with formulas and numerical calculations, there exist one very important nonnumerical conv
zheka24 [161]

Option B is the correct answer.

MKS system gives the following units:

Distance ----- meters

Mass ----- Kilograms

Time ----- seconds

meter is basic unit for length measurement. smaller units are centimeter, millimeter, micrometer, bigger units are kilometer and so on.

kilogram is the basic unit for mass. smaller unit is gram.

second is the basic unit for time. Greater units are minutes, hours, smallest unit are micro second and so on.

8 0
4 years ago
QUESTION 23
Nutka1998 [239]

Answer:

E = 3600 J

Explanation:

Given that,

Voltage, V = 115 V

Power of electric bulb, P = 60 W

We need to find the electric energy used in 1 minute. The electric energy use is given by :

E=P\times t\\\\E=60\ W\times 60\ s\\\\E = 3600\ J

Hence, the electrical energy is 3600 J.

8 0
3 years ago
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