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3 years ago

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Physics

1 answer:

Katyanochek1 [597]3 years ago

4 0

1. 2 way radio

2. radio waves

3. communication

4. convert the voltage from a transmitter into a radio signal

5. signal strength refers to the transmitter power output

6. sorry, not so sure.. Though it might be the waves.

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2Mg+2HCI TO 2MgCI+2H2

motikmotik

2Mg + 2HCl →2MgCl + 2H₂

Balanced is:

4Mg + 4HCl → 4MgCl + 2H₂

I Hope I Helped

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6 0

3 years ago

A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a

Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C = 50 - x cm

We know that ,electric field is given as

$E=K\dfrac{q}{r^2}$

$K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\$

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0

2 years ago

Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rath

Sophie [7]

Answer:

Change/ Potential

Explanation:

Work is the amount of energy required to perform an action that is for a force to cause a displacement.

From work-energy theorem, work done by body is equal to change in its kinetic energy.

Work of gravity is basically the potential energy stored in the body due to gravity. From the law of conservation of mechanical energy, increased kinetic energy comes from the change of the potential energy of the stone.

8 0

2 years ago

Projectile Motion—A tennis ball is thrown out a window 28 m above the ground at an initial velocity of 15.0 m/s and 20.0° below

NNADVOKAT [17]

Answer:

The distance will be x = 41.7 [m]

Explanation:

We must first find the components in the x & y axes of the initial velocity.

$(v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s]$

The acceleration is the gravity acceleration therefore.

g = 9.81 [m/s^2]

Now we can calculate how long it takes to fall.

$y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s]$

With this time we can find the horizontal distance that runs the projectile.

$x=(v_{o})_{x}*t\\x=14.09*2.96\\x=41.7[m]$

5 0

2 years ago

Using the law of conservation of angular momentum, estimate how fast a collapsed stellar core would spin if its initial spin rat

Nataly_w [17]

Answer:

$\omega_{f} = 1000000\,\frac{rev}{day}$

Explanation:

The law of conservation of angular momentum states that angular momentum remains constant when there is no external moment or forces applied to the system. Let assume that star can be modelled as an sphere, then:

$\frac{2}{5}\cdot M\cdot R_{o}^{2} \cdot \omega_{o} = \frac{2}{5}\cdot M\cdot R_{f}^{2} \cdot \omega_{f}$

The final angular speed is:

$\omega_{f} = \omega_{o}\cdot (\frac{R_{o}}{R_{f}})^{2}$

$\omega_{f} = (1\,\frac{rev}{day} )\cdot (\frac{10000\,km}{10\,km} )^{2}$

$\omega_{f} = 1000000\,\frac{rev}{day}$

3 0

2 years ago