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3 years ago

Pain in the squadron can be a sign of appemditicis

Physics

2 answers:

alexgriva [62]3 years ago

6 0

No it can also be a sign of bruising or a stomach infection

timofeeve [1]3 years ago

4 0

Yes. Or it can be a sign of intestinal gas, or of constipation. Or it can be a sign of an internal bruise caused by sleeping too long on the same side. Or it can be the result of the new squadron commander cancelling all scheduled leaves.

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The New England Merchants Bank Building in Boston is 152 mm high. On windy days it sways with a frequency of 0.20 HzHz , and the

kkurt [141]

Answer:

The total distance, side to side, that the top of the building moves during such an oscillation = 31 cm

Explanation:

Let the total side to side motion be 2A. Where A is maximum acceleration.

Now, we know know that equation for maximum acceleration is;

A = α(max) / [(2πf)^(2)]

So 2A = 2[α(max) / [(2πf)^(2)] ]

α(max) = (0.025 x 9.81) while frequency(f) from the question is 0.2Hz.

Therefore 2A = 2 [(0.025 x 9.81) / [((2π(0.2)) ^(2)] ] = 2( 0.245 / 1.58) = 0.31m or 31cm

3 0

3 years ago

A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, th

Iteru [2.4K]

<h2>Answer: 2T</h2>

According to the Third Kepler’s Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $R$ of its orbit.

This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):

$T^{2}=\frac{4\pi^{2}}{GM}R^{3}$ (1)

Where:

$G$ is the Gravitational Constant

$M=1.9(10)^{27}kg$ is the mass of planet X

$R$ is the radius of the orbit of the satellite around planet X

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=2\pi\sqrt{\frac{R^{3}}{GM}}$ (2)

Now, we are asked to find the period when tha mass of the planet is $\frac{1}{4}M$ . In order to do this, we have to rewrite equation (2) with this new value:

$T=2\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (3)

Solving:

$T=4\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (4)

On the other hand, if we multiply both sides of equation (2) by 2, we have:

$2T=4\pi\sqrt{\frac{R^{3}}{GM}}$ (5)

As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.

Hence, the answer is:

If Planet X had one-fourth as much mass, the orbital period of this satellite in an orbit of the same radius would be 2T.

3 0

3 years ago

2. A wave on a rope has a wavelength of 2.0 m and a frequency of 2.0 Hz. What is the speed of the

TEA [102]

Answer:

4 m/s or 4 meters per second.

Explanation:

In order to calculate the speed of wave, you multiply the wavelength in meters and the frequency of the Wave in Hertz. 2 times 2 equals 4. The wave speed is always in m/s considering that the wavelength is also in meters.

7 0

3 years ago

Consider three planets. All have the same mass as Earth, but with different radii (from largest to smallest: Planet 1, 2, 3). Fo

LuckyWell [14K]

Answer:

option C

Explanation:

given,

mass of the three planet is same

radius of the planets are

R₁ > R₂ > R₃

expression of escape velocity

$v = \sqrt{\dfrac{2GM}{R}}$

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

from the above expression we can clearly conclude that the escape velocity is inversely proportional to the radius of the Planet.

radius of planet increases escape velocity decreases.

Hence planet 3 has the smallest radius so the escape velocity of the third planet will be maximum.

The correct answer is option C

3 0

3 years ago

Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains

OLEGan [10]

Answer:

C = 771.35 J/kg°C

Explanation:

Here, e consider the conservation of energy equation. The conservation of energy principle states that:

Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container

Since,

Heat Given or Absorbed by a material = m C ΔT

Therefore,

m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃

where,

m₁ = Mass of Metal Piece = 2.3 kg

C = Specific Heat of Metal = ?

ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C

m₂ = Mass of Metal Container = 3.8 kg

ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C

m₃ = Mass of Water = 20 kg

C₃ = Specific Heat of Water = 4200 J/kg°C

ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C

Therefore,

(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)

C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J

C = 252000 J/326.7 kg°C

C = 771.35 J/kg°C

5 0

3 years ago