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beks73 [17]
3 years ago
10

On a map, a measurement of 1.0 cm represents a velocity of 100 m/s. If you were to draw a vector on the map that represents 80 m

/s, how long would the vector be?
A. 8.0 cm B. 1.25 cm C. 1.25 mm D. 8.0 mm
Physics
1 answer:
zimovet [89]3 years ago
5 0

Answer:

80 \frac{m}{s} = 8 millimeter

Hence, option D is correct

Explanation:

1 centimeter on map represents 100 \frac{m}{s}

10 millimeter on map represents 100 \frac{m}{s}.

So, 1 millimeter on map represents 10 \frac{m}{s}.

10 \frac{m}{s} = 1 millimeter

Thus,

80 \frac{m}{s} = 8 millimeter

Hence, option D is correct

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andreev551 [17]

Answer:

Polarization occurs when an electric field distorts the negative cloud of electrons around positive atomic nuclei in a direction opposite the field. Polarization P in its quantitative meaning is the amount of dipole moment p per unit volume V of a polarized material, P = p/V.

Explanation:

8 0
3 years ago
Read 2 more answers
(a) The electric potential due to a point charge is given by V = kq⁄r where q is the charge, r is the distance from q and k = 8.
LiRa [457]

Answer:

a)  [volts] = [N m / C],

b) The lines or surface that has the same potential are called equipotential

c) the equipotential lines must also be perpendicular to the electric field lines

Explanation:

a) find the units of the volt

the electric potential energy is

             V = k q / r

             V = [N m² / C²] C / m

              V = [N m / C]

The electric potential is defined as

             V = E .s

             V = [N / C] [m]

             V = [N m / C] = [volt]

we see that in the two expressions the same result is obtained therefore the volt is

            [volts] = [N m / C]

b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.

c) The electric potential is defined as the gradient of the electric field

             v =  - \frac{dE}{dx} i^

therefore the equipotential lines must also be perpendicular to the electric field lines

4 0
3 years ago
Un cubo de madera de densidad 0.780 g/cm³ mide 11.2 cm en un lado. Cuando se coloca en agua, ¿qué altura del bloque flotará sobr
Stolb23 [73]

Answer:

2.464 cm above the water surface

Explanation:

Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.

We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3  cm^3):

weight of the block = 0.78 * 11.2^3  gr

Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.

So the weight of the volume of water displaced is:

weight of water = 1 * 11.2^2 * x

we make both weight expressions equal each other for the floating requirement:

0.78 * 11.2^3 = 11.2^2 * x

then x = 0.78 * 11.2 cm = 8.736 cm

This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:

11.2 cm - 8.736 cm = 2.464 cm

6 0
3 years ago
The bending of light when it changes media is called
olchik [2.2K]
That's called "refraction".
3 0
3 years ago
Read 2 more answers
Devise an exponential decay function that fits the given​ data, then answer the accompanying questions. Be sure to identify the
7nadin3 [17]

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

P=P_0e^{-kh}

where,

P_0 =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

P=\dfrac{1}{3}P_0

\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}

Now

P=\dfrac{1}{2}P_0

ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft

The altitude will be 22145.27733 ft

P=0.02P_0

0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft

The elevation is 124984.76055 ft

6 0
3 years ago
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