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Sunny_sXe [5.5K]
2 years ago
13

Given three different locations on Earth's surface, where will the weight of a person be greatest? in New York City, which is ab

out 6369 km from the center of the Earth at the South Pole, which is about 6356 km from the center of the Earth on the Equator line, which is about 6378 km from the center of the Earth the weight is the same at all three locations
Physics
1 answer:
zhenek [66]2 years ago
3 0

Answer:

B. South Pole.

Explanation:

In order to answer this question, we simply have to refer to the laws of the equations of gravitational mechanics.

The equation given by Newton tells us that

F = \frac {Gm_1m_2} {r^2}

In the case where we compare a specific place where the Force of Gravity is greater or lesser, we focus on the term assigned to the Planet's Radius.

In the case of G, m_1, m_2, we understand that they are constant.

We can easily notice that the more the Radius (Height seen from a viewer on the ground), the lower the force will be.

<em>In other words, the smaller the radius in which the measurement is made with respect to the center of the earth, the greater the gravitational force.</em>

In that order of ideas the smallest radio has South Pole, which is about 6356 km from the center of the Earth on the Equator line

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netineya [11]

Answer:

The activation energy for this reaction, Ea = 159.98 kJ/mol

Explanation:

Using the Arrhenius equation as:

ln\frac {K_2}{K_1}=-\frac {E_a}{R}\times (\frac {1}{T_2}-\frac {1}{T_1})

Where, Ea is the activation energy.

R is the gas constant having value 8.314 J/K.mol

K₂ and K₁ are the rate constants

T₂ and T₁ are the temperature values in kelvin.

Given:

K₂ = 8.66×10⁻⁷ s⁻¹ , T₂ = 425 K

K₁ = 3.61×10⁻¹⁵ s⁻¹ , T₁ = 298 K

Applying in the equation as:

ln\frac {8.66\times 10^{-7}}{3.61\times 10^{-15}}=-\frac {E_a}{8.314}\times (\frac {1}{425}-\frac {1}{298})

Solving for Ea as:

Ea = 159982.23 J /mol

1 J/mol = 10⁻³ kJ/mol

Ea = 159.98 kJ/mol

7 0
2 years ago
The inclined plane in the figure above has two sections of equal length and different roughness. The dashed line shows where sec
saveliy_v [14]

The static friction exerted on the block by the incline is \mu _ s _1 Mgcos \ \theta.

The given parameters;

  • <em>mass of the block, = M</em>
  • <em>coefficient of static friction in section 1, = </em>\mu_s_1<em />
  • <em>angle of inclination of the plane, = θ</em>

<em />

The normal force on the block is calculated as follows;

Fₙ = Mgcosθ

The static friction exerted on the block by the incline is calculated as follows;

F_s = \mu_s F_n\\\\F_s = \mu _s_1(Mg cos\ \theta)\\\\F_s = \mu _s_1 Mgcos\ \theta

Thus, the static friction exerted on the block by the incline is \mu _ s _1 Mgcos \ \theta

Learn more here:brainly.com/question/17237604

3 0
2 years ago
Voices of swimmers at a pool travel 400 m/s through the air and 1,600 m/s underwater. The wavelength changes from 2 m in the air
frosja888 [35]

The frequency of the wave has not changed.

In fact, the frequency of a wave is given by:

f=\frac{v}{\lambda}

where v is the wave's speed and \lambda is the wavelength.

Applying the formula:

- In air, the frequency of the wave is:

f=\frac{400 m/s}{2 m}=200 Hz

- underwater, the frequency of the wave is:

f=\frac{1600 m/s}{8 m}=200 Hz

So, the frequency has not changed.

3 0
2 years ago
Read 2 more answers
How much power does it take to lift a 24 kg gift box 6m above the floor in 4 s?
Mrac [35]

Answer:

<h2>5.6kW</h2>

Explanation:

Step one:

given

mass m= 24kg

distance moved= 6m

time taken= 4seconds

Step two:

Required

power

but work done is the force applied at a distance, and the power is the work done time the time taken

Work done= F*D

F=mg

W= mg*D

W=24*9.81*6

W=1412.6J

Power P= work * time

P=1412.6*4

p=5650.5W

P=5.6kW

3 0
1 year ago
The air surrounding an airplane in flight exerts a drag force that acts opposite to the airplane's motion. When an Airbus A380 i
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Answer:

4.32\cdot 10^5 hp, 3.22\cdot 10^8 W

Explanation:

The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.

Therefore, we can write:

4F_T - F_d = 0

where

F_T = 322,000 N is the thrust force generated by each engine of the jet

F_d is the drag force

Solving for Fd,

F_d = 4 F_T = 4(322,000)=1.288\cdot 10^6 N

The velocity of the jet is

v=250 m/s

So, the rate at which the drag force does work (which is the power) is

P=F_d v

and substituting

F_d = 1.288\cdot 10^6 N\\v = 250 m/s

we find

P=(1.288\cdot 10^6)(250)=3.22\cdot 10^8 W

Converting into horsepower,

P=\frac{3.22\cdot 10^8}{746}=4.32\cdot 10^5 hp

4 0
2 years ago
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