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2 years ago

Please help me, it's a simple equation

Physics

1 answer:

castortr0y [4]2 years ago

3 0

Because the elevator moves at a constant speed, it's in equilibrium and the net force acting on it is zero. Then the tension in the cable exactly equals the magnitude of the elevator's weight, which is

(3000 kg) (9.80 m/s²) = 29,400 N

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A 2 kg ball is dropped above the surface of Planet X. If the gravitational field strength at the surface of Planet X is 5 N/kg,

Trava [24]

Given data:

* The mass of the ball is 2 kg.

* The gravitational field strength at the surface of planet X is 5 N/kg.

Solution:

The weight of the ball on the planet X is,

$W=ma$

where m is the mass of ball, a is the gravitational field strength,

Substituting the known values,

$\begin{gathered} W=2\times5 \\ W=10\text{ N} \end{gathered}$

Thus, the weight of the ball on the surface of planet X is 10 N.

3 0

1 year ago

The gravity of earth depends on the _____ from its ____ . as this _____increase, the magnitude of the gravity _____.

Anna [14]

<span>The gravity of earth depends on the magnetism from its core. as this magnetism increases, the magnitude of the gravity increases.</span>

3 0

3 years ago

A piano tuner is using a 392 Hz tuning fork to tune the wire for a G-Natural note. She hears 4 beats per second. What are the tw

inysia [295]

A beat is an interference pattern between two sounds of slightly different frequencies, perceived as a periodic variation in volume whose rate is the difference of the two frequencies. Frequency beat is equal to,

$f_{beat} =| f_2\pm f_1 |$

The reference frequency in our case would be 392Hz, and since there is the possibility of the upper and lower range for the amount of beats per second that the two possible frequencies are heard would be

$f_{beat} =|392+4|= 396Hz$

$f_{beat} =|392-4|=388Hz$

Therefore the two possible frequencies the piano wire is vibrating at, would be 396Hz and 388Hz

5 0

3 years ago

A rope of length L has circular cross-sectional area A and density rho = m/V , where m is the mass of the rope and V = A · L is

hram777 [196]

Answer: µ = ρ¹ * A¹

Where x=1 and y=1

Explanation: According to the question, the mass per unit length (µ) is related to the density (ρ) and area A are related by the formulae below

µ = ρ * A

The dimension for each of these quantities is given below

Since µ is mass per unit length, unit is Kg/m and the dimension is ML^-1

ρ is density with unit kg/m³ and the dimension is ML^3

A is area with unit m², thus the dimension is M^2

Note that using dimensional analysis means we will be using the 3 fundamental quantities (mass, length and time) in our analysis.

Their dimensions below

Mass = M

Length = L

Time = T

Since the mass per unit length is related to density and area, we have a mathematical equation to provide a solution as shown below

µ = ρ^x * A^y.

By getting the power of x and y we will be able to get the formula that relates the quantities.

This is done by slotting in the dimensions of the respective quantities.

ML^-1 = (ML^-3)^x * (L²) ^y

By using law of indices on the right hand side of the equation, we have that

ML^-1 = (M^x * L^-3x) * (L^2y)

Also applying law of indices on the right hand side, we have that

ML^-1 = (M^x) * (L^-3x +2y)

The next step is to relate equal variables on both sides

For the M variable

M¹ = M^x which results to

x = 1

For the L variable

L^-1 = L^-3x+2y which results to

-1 = - 3x +2y

But x = 1

We have that

-1 = - 3(1) +2y

-1 = - 3 + 2y

-1 +3= 2y

2 = 2y

y = 1

Thus x=1 and y=1 and the formulae that relates the quantities is

µ = ρ¹ * A¹

3 0

3 years ago

What is the molar mass of an ideal gas if a 0.800 g sample of this gas occupies a volume of 200. mL at 50.0 oC and 720. mm Hg?

Paladinen [302]

PV=nRT
(720/760)(0.200)=(0.800/x)(0.08206)(323.15)
(0.1894736842)=(0.800/x)(0.08206)(323.15)
.0071451809=(0.800/x)
x=MM=111.9635758 g/mol

7 0

3 years ago