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3 years ago

Describe a radio and its function

Physics

2 answers:

cupoosta [38]3 years ago

8 0

Radio is usually used to hear news or music

Lubov Fominskaja [6]3 years ago

3 0

Answer:

Radio, sound communication by radio waves, usually through the transmission of music, news, and other types of programs from single broadcast stations to multitudes of individual listeners equipped with radio receivers. ... About 1945 the appearance of television began to transform radio's content and role.

Explanation:

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3 years ago

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A computational model predicts the maximum potential energy a roller coaster car can have given its mass and its speed at the lo

andreyandreev [35.5K]

Answer:

The prediction for its maximum potential energy is 109,375 J

Explanation:

Given;

mass of the coaster car, m = 350 kg

speed of the coaster car at the lowest point, v = 25 m/s

The coaster car will have maximum kinetic energy at the lowest point and based on law of conservation of mechanical energy, the maximum kinetic energy of the coaster car at the lowest point will be equal to maximum potential energy at the highest point.

$K.E_{max} = P.E_{max}$

$K.E_{max} = \frac{1}{2} mv^2\\\\K.E_{max} = \frac{1}{2} (350)(25)^2\\\\K.E_{max} =109,375 \ J$

$Thus, P.E_{max} = 109,375 \ J$

Therefore, the prediction for its maximum potential energy is 109,375 J

3 0

3 years ago

At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

lbvjy [14]

Answer:

$F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

$|F_T|=2\sqrt{34}k\frac{Q^2}{L}$

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

$F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]$

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

$F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

and the magnitude is:

$|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}$

the direction is:

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

4 0

3 years ago