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sergij07 [2.7K]

3 years ago

11

Which of the following is most likely to be attracted to a magnet? A. A piece of glass B .A piece of plastic C .A stainless-stee

l paper clip D .A wet newspaper

2 answers:

Law Incorporation [45]3 years ago

7 0

C. A stainless steel paper clip

GuDViN [60]3 years ago

3 0

It would be C A stainless steel paperclip because it is metal A,B,D are not magnetic because they are not metal.

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Differentiate between step up and step down transformer

Sav [38]

Answer:

The main difference between the step-up and step-down transformer is that the step-up transformer increases the output voltage, while the step-down transformer reduces the output voltage.

5 0

2 years ago

A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2,

Stella [2.4K]

Answer:

the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is $P = 104.04 \hat{i} -314.432 \hat{j}$

Explanation:

The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;

the peripheral velocity that is directed downward $(-V_y)$ along the y-axis

the linear velocity $(V_x)$ that is directed along the x-axis

Now;

$V_x = \frac{d}{dt}(12t^3+2) = 36 t^2$

$V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s$

Also,

$-V_y = R* \omega$

where $\omega$ (angular velocity) = $\frac{d\theta}{dt} = \frac{d}{dt}(8t^4)$

$-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s$

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is $P = 104.04 \hat{i} -314.432 \hat{j}$

3 0

3 years ago

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

2 years ago

How do you find the change in potential energy

ratelena [41]

P.E = mgh

This is the formula for potential energy.

This is where m is mass, g is the acceleration due to gravity, and h is height.

All you have to do is multiply all these numbers together.

3 0

3 years ago

HEY CAN ANYONE PLS ANSWER DIS RQ!!!!

Bezzdna [24]

Answer:

Negative

Explanation:

If you have to move the decimal point to the right to get the original number, the exponent will be a positive number, if you have to move the decimal point to the left to get the original number, the exponent will be a negative number.

Hope this helped!

8 0

2 years ago