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2 years ago

What is the new kinetic energy of the 1900 kg ship on the right moving at 4 m/s?

Physics

2 answers:

Kay [80]2 years ago

8 0

Answer:

15,200J

Explanation:

The equation is 1/2 x m x v², where m= mass and v= velocity.

Elis [28]2 years ago

7 0

Answer:

soi nuevo

Explanation:me regalan puntos

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A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

UNO [17]

Answer:

$P_{avg} = 6.283*10^{-9} \ W$

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

$\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})$

where;

$E= \frac{d \phi}{dt}$

$E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t$

$E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}$

Replacing our values into above equation; we have:

$E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}$

$E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}$

$E_{rms} = 6.356*10^{-4} \ V$

Then the $P_{avg$ is calculated as:

$P_{avg} = \frac{E^2}{R}$

$P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}$

$P_{avg} = 6.283*10^{-9} \ W$

6 0

3 years ago

stephen buys a new moped . he travels 4km south and then 6km east. how far does he need to go to get back where he started??

stiks02 [169]

Answer:

My answer is 7.2 km

Explanation:

When Stephen goes to the south and then to the east, he is drawing a right triangle, where the 4 km and 6 km sides are the cathetus of a right triangle.

Then we use the Pithagorean theorem to solve this problem. We need to find the hypotenuse.

c² = a² + b²

c² = 4² + 6²

c² = 16 + 36

c² = 52

c = 7.2 km

7 0

3 years ago

the declaration of Independence discusses the protection of natural rights. These rights include the right life liberty and the

nikdorinn [45]

Answer:

i can't understand ur question

4 0

2 years ago

If a soap bubble is 115 nm thick, what wavelength is most strongly reflected at the center of the outer surface when illuminated

sp2606 [1]

Answer:

$611.8$ nm

Explanation:

$n_{oil}$ = Index of refraction of soap bubble = 1.33

$t$ = thickness of the soap bubble = 115 nm = 115 x 10⁻⁹ m

$\lambda$ = wavelength of light = ?

$m$ = order = 0

For reflection , the necessary condition is

$2 n_{oil} t = (m + 0.5) \lambda$

$2 (1.33)(115\times 10^{-9})= (0 + 0.5) \lambda$

$\lambda = 6.118\times 10^{-7}$

$\lambda = 611.8$ nm

7 0

2 years ago

If a car was cruising down the highway at 80 km/hr, how fast would the people inside the car be traveling?

wolverine [178]

Because the people in the car are attached to the vehicle, the people inside the vehicle are going the same speed as the vehicle.

Hope this helps! :)

6 0

2 years ago

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