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Juli2301 [7.4K]
2 years ago
9

There is more than one answer

Physics
2 answers:
Sveta_85 [38]2 years ago
5 0

Explanation:

the answers are the first 3.

Soloha48 [4]2 years ago
3 0

Answer:

4.6

Explanation:

Ans is 4.6 b coz the difference b/w 4.6 and the other values is comparitively the largest.

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Is this right? plzz anwser soon
-Dominant- [34]

Answer:

yes

Explanation:

6 0
2 years ago
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I need help it is about physics
VMariaS [17]

Answer:

a force

Explanation:

Because if we apply force then only an object can slow down, speed up or change direction

6 0
3 years ago
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Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
Can you please answer these ASAP!
SVEN [57.7K]

Answer:

Let's explain this briefly.

Suppose that we have a piece of ice (this is, solid water) now we give energy to the piece of ice, so the temperature of the ice increases. There is a point where the piece of ice will start a change of phase, at this point the temperature of the ice stops increasing because all the energy we give to the ice is used in the change of phase.

Once we have a complete change of phase, the temperature can increase again, and now we will have liquid water.

If we keep increasing the temperature we will see this happen again, when we have the transition from liquid to gas.

(and a similar thing happen when we have a material in a given phase and we remove heat from the material).

In the images we can see the different changes of phase of water.

1) In the first image we can see the circle in a part where the temperature is constant, so the temperature does not change in this part, which means that there is a change of phase happening.

2) Here we have the circle in a diagonal line, so here the temperature is changing, meaning that we have an increase of temperature in this region.

3) Here we want to know what the x-axis represents, this should rerpesent the energy that is being given to the material (so in some parts we see that the temperature increases and in other parts we see that the material changes of phase)

Then here the correct option is heat over time.

4) The freezing point is the temperature in which the change of phase from liquid to solid happens (or solid to liquid).

In the graph we can see that this change of phase happens at the temperature T = -210°C

Then the correct option is -210°C (The last option)

4 0
2 years ago
For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate
faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue=h_1=0.200 m

Metal volume flow  rate,Q=0.03m^3/min

Q=\frac{0.03}{60}=5\times 10^{-4}m^3/s because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

d_=0.030 m

h_2=0

A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}

A_1=7.065\times 10^{-4} m^2

v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}

v_1=0.708 m/s

Pressure at the top and bottom of the sprue is atmospheric

h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}

Substitute the values

0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}

v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264

v_2=\sqrt{4.421264}=2.1 m/s

Q=A_2v_2

5\times 10^{-4}=A_2\times 2.1

A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2

Reynolds number=\frac{v_2D\rho}{\eta}

\eta=0.004 N.s/m^2

\rho=2700 kg/m^3

Substitute the values then we get

Reynolds number=\frac{2.1\times 0.03\times 2700}{0.004}

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0
3 years ago
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