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3 years ago

The diameter of a steel rod is 56.47 ± 0.02mmWhat does it mean....... please

Physics

1 answer:

mamaluj [8]3 years ago

8 0

Answer:

The rod is nominally 56.47mm in diameter but, due to manufacturing tolerances, there could be some variation.

The supplier is guaranteeing that the rod will not be smaller than 56.45mm nor will it be larger than 56.49mm in diameter.

Explanation:

It's means that the range of diameter of steel rod during its measurements will fall in between of 56.47–0.02mm= 56.45mm to 56.47+0.02mm= 56.49

+/- 0.02mm is the basically called as tolerance level.

Hope you get it.

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A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

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3 years ago

If the pendulum took longer to complete one oscillation, how would the graph change?

pickupchik [31]

We don't know what kind of graph it is.

For example, it might be a graph of the pendulum's distance from center,

angle from center, speed, acceleration, total distance swung since it was

started, mass, weight, temperature, etc.

If the graph shows the pendulum's distance from center, angle from center,

speed, or acceleration, then the graph will look like a wave, with the period

of the wave being the period of the pendulum's oscillation. If the pendulum

took longer to complete one oscillation, that means its PERIOD increased,

and the distance between the peaks of the graph would be longer.

If it was a graph of total distance the pendulum swung since it was started,

the graph wouldn't look like a wave, just a steadily rising wiggle line. If the

pendulum took longer to complete one oscillation, the wiggles in the line

would be farther apart, and the average slope of any large section of the

line would be less.

If it was a graph of the pendulum's mass, weight, temperature, cost, etc.,

then the graph would be a horizontal line, and nothing that might change

the period of oscillation would have any effect on the graph.

7 0

3 years ago

Read 2 more answers

A polarized light that has an intensity I0 = 60.0 W/m² is incident on three polarizing disks whose planes are parallel and cente

nikitadnepr [17]

Answer:

The transmitted intensity through all polarizers is $I_3 =41.31 W/m^2$

Explanation:

According to Malu's law the intensity of a polarized light having an initial intensity $I_0$ is mathematically represented as

$I = I_0cos^2 \theta$

Now considering the polarizer(The polarizing disk) the equation above becomes

$I = I_0 (cos^2 \theta)^n$

Where n is the number of polarizers

Substituting $60.0W/m^2$ for the initial intensity 3 for the n and 20° for the angle of rotation

$I_3 = 60 (cos^220)^3$

$=41.31 W/m^2$

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3 years ago

Heres a random question just for fun and out. of boredom ok so who started the nasa program? and why? what made him want to do t

Pachacha [2.7K]

Eisenhower started the NASA project to develop technology for military application.

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3 years ago

A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and

Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

$q_{out}=q_{2}-q_{1}$

$q_{out}=-6.43-(-4.00)$

$q_{out}=-2.43\ \mu C$

The charge on the inner surface is q.

$q+(-2.43)=-6.43$

$q=-6.43+2.43= 4.00\ \mu C$

We need to calculate the electric field outside the shell

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}$

$E=33927618.73\ N/C$

$E=3.39\times10^{7}\ N/C$

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

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3 years ago

The diameter of a steel rod is 56.47 ± 0.02mmWhat does it mean....... please ​

The diameter of a steel rod is 56.47 ± 0.02mmWhat does it mean....... please