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guapka [62]
3 years ago
9

Kim has a metal casting company which makes commemorative coins. She has 0.12 cubic meters of silver which she needs to make int

o 1.07e5 coins of thickness 2 mm. If she uses all of the silver, what is the diameter in centimeters of each coin
Physics
1 answer:
a_sh-v [17]3 years ago
5 0

Answer:

The coin has a diameter of 2.67 cm

Explanation:

First, we need to find the volume of each coin by dividing the total volume of silver by the number of coins. We have also to do a conversion of units in terms of centimeters as follows:

V=0.12 m^3\times (\frac{100cm}{1m})^3=12000\ cm^3\\V_c=\frac{120000\ cm^3}{1.07\times10^5}= 1.121 cm^3

Then, we define the coin as a tiny cilinder to determine its diameter. In that order we use the cilinder's volumen equation as follows:

V=\pi r^2h\\r = \sqrt \frac{V}{\pi h}= \sqrt\frac{1.121 cm^3}{\pi \times 0.2cm}=1.336 cm

Finally, we know that the diameter is twice the radius, therefore the diameter of each coin is 2.67 cm.

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Here is a force diagram of an object in water. The weight of the object is 15N and the buoyancy force is 17N. Will the object fl
lawyer [7]

Answer:

Object will float.

Explanation:

Total force on the body = Weight of body + Buoyancy force on body.

 Weight of body = 15 N downwards = 15 N

 Buoyancy force on body = 17 N upwards = -17 N

 Total force on body = 15 - 17 = -2 N = 2 N upwards

 So, the body will float.

Object will float.

8 0
3 years ago
How much work is done when a porcelain vase with mass of 2.5 kg is lifted up 3 m to a shelf? (HINT- Weight can be used as Force
alexdok [17]
Work = force x distance
F= 2.5
D= 3
Work = 2.5 x 3 =7.5
Work = 7.5 J
J=Jules (Jules is the unit uses to calculate work)
8 0
3 years ago
An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _
Viefleur [7K]
Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
where 
f=5 cm is the focal length
p=10 cm is the distance of the object from the mirror
q is the distance of the image from the mirror.

Rearranging the equation, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}
so, the distance of the image from the mirror is q=10 cm.

1b) The image height is given by the magnification equation:
\frac{h_i}{h_o}=- \frac{p}{q}
where h_i is the heigth of the image and h_o=1 cm is the height of the object. By rearranging the equation and using p and q, we find
h_i=-h_o  \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

3) As before, we find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

And now we can use the magnification equation to find the image height:
\frac{h_i}{h_o}=- \frac{p}{q}
Rearranging it, we find
h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm
and the negative sign means the image is inverted.
5 0
3 years ago
A 1.80-m string of weight 0.0126 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
Veseljchak [2.6K]

Answer:

W = 0.135 N

Explanation:

Given:

- y (x, t) = 8.50*cos(172*x -2730*t)

- Weight of string m*g = 0.0126 N

- Attached weight = W

Find:

The attached weight W given that Tension and W are equal.

Solution:

The general form of standing mechanical waves is given by:

                            y (x, t) = A*cos(k*x -w*t)  

Where k = stiffness and w = angular frequency

Hence,

                           k = 172 and w = 2730

- Calculate wave speed V:

                            V = w / k = 2730 / 172 = 13.78 m/s

- Tension in the string T:

                            T = Y*V^2

where Y: is the mass per unit length of the string.

- The tension T and weight attached W are equal:

                           T = W = Y*V^2 = (w/L*g)*V^2

                            W = (0.0126 / 1.8*9.81)*(13.78)^2

                            W = 0.135 N

4 0
3 years ago
A spherically spreading EM wave comes from an 1800-W source. At a distance of 5.0 m, what is the intensity, and what is the rms
Aleonysh [2.5K]

Explanation:

It is given that,

Power of EM waves, P = 1800 W

We need to find the intensity at a distance of 5 m. Also, the rms value of the electric field.

Intensity,

I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{1800}{4\pi\times (5)^2}\\\\I=5.72\ W/m^2

The formula that is used to find the rms value of the electric field is as follows :

I=\epsilon_o cE^2_{rms}

c is speed of light and \epsilon_o is permittivity of free space

So,

E_{rms}=\sqrt{\dfrac{I}{\epsilon_o c}}\\\\E_{rms}=\sqrt{\dfrac{5.72}{8.85\times 10^{-12}\times 3\times 10^8}}\\\\E_{rms}=46.41\ V/m

Hence, this is the required solution.

4 0
3 years ago
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