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Andreas93 [3]
3 years ago
15

The Stronger the force that is exerted on an object,the ____ the change in velocity will be.

Physics
2 answers:
sesenic [268]3 years ago
7 0

Answer:

answer is stronger the change in velocity

leonid [27]3 years ago
5 0

osisisksj

#sorryneedpoints

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7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf
Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to  1 μV/m

Intrinsic impedance of air = 120 π  Ω

Intrinsic impedance of  water = \dfrac{120\pi}{\epsilon_r}

Using equation to solve the problem

  E(z) = E_0 e^{-\alpha\ z}

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}

\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}

\alpha =21.07\ Np/m

now ,

  1 \times 10^{-6} = 20 e^{-21.07\times z}

  e^{21.07\times z}= 20\times 10^{6}

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

5 0
3 years ago
Within a vacuum, the property to all electromagnetic waves is their
bezimeni [28]
The answer is D) Velocity
7 0
4 years ago
What do you need to calculate the mechanical advantage of a block and tackle
Ira Lisetskai [31]
In a block and tackle, some friction in the pulleys will reduce the mechanical advantage of the machine. To include friction in a calculation of the mechanical advantage of a block and tackle, divide the weight of the object being lifted by the weight necessary to lift it.

Hope this helps
7 0
3 years ago
Read 2 more answers
A ball with a mass of 5.0 g is moving at a speed of 2.0 m/s. Would doubling the mass or doubling the speed have a greater effect
Arte-miy333 [17]

Answer:

Yes

Explanation:

Kinetic energy is K.E1/2mv2 so that means it is directly proportional to mass and velocity.

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3 years ago
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When a voltage difference is applied to a piece of metal wire, a current flows through it. if this metal wire is now replaced wi
Nikitich [7]

The resistance of the cylindrical wire is R=\frac{\rho l}{A}.

Here R is the resistance, l is the length of the wire and A is the area of cross section. Since the wire is cylindrical A=\frac{\pi d^2}{4} .

Comparing two wires,

R_1=\frac{\rho_1 l}{A_1} \\ R_2=\frac{\rho_2 l}{A_2}

Dividing the above 2 equations,

\frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2}  \frac{A_2 }{A_1}  \\ \frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2}  \frac{d_2^2 }{d_1^2}  \\

Since d_2=2d_1

The above ratio is

\frac{R_1}{R_2}=\frac{1.68(10^{-8})  }{1.59(10^{-8}) } (4)\\ \frac{R_1}{R_2}=4.2264

We also have,

\frac{E/R_1}{E/R_2} =\frac{I_1}{I_2} \\ I_2=\frac{R_1}{R_2}I_1 \\ I_2=4.23I_1

The current through the Silver wire will be 4.23 times the current through the original wire.

8 0
3 years ago
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