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-BARSIC- [3]
3 years ago
9

A sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is 0.0055 m/s2 less than that at sea leve

l (gsealevel = 9.83 m/s2).
Physics
1 answer:
pshichka [43]3 years ago
7 0
Gs*rs^2 = gm*rm^2 
<span>rm = rs*√gs/gm </span>
<span>rm = 6370*√9.83/(9.83-0.009) = 6372.92 </span>
<span>mountain observatory is placed at an altitude worth 2920 m asl

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
</span>
You might be interested in
A sound from a source has an intensity of 270 db when it is 1 m from the source. what is the intensity of the sound when it is 3
Lina20 [59]
Sound intensity is inversely proportional to the square of the distance between the source and the receiver.
That is 
I = k/r^2
 where
 k = constant
 r = radius

When r=1,  the intensity is I₁ = k/1 = k
When r=3, the intensity I₂ = k/3² = k/9
Therefore
 I₂ = I₁ /9

In decibels,
I = 10 log₁₀(I/I₀)
where I₀ = reference intensity

When r=1,
10 log₁₀ (I₁/I₀) = 270

When r =3,
10 log₁₀ (I₂/I₀) = 10 log₁₀ [(I₂/I₁)*(I₁/I₀)]
                     = 10 log₁₀ [(1/9)*(I₁/I₀)]
                     = 10 log₁₀(1/9) + 270
                     = 260.5

Answer: 260.5 dB (nearest tenth)
4 0
3 years ago
If an object has a kinetic energy of 30 j and mass of 34kg how fast is the object moving ?
cricket20 [7]

Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.

Ek = mv^2 / 2 — multiply both sides by 2

2Ek = mv^2 — divide both sides by m

2Ek / m = V^2 — switch sides

V^2 = 2Ek / m — plug in values

V^2 = 2*30J / 34kg

V^2 = 60J/34kg

V^2 = 1.76 m/s — sqrt of both sides

V = sqrt(1.76)

V = 1.32m/s (roughly)

3 0
3 years ago
Find the energy released when there is a decrease of 0.3kg of material in a nuclear reaction
MrRissso [65]

Answer:

E = 2.7 x 10¹⁶ J

Explanation:

The release of energy associated with the mass can be calculated by Einstein's mass-energy relation, as follows:

E = mc^2

where,

E = Energy Released = ?

m = mass of material reduced = 0.3 kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

E = (0.3\ kg)(3\ x\ 10^8\ m/s)^2

<u>E = 2.7 x 10¹⁶ J</u>

4 0
2 years ago
A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s
Paraphin [41]

Answer:

\omega = 0.016\,\frac{rad}{s}

Explanation:

The rotation rate of the man is:

\omega = \frac{v}{R}

\omega = \frac{0.80\,\frac{m}{s} }{5\,m}

\omega = 0.16\,\frac{rad}{s}

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega

The final angular speed is:

\omega = 0.016\,\frac{rad}{s}

3 0
3 years ago
A copper telephone wire has essentially
Lunna [17]

Answer:

128.21 m

Explanation:

The following data were obtained from the question:

Initial temperature (θ₁) = 4 °C

Final temperature (θ₂) = 43 °C

Change in length (ΔL) = 8.5 cm

Coefficient of linear expansion (α) = 17×10¯⁶ K¯¹)

Original length (L₁) =.?

The original length can be obtained as follow:

α = ΔL / L₁(θ₂ – θ₁)

17×10¯⁶ = 8.5 / L₁(43 – 4)

17×10¯⁶ = 8.5 / L₁(39)

17×10¯⁶ = 8.5 / 39L₁

Cross multiply

17×10¯⁶ × 39L₁ = 8.5

6.63×10¯⁴ L₁ = 8.5

Divide both side by 6.63×10¯⁴

L₁ = 8.5 / 6.63×10¯⁴

L₁ = 12820.51 cm

Finally, we shall convert 12820.51 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

12820.51 cm = 12820.51 cm × 1 m / 100 cm

12820.51 cm = 128.21 m

Thus, the original length of the wire is 128.21 m

5 0
3 years ago
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