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Bess [88]
3 years ago
7

White dwarfs are too small to see with telescopes true or false

Physics
1 answer:
trasher [3.6K]3 years ago
5 0

Answer:

False

Explanation:

A white dwarf star that is easy to locate and see with small telescopes.

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a street light is mounted at the top of a 15 foot pole. A man 6 ft tall walks away from the pole wit a speed of 7 ft/s along a s
Mariulka [41]

Answer:

16.3 ft/s

Explanation:

Let d=distance

and

x = length of shadow.

Therfore,

x=(d + x)

 = 6/15

So,

    15x = 6x + 6d

     9x = 6d.

x = (2/3)d.

As we know that:

dx=dt

   = (2/3) (d/dt) 

Also,

Given:

d(d)=dt

     = 7 ft/s

Thus,

d(d + x)=dt

           = (7/3)d (d/dt)

Substitute, d= 7  

d(d + x) = 49/3 ft/s.

Hence,

d(d + x) = 16.3 ft/s.

4 0
3 years ago
A scientist is designing a device that will mimic Earth's atmosphere by
mafiozo [28]

Answer:

1. Ultraviolet light (UV)

2. X-rays

3. Gamma-rays

Explanation:

Though there are different types of energy or electromagnetic waves with varying wavelengths, including the likes of Gamma X-rays, ultraviolet light, visible light, infrared radiation, and microwave radiation.

What is more certain is that the atmosphere blocked the high-energy waves from getting to the earth surface or biosphere such as Ultraviolet light (UV), X-rays and Gamma-rays

8 0
3 years ago
Question 2.
choli [55]

Explanation:

1 inch = 25.4 mm

1 foot = 12 inches

1 mile = 5260 feet

1 cm = 0.01 m or 10 mm

Now Sammy's height is 5 feet and 5.3 inches.

(a) We need to find Sammy's height in inches.

Since, 1 foot = 12 inches

5 feet = 5 × 12 inches = 60 inches

Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches

Sammy's height is 65.3 inches.

(b) We need to find Sammy's height in feet.

Since, 1 foot = 12 inches

1\ \text{inch}=\dfrac{1}{12}\ \text{feet}

So,

5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}

5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet

Sammy's height is 5.44 feet.

3 0
3 years ago
A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1
balandron [24]

Answer:

The answer is "143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}"

Explanation:

For point a:

Energy balance equation:

\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\

W=0\\\\Q=0\\\\m_e=0

From the above equation:

\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\

because the rate of air entering the tank that is h_i constant.

\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\

Since the tank was initially empty and the inlet is constant hence, m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\

Interpolate the enthalpy between T = 300 \ K \ and\ T=295\ K. The surrounding air  

temperature:

T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

Substituting the value from ideal gas:

\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}

Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

Interpolate

\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}

Substitute values from the table.

 \frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\

For point c:

 Entropy is given by the following formula:

\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}

5 0
2 years ago
When a warm air mass is trapped between two cooler air masses, it is called a/an
Semenov [28]
The answer is C. An occluded front.
8 0
3 years ago
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