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3 years ago

7

A 11 g plastic ball is moving to the left at 29 m/s. How much work must be done on the ball to cause it to move to the right at

29 m/s ? Express your answer using two significant figures.

1 answer:

Rasek [7]3 years ago

8 0

Answer:

4.25 J

Explanation:

Given that

mass of plastic ball = 11 g

Mass of plastic ball = 0.011 kg

velocity of ball = 29 m/s

We know that from work power energy theorem

$W_{all}=Change\ in\ kinetic\ energy\ of\ system$

We know that kinetic energy of moving mass given as

$KE=\dfrac{1}{2}mv^2$

Now by pitting the values

$KE=\dfrac{1}{2}mv^2$

$KE=\dfrac{1}{2}\times 0.011\times 29^2$

KE= 4.25 J

So the work done on the ball is 4.25 J

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A coating is being applied to reduce the reflectivity of a pane of glass to light with a wavelength of 522 nm incident near the

fredd [130]

Answer:

t = 94.91 nm

Explanation:

given,

wavelength of the light = 522 nm

refractive index of the material = 1.375

we know the equation

c = ν λ

where ν is the frequency of the wave

c is the speed of light

$\nu= \dfrac{c}{\nu\lambda}$

$\nu = \dfrac{3\times 10^8}{522 \times 10^{-9}}$

ν = 5.75 x 10¹⁴ Hz

the thickness of the coating will be calculated using

$t = \dfrac{\lambda}{4\mu_{material}}$

$t = \dfrac{522 \times 10^{-9}}{4\times 1.375}$

t = 94.91 nm

the thickness of the coating will be equal to t = 94.91 nm

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3 years ago

In the figure, a weightlifter's barbell consists of two identical small but dense spherical weights, each of mass 50 kg. These w

kondaur [170]

The moment of inertia is $24.8 kg m^2$

Explanation:

The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.

The moment of inertia of the rod about its centre is given by

$I_r = \frac{1}{12}ML^2$

where

M = 24 kg is the mass of the rod

L = 0.96 m is the length of the rod

Substituting,

$I_r = \frac{1}{12}(24)(0.96)^2=1.84 kg m^2$

The moment of inertia of one ball is given by

$I_b = mr^2$

where

m = 50 kg is the mass of the ball

$r=\frac{L}{2}=\frac{0.96}{2}=0.48 m$ is the distance of each ball from the axis of rotation

So we have

$I_b = (50)(0.48)^2=11.5 kg m^2$

Therefore, the total moment of inertia of the system is

$I=I_r + 2I_b = 1.84+ 2(11.5)=24.8 kg m^2$

Learn more about inertia:

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3 years ago

A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to

jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

b) Current at the center of the circular coil $=2.48\times 10^{5}\ A$

c) Current near the center of a solenoid $236.8\ A$

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

Conversion $1\times 10^6 A = 1\ MA$ ,and $2cm=\frac{2}{100}=0.02\ m$

$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

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3 years ago

Why is the Mid-Atlantic Ridge an ideal place for SWARM to collect electrical conductivity data?

ElenaW [278]

Answer: Mid-ocean ridges are geologically important because they occur along the kind of plate boundary where new ocean floor is created as the plates spread apart. Thus the mid-ocean ridge is also known as a "spreading center" or a "divergent plate boundary." The plates spread apart at rates of 1 cm to 20 cm per year.

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3 years ago

Read 2 more answers

Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through

Ad libitum [116K]

Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,

d l = d x

in this situation. We can therefore write

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

4 0

3 years ago