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Sindrei [870]
3 years ago
14

Which image illustrates why dark clothing helps to keep you warm on a cool, sunny day?

Physics
2 answers:
xeze [42]3 years ago
7 0

Answer:

Explanation:

I suppose it has to do with the way the diagram is drawn. The heat does not reflect which makes both A and B incorrect.

C would have nothing to do with either reflection or refraction.

That only leaves D which is the answer.

strojnjashka [21]3 years ago
5 0

Answer: D

Explanation:

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The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
natulia [17]

Answer:

15.88°C I am not 100% sure this is right but I am 98% sure this IS right

7 0
3 years ago
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
HOW DO PROTONS, NEUTRONS AND ELECTRONS DIFFER?
aliina [53]
Protons are positively charged and neutrons are neutral whereas electrons are negatively charged.
6 0
3 years ago
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A small grinding wheel has a moment of inertia of 4.0*10-5kgm2. What net torque must be applied to the wheel for its angular acc
kvv77 [185]

Hi there!

We can use the rotational equivalent of Newton's Second Law:

\huge\boxed{\Sigma \tau = I \alpha}

Στ = Net Torque (Nm)

I = Moment of inertia (kgm²)

α = Angular acceleration (rad/sec²)

We can plug in the given values to solve.

\Sigma \tau = (4 * 10^{-5})(150) = \boxed{0.006 Nm}

4 0
3 years ago
Explain any two factors that affect the pressure at a point inside a liquid​
vredina [299]
The density of the body and the height or the depth of the body since the formula of liquid pressure is density x height gravity
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2 years ago
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