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IrinaK [193]
3 years ago
7

A wave has a frequency of 875 hz and a wavelength of 352 m. At what speed is this wave traveling 

Physics
1 answer:
Sonbull [250]3 years ago
5 0

Answer:

308,000 or 30.8×10^3

Explanation:

v=f×lamda

v is ?, f is 875Hz, lamda is 352m

v=875×352

v=308,000

v=30.8×10^3 m/s

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An object starts from rest and uniformly accelerates at a rate of 1.25 m/s2 for 7.0 seconds.
stich3 [128]

Explanation:

Since its accelerating, the velocity vs time graph is linear

For displacement we need initial velocity (which is zero because it starts from rest) and final velocity (which is calculatee thro acceleration formula

A= (vf - vi)/t

a= vf-0/t

1.25=vf / 7

1.25*7=vf

8.75 = vf

Now for displacement plug all the values in

X = 1/2(vf-vi)/t formula

The displacement (x) is 30.625 m

For part 3, we know new displacement that is 22m , the final and initial velocities are the same so just plug in the values for same formula above

The answer is t = 5.02

Im pretty sure all the answers are correct

8 0
2 years ago
"Pluto Is Changing" talks about sheets of frozen nitrogen on Pluto. Tell what is happening to these sheets of ice
drek231 [11]
What is Pluto is changing exactly <span />
7 0
3 years ago
How fast should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?A)
Elan Coil [88]

Answer:

Option (D) is correct.

Explanation:

Let the speed is v.

\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c

Option (D) is correct.

4 0
3 years ago
Ch 27 HW Exercise 27.12 10 of 20 Constants A horizontal rectangular surface has dimensions 2.80 cm by 3.15 cm and is in a unifor
deff fn [24]

Answer:

Magnetic field, B = 0.88 T

Explanation:

It is given that,

The dimension of rectangular surface is 2.80 cm by 3.15 cm. The area of rectangular surface is, A=8.82\ cm^2=0.000882\ m^2

Angle between the uniform magnetic field and the horizontal, \theta=31

Magnetic flux, \phi=4\times 10^{-4}\ Wb

Let B is the magnitude of magnetic field in which the rectangular surface is placed. It is given by :

\phi=BA\ cos\theta

\theta is the angle between magnetic field and the area

Here, \theta=90-31=59^{\circ}

B=\dfrac{\phi}{A\ cos\theta}

B=\dfrac{4\times 10^{-4}}{0.000882\times cos(59)}

B = 0.88 T

So, the magnitude of magnetic field is 0.88 T. Hence, this is the required solution.

7 0
3 years ago
If an object has a mass of 8 kg, what is its approximate weight on Earth?
Fittoniya [83]

Answer:

10^24 kg

Explanation:

3 0
3 years ago
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