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kobusy [5.1K]
4 years ago
14

If 320 J of work is done on a spring with a spring constant of 730 N/m, how far will it stretch?

Physics
1 answer:
Lubov Fominskaja [6]4 years ago
6 0

Answer:

0.94 m

Explanation:

EE = ½ kx²

320 J = ½ (730 N/m) x²

x = 0.94 m

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5. What is the density of 4.5 mL of a liquid that has a mass of 1.3 grams?
antoniya [11.8K]

Answer:

A. 0.289g/mL

Explanation:

Using the equation for density which is d = m/v  or density = mass/volume, we input 1.3g/4.5mL and get 0.289g/mL.

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A car accelerates from rest at 3.00 m/second squared What is the velocity after 5 secs ? What is the displacement after 5 second
andrey2020 [161]

Explanation:

after 5 seconds, the velocity is (5s)(3m/s²) = 15m/s

The displacement after 5s is

x=vo + (1/2)at²

x = 0 + (1/2)(3m/s²)(5s)(5s)

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Please help me!!!!!!!!!!Which is true about the actual mechanical advantage of a machine?
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What is the wavelength of a wave if its frequency is 256 Hz and speed of the wave is 350 m/s?
DiKsa [7]

Answer:

Explanation:

The equation for this is

f=\frac{v}{\lambda} where f is the frequency, v is the velocity, and lambda is the wavelength. Filling in:

256=\frac{350}{\lambda} and

\lambda=\frac{350}{256} which means that

the wavelength is 1.37 m, rounded to the correct number of significant digits.

7 0
3 years ago
A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th
Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

E=-\frac{d}{dt}(BACOS\alpha )\\

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using \pi r^{2} \\ where r is the radius of 5cm or 0.05m

A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\

since we no that the angle is at 0^{0}

we determine the magnitude of the magnetic filed

B=0.5e^{-t} \\t=2s

E=-(0.5e^{-2} * 0.00785)

E=-0.000532v\\

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A

I=2.6mA

6 0
4 years ago
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