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3 years ago

Which of the following forces can change the motion of an object? A. push B. shove C. pull D. all of these

Physics

1 answer:

Goshia [24]3 years ago

3 0

Answer: The answer is D all of these sorry if i am wrong

Explanation:

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An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average mag

Afina-wow [57]

Answer:

The induced emf in the loop is $7.35\times 10^{-4}\ V$

Explanation:

Given that,

Length of the wire, L = 1.22 m

It changes its shape is changed from square to circular. Then the side of square be its circumference, 4a = L

4a = 1.22

a = 0.305 m

Area of square, $A=a^2=(0.305)^2=0.0930\ m^2$

Circumference of the loop,

$C=2\pi r=L\\\\r=\dfrac{L}{2\pi}\\\\r=\dfrac{1.22}{2\pi}=0.194\ m$

Area of circle,

$A'=\pi r^2\\A'=\pi (0.194)^2\\\\A'=0.118\ m^2$

The induced emf is given by :

$\epsilon=\dfrac{\d\phi}{dt}\\\\\epsilon=\dfrac{\d(BA)}{dt}\\\\\epsilon=B\dfrac{A'-A}{t}\\\\\epsilon=0.125 \times \dfrac{0.118-0.0930}{4.25}\\\\\epsilon=7.35\times 10^{-4}\ V$

So, the induced emf in the loop is $7.35\times 10^{-4}\ V$

8 0

3 years ago

011 10.0 points

Sedbober [7]

<h2>The temperature of the air is 66.8° C</h2>

Explanation:

From the Newton's velocity of sound relationship , the velocity of sound is directly proportional to the square root of temperature .

In this case The velocity of sound = frequency x wavelength

= 798 x 0.48 = 383 m/sec

Suppose the temperature at this time = T K

Thus 383 ∝ $\sqrt{T}$ I

The velocity of sound is 329 m/s at 273 K ( given )

Thus 329 ∝ $\sqrt{273}$ II

Dividing I by II , we have

$\frac{383}{329}$ = $\sqrt{\frac{T}{273} }$

or $\frac{T}{273}$ = 1.25

and T = 339.8 K = 66.8° C

4 0

4 years ago

Una banda transportadora se utiliza para cargar canastos en un avión de carga, si µs = 0.5. ¿Cuál es el ángulo máximo de elevaci

Sonja [21]

No se creo es 45.7 de nada

7 0

3 years ago

An airplane is pushed 22 degrees eastward off its northward course by a jet stream traveling at a speed of 136.73 km/hr. The new

uranmaximum [27]

As the speed of airplane is change due to jet stream

So the net speed is given as

$v_{net} = v_{plane} + v_{stream}$

now we can rearrange it as

$v_{plane} = v_{net} - v_{stream}$

now by the formula of vector difference we have

$v_{plane} = \sqrt{v_{net}^2 + v_{stream}^2 - 2v_{net}v_{stream}cos\theta}$

now plug in all values

$v_{plane} = \sqrt{365^2 + 136.73^2 - 2* 365* 136.73*cos22}[tex]v_{plane} = 243.7 km/hr$

so above is the speed of the plane

3 0

3 years ago

g Using the absorbance vs wavelength spectrum where the 1st peak is slightly less intense than the 2nd peak, determine the wavel

Andre45 [30]

Answer:

The wavelength range is always used to know the probable material present

Explanation:

The wavelength variation from with the concentration shows the type of material in as much the Spectrometer is well initialized before running the sample. The peaks interval may have effect on band gap.

5 0

3 years ago