A 0.15-m straight wire moves with a constant velocity of 7.0 m/s perpendicularly

A large rock of mass me materializes stationary at the orbit of Mercury and falls into the sun. Itf the Sun has a mass ms and ra

son4ous [18]

Answer:

The answer is $v = \sqrt{2G\frac{M_s}{R^2}(R-r_s)}$ .

Explanation:

From the law of gravity,

$F = G \frac{Mm}{r^2}$

considering F as a conservative force, $F = - \nabla U$ ,

the general expression for gravitational potential energy is

$U = -G \frac{Mm}{r}$ ,

where G is the gravitational constant, M and m are the mass of the attracting bodies, and r is the distance between their centers. The negative sign is because the force approaches zero for large distances, and we choose the zero of gravitational potential energy at an infinite distance away.

However, as the mass of the Sun is much greater than the mass of the rock, the gravitational acceleration is defined as

$g = -G \frac{M}{r^2}$ ,

(the negative sign indicates that the force is an attractive force), and the potential energy between the rock and the Sun is

$U = g M_e R$ ,

which is actually the total energy of the system, because the rock materializes stationary at this point (there is no radial kinetic energy).

When the rock hits the surface of the Sun, almost all potential energy is converted to kinetic energy, but not all because the Sun is not a puntual mass. So the potential energy converted to kinetic energy is

$U_p = g M_e(R- r_s)$ ,

then, the kinetik energy when the rock hits the surface is

$U_k =\frac{1}{2}M_e v^2 = g M_e(R- r_s)$ ,

so

$v = \sqrt{2g(R-r_s)}$

where g is the gravitational acceleration generated by the Sun at R,

$g = G \frac{M_s}{R^2}$ .

8 0

2 years ago

explain why earths acceleration is usually very small compared to the acceleration of the object the earth interact with

Marianna [84]

Answer:

F-ma

Explanation:

If you are speaking of objects like satellites, etc. then their mass is much less than that of the Earth. A good approximation is Newton's first law of motion:

Force = Mass × Acceleration

often written:

F = m a

The gravitational force is the same between the Earth and the object - only the mass differs. So the acceleration is inversely proportional to the mass.

6 0

2 years ago

A beaker has a mass of 125g. What is the mass of this beaker in decigrams

Anit [1.1K]

1250 decigrams
1 gram = 10 decigrams

4 0

3 years ago

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$

where k=0,1,2

for minimum frequency $f_1$ , k=1

${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

So the minimum frequency is 702.22 Hz

5 0

2 years ago

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th

vlada-n [284]

Answer:

Explanation:

The process is isothermic, as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303 log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv = m / M RT

v / m = RT / p M

specific volume = RT / p M

option B is correct.

5 0

3 years ago