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3 years ago

the olympic swimmer swims to the end of the 50 m pool and back 4 times. calculate the distance covered.

Physics

1 answer:

Ludmilka [50]3 years ago

7 0

Answer:

400 m

Explanation:

The swimmer swims 50 meters to one end of the pool but has to swim back therefore you double 50 which would be 100 meters. Then you have to multiply 100 by 4 since the swimmer did it 4 times.

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What is the temperature of a 3.72 mm cube (e=0.288) that radiates 56.6 W?

blsea [12.9K]

Answer:

The temperature is 2541.799 K

Explanation:

The formula for black body radiation is given by the relation;

Q = eσAT⁴

Where:

Q = Rate of heat transfer 56.6

σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)

A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²

e = emissivity = 0.288

T = Temperature

Therefore, we have;

T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴

T = 2541.799 K

The temperature = 2541.799 K.

7 0

3 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago

The earth has a vertical electric field at the surface, pointing down, that averages 119 N/C . This field is maintained by vario

velikii [3]

Answer:

$q=5.37*10^{5}C$

Explanation:

If we assume that the Earth is a spherical conductor, according to Gauss's Law, the electric field is given by:

$E=\frac{kq}{r^2}$

Here k is the Coulomb constant, the excess charge on the Earth's surface and r its radius. Solving for q:

$q=\frac{Er^2}{k}\\q=\frac{119\frac{N}{C}(6.371*10^6m)^2}{8.99\frac{N\cdot m^2}{C^2}}\\q=5.37*10^{5}C$

5 0

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A car veers off course and runs straight into a brick wall. This is an example

ValentinkaMS [17]

Short time large force

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Professor blossom expects her students to be on time for class holds, them for the entire class period and has firm deadlines se

defon

Answer:

A dictatorship or tyranny.

Explanation:

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3 years ago