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3 years ago

Why does the gravitational field strength increase with latitude

1 answer:

4 0

The second major reason for the difference in gravity at differentlatitudes is that the Earth's equatorial bulge (itself also caused by centrifugalforce from rotation) causes objects at the Equator to be farther from the planet's centre than objects at the poles.

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A sample of helium has a volume of 12.7 m3. The temperature is raised to 323 K at which time the gas occupies 32.5 m3? Assume pr

jasenka [17]

Answer: The original temperature was

$T_{1}=126.51K$

Explanation:

Let's put the information in mathematical form:

$V_{1}=12.7m^{3}$

$T_{1}=?$

$V_{2}=32.5m^{3}$

$T_{2}=323K$

$P_{1}=P_{2}=3atm$

If we consider the helium as an ideal gas, we can use the Ideal Gas Law:

$PV=nRT$

were <em>R</em> is the gas constant. And <em>n</em> is the number of moles (which we don't know yet)

From this, taking $R=0.08205746\frac{atm.l}{mol.K}$ , we have:

$n=\frac{P_{2}V_{2}}{RT_{2}}$

⇒ $n=3.67mol$

Now:

$T_{1}=\frac{P_{1}V_{1}}{nR}$

⇒ $T_{1}=126.51K$

7 0

3 years ago

Read 2 more answers

MARKING BRAINLIST | Which situation below would have the STRONGEST gravitational force between them?

maks197457 [2]

Case d) has the strongest gravitational force

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

a) For this pair of objects:

m1 = 10 kg

m2 = 2 kg

r = 30 km = 30,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(10)(2)}{30000^2}=1.48\cdot 10^{-18}N$

b) For this pair of objects:

m1 = 10 kg

m2 = 10 kg

r = 30 km = 30,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(10)(10)}{30000^2}=7.41\cdot 10^{-18}N$

c) For this pair of objects:

m1 = 2 kg

m2 = 2 kg

r = 10 km = 10,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(2)(2)}{10000^2}=1.33\cdot 10^{-17}N$

d) For this pair of objects:

m1 = 10 kg

m2 = 10 kg

r = 10 km = 10,000 m

So the gravitational force is

$F=(6.67\cdot 10^{-11})\frac{(10)(10)}{10000^2}=6.67\cdot 10^{-17}N$

Therefore, the strongest gravitational force is in case d).

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

3 years ago

Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence

Keith_Richards [23]

Answer:

1. Recollapsing universe

2. Critical universe

3. Coasting universe

Explanation:

Recollapsing universe has dark matter density greater than critical density. While critical universe has its matter density equal to the critical sensity. Coasting universe on the other hand has much smaller matter density compared to critical density.

Note that the critical density is approximately 10^-20 grams/cm3

3 0

3 years ago

Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun

Dominik [7]

Answer:

a. $K_{Axis}=2.574x10^{29}J$

b. $K_{Orbit}=2.6577x10^{33}J$

Explanation:

$K_{Axis}=\frac{1}{2}I*w^2$

$I_{Sphere}=\frac{2}{5}*m*r^2$

$w=\frac{2\pi }{T}$ , $T=24hrs*\frac{3600s}{1hr} =86400s$

radius earth = 6371 km

mass earth = 5,972*10^24 kg

$K_{Axis}=\frac{1}{2}*\frac{2}{5}*m*r^2*(\frac{2\pi}{T})^2$

$K_{axis}=\frac{4\pi^2}{5}*5.98x10^{24}kg*(6.38x10^6m)^2*(\frac{1}{86400s})^2$

$K_{Axis}=2.574x10^{29}J$

$T=1year*\frac{365day}{1year}*\frac{24hr}{1day}*\frac{3600s}{1hr}=31536000s$

$K_{Orbit}=\frac{1}{2}*I*w$

$I=m*r^2$

$K_{Orbit}=\frac{1}{2}*m*r^2*(\frac{2\pi}{T})^2$

$K_{Orbit}=\frac{4\pi^2}{5}*5.98x10^{24}*6.38x10^6m*(\frac{1}{31.536x10^6s})^2$

$K_{Orbit}=2.6577x10^{33}J$

4 0

3 years ago

An elaborate pulley consists of four identical balls at the ends of spokes extending out from a rotating drum. A box is connecte

Klio2033 [76]

Answer:

its speed will be less than V

Explanation:

When the ball falls a distance d, its final kinetic energy plus rotational kinetic energy of the drum equals its initial potential energy.

K = U

With its speed V at the end of d, we have

1/2mV² + 1/2Iω² = mgd where I = rotational inertia of drum and balls, ω = angular speed of drum and balls and m = mass of box

1/2mV² + 1/2Iω² = mgd

1/2mV² = mgd - 1/2Iω²

V² = [2(mgd - 1/2Iω²)/m]

V = √[2(mgd - 1/2Iω²)/m]

When the four balls are moved inward closer to the drum, their rotational inertia increases and also its angular speed which thus causes an increase in rotational kinetic energy. But, since the box still falls the same distance of d, its final kinetic energy plus rotational kinetic energy of the drum plus balls still equals its initial potential energy

K = U

I' = new rotational inertia of drum and balls, ω' = new angular speed of drum and balls

With its new speed is now V' at the end of d,

1/2mV'² + 1/2I'ω'² = mgd

1/2mV'² = mgd - 1/2I'ω'²

V² = [2(mgd - 1/2I'ω'²)/m]

V' = √[2(mgd - 1/2I'ω'²)/m]

Since I' and ω' increase, the rotational kinetic energy of the drum and balls (1/2I'ω'²) increases. Thus, the difference (mgd - 1/2I'ω'²) < (mgd - 1/2Iω²) which implies that the kinetic energy of the box decreases. Hence, since its kinetic energy decreases, its speed V' also decreases.

So, V' < V

So, its speed will be less than V.

3 0

3 years ago