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serious [3.7K]
2 years ago
12

1. If point Q is reflected across x = 1, what are the coordinates of its reflection image?

Physics
2 answers:
QveST [7]2 years ago
7 0
If the original coordinates of point Q are (A, B) and
the point is reflected across the line  x=1, then the
coordinates of the reflected point Q' are

Q' (2-A, B) .
Tresset [83]2 years ago
6 0

Answer:

Is there a picture of the graph? To figure this out remember the rule. (x,y) reflected across the x-axis would make it(x,-y)

You might be interested in
The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o
matrenka [14]

Answers:

(a) 0.0073kg

(b) Volume gold: 3.79(10)^{-7}m^{3}, Volume cupper: 7.6(10)^{-8}m^{3}

(c) 17633.554kg/m^{3}

Explanation:

<h2>(a) Mass of gold </h2><h2 />

We are told the total mass M of the coin, which is an alloy  of gold and copper is:

M=m_{gold}+m_{copper}=7.988g=0.007988kg   (1)

Where  m_{gold} is the mass of gold and m_{copper} is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats K and mass is:

K=24\frac{m_{gold}}{M}   (2)

Finding {m_{gold}:

m_{gold}=\frac{22}{24}M   (3)

m_{gold}=\frac{22}{24}(0.007988kg)   (4)

m_{gold}=0.0073kg   (5)  This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density \rho of an object is given by:

\rho=\frac{mass}{volume}

If we want to find the volume, this expression changes to: volume=\frac{mass}{\rho}

For gold, its volume V_{gold} will be a relation between its mass m_{gold}  (found in (5)) and its density \rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}:

V_{gold}=\frac{m_{gold}}{\rho_{gold}}   (6)

V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}   (7)

V_{gold}=3.79(10)^{-7}m^{3}   (8)  Volume of gold in the coin

For copper, its volume V_{copper} will be a relation between its mass m_{copper}  and its density \rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}:

V_{copper}=\frac{m_{copper}}{\rho_{copper}}   (9)

The mass of copper can be found by isolating m_{copper} from (1):

M=m_{gold}+m_{copper}  

m_{copper}=M-m_{gold}  (10)

Knowing the mass of gold found in (5):

m_{copper}=0.007988kg-0.0073kg=0.000688kg  (11)

Now we can find the volume of copper:

V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}   (12)

V_{copper}=7.6(10)^{-8}m^{3}   (13)  Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density \rho_{coin will be a relation between its total mass M and its total volume V:

\rho_{coin}=\frac{M}{V} (14)

Knowing the total volume of the coin is:

V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3} (15)

\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}} (16)

Finally:

\rho_{coin}=17633.554kg/m^{3}} (17)  This is the total density of the British sovereign coin

6 0
3 years ago
A driver averaged 64 mph and took 3½ hours to travel from st. Louis to chicago. Based on this, what is the distance between st.
sammy [17]

Average speed of the driver is given as

v = 64 mph

if he moved for total time t = 3.5 hours

so the distance between the tow is given as

d = v* t

d = 64 * 3.5

d = 224 miles

so the distance between St. Louis and Chicago is 224 miles

6 0
3 years ago
A force is a push or a pull. Under the right circumstances, anything can exert a force.
Sliva [168]

Answer:

Explanation:

the force that you applied caused the notebook to move.

the force applied on the notebook by the table causes it to stop moving.

this is because after sometime the book uses up the force and later the force you applied is less than that of the force by the table.

7 0
3 years ago
FILL IN THE BLANKS here on earth, the pull of gravity on a mass of 1 kg is ......... newtons​
Aleks [24]

Answer:

9.8 Newton

Explanation:

At average gravity on earth (conventionally, g=9.80665m/s2),

a kilogram mass exerts a force of about 9.8 newton

I hope this answer helps you

4 0
2 years ago
If a stone dropped into a well reaches the water's surface after 3.0 seconds, how far did the stone drop before hitting the wate
Ulleksa [173]
<span>s= 0.5 at^2
</span>
0.5 x 9.81 x 3^2 = 44m
8 0
3 years ago
Read 2 more answers
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