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igomit [66]
3 years ago
9

Someone answer these questions please???

Physics
2 answers:
chubhunter [2.5K]3 years ago
8 0
Yeah sure someone else in answering rn
Fofino [41]3 years ago
4 0

Answer:

a. same

b. less

c. less

Explanation:

you can use the position, time, acceleration equation

xf=xi + vi*t+(at^2)/2

you can say the final distance will greater for X than Y

using acceleration for gravity a=9.81m/s^2

a.  both X and Y will have the same acceleration due to gravity

b. to solve for final speed vf=vi+gt

since the initial velocity is 0 and X will take longer to fall, Y will hit the ground with less speed

c. as mentioned before, X will take longer to fall; therefore Y will have less time of descent.

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Answer:I would prefer ginger with lemon and honey tea...Stay hydrated,Rest,

Explanation:

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3 years ago
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A spring has a spring constant of 330 n/m. how far is the spring compressed when 150 newtons of force are used?
vitfil [10]

Answer:

b. 0.45 meters

Explanation:

Given the following data;

Spring constant, k = 330 N/m

Force = 150 N

To find the extension of the spring;

Mathematically, the force exerted on a spring is given by the formula;

Force = spring constant * extension

Substituting into the formula, we have;

150 = 330 * extension

Extension, e = 150/330

Extension, e = 0.45 meters

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2 years ago
What is TRUE about a capitalist society?
Makovka662 [10]

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3 years ago
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A certain car traveling at 97 km/h can stop in 47 m on a level road find the coefficient of friction
IrinaVladis [17]

The coefficient of friction between the road and the car's tire is determined as 0.78.

<h3>Acceleration of the car</h3>

The acceleration of the car is calculated as follows;

v² = u² - 2as

0 = u² - 2as

a = u²/2s

where;

  • u is the initial velocity = 97 km/h = 26.94 m/s

a = (26.94)²/(2 x 47)

a = 7.72 m/s²

<h3>Coefficient of friction</h3>

μ = a/g

μ = (7.72)/9.8

μ = 0.78

Learn more about coefficient of friction here: brainly.com/question/14121363

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5 0
1 year ago
A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin
Alexxx [7]

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

\epsilon = \frac{NBA}{t}

Where,

\epsilon  =electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

R = R_c + R_d

And that the current can be expressed as function of charge and time, then

I = \frac{q}{t}

Equation Faraday's law and Ohm's law we have,

V = \epsilon

IR = \frac{NBA}{t}

(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}

Re-arrange for Magnetic Field B, we have

B = \frac{q(R_c+R_d)}{NA}

Our values are given as,

R_c = 58.7\Omega

R_d = 45.5\Omega

N = 120

q = 3.53*10^{-5}C

A = 3.21cm^2 = 3.21*10^{-4}m^2

Replacing,

B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}

B = 0.095T

Therefore the magnetic field in the System is 0.095T

3 0
3 years ago
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