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2 years ago

What did copernicus say about the motion of the sun?.

Physics

1 answer:

eimsori [14]2 years ago

8 0

the sun was stationary in the center of the universe and the earth revolved around it....

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5. Granny Goodwitch boards a bus at the local shopping center. The bus travels an

irina [24]

As BBC ztrrtv B 4 C w.a)z

5 0

2 years ago

Consider the following cyclic process carried out in two steps on a gas. Step 1: 44 J of heat is added to the gas, and 20. J of

notka56 [123]

Answer:37 J

Explanation:

Given

Step :1

Heat added Q=44 J

Work done=-20 J

$\Delta E_1=Q+W=44-20=24 J$

Step :2

Heat added Q=-61 J

work done $W_2$

$\Delta E_2=Q+W_2$

$\Delta E_2=61+W_2$

$\Delta E_1+\Delta E_2=0$

as the process is cyclic

$44-20-61+W_2=0$

$W_2=37 J$

work done in compression is 37 J

3 0

3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

1 year ago

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op

Juliette [100K]

Answer:

The fundamental frequency of can is 2.7 kHz.

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

$f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz$

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

7 0

3 years ago

Which scenario will drain the battery in the circuit diagram fastest: three 20 W bulbs in positions one, two, and three or 80 W

kakasveta [241]

Answer:

See the explanation below.

Explanation:

If you connect the three bulbs 20 W each will have a total power of 60W.

Now we need to understand to assign the meaning of the word gap, that is, if the circuit is open at that point or if there is no bulb connected at that point.

If the circuit is open at Point 2, there will be no current in the circuit, so the battery will drain faster with the three bulbs 20W.

In the second event, where gap means that there are no bulbs connected at that point, it means that you have two bulbs connected in series of 80W each.

In this case the bulbs will consume 160W thus drain the battery faster than the three 20W bulbs connected in series.

3 0

3 years ago