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Alja [10]
3 years ago
12

A snowball that will be used to build a snowman is at the top of a only hill. If the

Physics
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

h = 24.11 m

Explanation:

Given that,

The potential energy of the snowball is 520 J

The mass of the snowball is 2.2 kg

We need to find the height of the hill. The potential energy of an object is given by the formula as follows :

E=mgh

g is acceleration due to gravity

h is height of the hill

h=\dfrac{E}{mg}\\\\h=\dfrac{520}{2.2\times 9.8}\\\\h=24.11\ m

So, the height of the hill is 24.11 m.

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In diving to a depth of 248 m, an elephant seal also moves 296 m due east of his starting point. What is the magnitude of the se
Vinvika [58]

Answer:

The displacement is 386.16m

Explanation:

A seal dives to a depth of 248m. To find displacement, we must calculate the resultant vectors which will give us the displacement

R= sqrt(vector1+vector2)

Since this is a right angle triangle

R= sqrt(248^2 + 296^2)

R= sqrt(149120)

R= 386.16m

Displacement = 386.16m

4 0
3 years ago
Choose a substance you're familiar with. what are its physical and chemical properties? How would you measure its density? What
Gekata [30.6K]
<span>you can look at magnesium, it can react with oxygen to form oxides. (chemical) it is malleable and a solid at room temperature. (physical)

</span><span>to measure its density, the mass and volume can be worked out and from this density too. look up the equation, it is quite easy :) 
</span><span>physical changes -- it can be melted, and oxidized 

</span><span>the chemical changes of oxidation magnesium looses electrons to form oxides, this is a chemical reaction- chemical change..--- use to get the density use (rho) or density D = M/V</span>
6 0
2 years ago
Two lasers, one red (with wavelength 633.0 nmnm) and the other green (with wavelength 532.0 nmnm), are mounted behind a 0.150-mm
Ratling [72]

(a) The screen  is 3.20m from the split.

(b) The closest minima for green, distance Δy = 0.45 cm.

When a wave hits a barrier or opening, numerous events are referred to as diffraction. It is described as the interference or bending of waves via an aperture or around the corners of an obstruction into the area that forms the geometric shadow of the obstruction or aperture.

(a)Equation of minima = sinθ  = mλ/α

Given, m = 3, λ = 6.33X10⁻⁷, α = 0.00015

Putting the values in formula to get θ.

  θ = sin⁻¹ ( \frac{3 X 6.33X10^{-7} }{0.00015} ) = 0.01266 rad

triangle need to be drawn to find relationship between θ, y$ and L

tan(θ) = y/L  where; y = 4.05 cm

L = y/tan(θ) = 3.20

Hence, the screen is 3.20m from the split.

(b) Find the closest minima for green

minima equation is sinθ  = mλ/α where, m = 4 (minima with smallest distance)

sinθ  = 4λ/α

θ = sin⁻¹ (\frac{4X6.33X10^{-7} }{0.00015}) = 0.01688 rad

Calculate L using

tanθ = y/L

  L = 4.5 cm

From equation subtract y₃ from y:

                 4.50 cm - 4.05 cm = 0.45 cm

Hence, distance Δy = 0.45 cm.

Learn more about the Diffraction with the help of the given link:

brainly.com/question/12290582

#SPJ4

I understand that the question you are looking for is "Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the ot

Question

Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the other side of the slit is a white screen. When the red laser is turned on, it creates a diffraction pattern on the screen.

a. The distance y3,red from the center of the pattern to the location of the third diffraction minimum of the red laser is 4.05 cm. How far L is the screen from the slit? Express this distance L in meters to three significant figures.

b. With both lasers turned on, the screen shows two overlapping diffraction patterns. The central maxima of the two patterns are at the same position. What is the distance Δy between the third minimum in the diffraction pattern of the red laser (from Part A) and the nearest minimum in the diffraction pattern of the green laser?

5 0
1 year ago
calculate the frequency and time period of sound wave of 35 m wave length propagating at a speed of 3500m/s​
zheka24 [161]

Answer:

Frequency = 100 Hz

Time period = 0.01 sec

Explanation:

f= 3500/35 = 100

T = 1/f = .01

5 0
2 years ago
Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.
vlabodo [156]
GAVNSTVVHUYCZ swffggff
4 0
2 years ago
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