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2 years ago

A 1938 nickel has a diameter of 21.21 mm, a thickness of 1.95 mm, and weighs 0.04905 N. What is its density?

1 answer:

4 0

Density = mass/volume
Mass of nickel = 0.04905/9.81=0.005kg
Volume of nickel = ( $\pi (\frac{0.02121}{2} )^{2} *0.00195$
=0.000002755
Density = 0.005/0.000002755 =1814.279 = 1810kg $m^{-3}$

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At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the le

Nana76 [90]

Answer:

Acceleration=24.9ft^2/s^2

Angular acceleration=1.47rads/s

Explanation:

Note before the ladder is inclined at 30° to the horizontal with a length of 16ft

Hence angular velocity = 6/8=0.75rad/s

acceleration Ab=Aa +(Ab/a)+(Ab/a)t

4+0.75^2*16+a*16

0=0.75^2*16cos30°-a*16sin30°---1

Ab=0+0.75^2sin30°+a*16cos30°----2

Solving equation 1

(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s

Also from equation 2

Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2

6 0

2 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

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2 years ago

A 5.09 × 1014-hertz electromagnetic wave is traveling through a transparent medium. The main factor that determines the speed of

sergiy2304 [10]

We are given an electromagnetic wave with a frequency of 5.09 x 10^14 Hz and travelling through a transparent medium. If the medium was vacuum, the speed of the wave would be equal to the speed of light. Otherwise, the main factor that would determine the speed of the wave is its wavelength.

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2 years ago

Match each label to the boundary it describes. convergent boundary new crust forms transform boundary crust submerges into the m

Paraphin [41]

The answers would be:

CONVERGENT boundary - Crust submerges into the mantle

TRANSFORM boundary - neither forms nor submerges

DIVERGENT boundary - new crust forms

If you'd like to know more about the different boundaries, read on:

Convergent boundaries occur when two plates move TOWARDS each other. The event where crust submerges into the mantle is called subduction and this occurs when an oceanic plate and a continental plate collide. The oceanic plate is more dense and thinner than the continental plate, so it slides under it.

Transform boundaries occur when two plates slide against each other. They move slide side by side, so nothing is formed nor do they go under each other. Although, this type of boundaries create strong earthquakes.

Lastly, divergent boundaries occur when two plates move apart. The separation creates a way for magma to come up. New crust is formed when the magma that seeps out is cooled by its cooler surroundings. This is observed in the mid oceanic ridge.

7 0

2 years ago

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A piston–cylinder device contains 0.15 kg of air initially at 2 MPa and 350°C. The air is first expanded isothermally to 500 kPa

Paraphin [41]

Answer:

<h2>jeusYgwyhedswusjsj</h2>

Explanation:

sjauajshsuyeueueheysb

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2 years ago