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Montano1993 [528]
2 years ago
8

The moon lies roughly 384,000 km from earth and the sun lies 150,000,000 km away. if both have the same angular size, as seen fr

om earth, how many times larger than the moon is the sun?
Physics
1 answer:
xxMikexx [17]2 years ago
3 0
Moon diameter - 2,160 miles Sun diameter - 864,000 miles 
So, in terms of diameter the Sun is 400 times bigger then the Moon. If we divide 150,000,000 km by 384,000 km we get 390.625 almost the same number. 
The Sun and the Moon have sizes which vary a small amount as seen from Earth. 
The Sun appears largest about January 4th and smallest around July 4th. 
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A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.
Sedaia [141]

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

7 0
2 years ago
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0
2 years ago
A test car and its driver, with a combined mass of 600 kg, are moving along a straight,horizontal track when a malfunction cause
ANEK [815]

Answer:

The two of the following measurements, when taken together, would allow engineers to find the total mechanical energy dissipated during the skid

B. The contact area of each tire with the track.

C. The co-efficent of static friction between the tires and the track.

D. The co-efficent of static friction between the tires and the track.

Explanation:

4 0
2 years ago
The temperature of a heat engine is 500k some of the heat generated by the engine flows to the surroundings which are at a temp
Ierofanga [76]

1-125/500)x100=efficiency

1-1/4)x100 =75pc

3 0
1 year ago
In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc
Vesna [10]

Answer:

Explanation:

Let T be the tension in the swing

At top point mg-T=\frac{mv^2}{r}

where v=velocity needed to complete circular path

r=distance between point of  rotation to the ball center=L+\frac{d}{2} (d=diameter of ball)

Th-resold velocity is given by mg-0=\frac{mv^2}{r}

To get the velocity at bottom conserve energy at Top and bottom

At top E_T=mg\times 2L+\frac{mv^2}{2}

Energy at Bottom E_b=\frac{mv_0^2}{2}

Comparing two as energy is conserved

v_0^2=4gr+gr

v_0^2=5gr

v_0=\sqrt{5gr}

v_0=\sqrt{5g\left ( \frac{d}{2}+L\right )}

6 0
2 years ago
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