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Naddik [55]
3 years ago
12

A rock is dropped from the edge of a cliff into a pool of water. Assume free-fall acceleration is 10 m/s per second, and air res

istance can be neglected. If the rock falls for 2.0 seconds before it hits the water, the height of the cliff above the pool is ___________.
Physics
1 answer:
il63 [147K]3 years ago
7 0

Answer:

20 m

Explanation:

From the equation of motion,

S = ut+1/2gt²................................. Equation 1

Where S = Height, u = initial velocity, t = time, g = acceleration due to gravity.

Note: Because the rocked is being dropped from a height, acceleration due to gravity is positive (g), and initial velocity (u) is negative

Given: t = 2.0 s, g = 10 m/s², u = 0 m/s (dropped from height)

Substituting into equation 1

S = 0(2) + 1/2(10)(2)²

S = 5(4)

S = 20 m

Hence the height of the the cliff above the pool is 20 m

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effort

Explanation:

effort is the answer

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____ is the detection of halo objects like brown dwarfs when their gravitational fields concentrate the light of distant sources
maksim [4K]

Answer:

Microlensing.

Explanation:

This techniques is called Microlensing.

Microlensing is a method of gravitational lensing where light from a backdrop point of origin is curved to develop distorted, numerous and/or lightened images by the gravity field of a foreground lens.

This method is very effective in discovering planets that are far-far from earth.It is actually an astronomical effect that was predicted by Albert Einstein's general theory of relativity.

7 0
3 years ago
Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27
densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

\rho=n\times\dfrac{m}{V_{c}\times N_{A}}

n=\dfrac{\rho\timesV_{c}\times N}{m}....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100

\rho=5.98

We calculate the mass of the alloy

m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100

m=111.15

Put the value into the equation (I)

n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}

n=1.99\approx 2\ atoms/cell

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0
4 years ago
Determine the value of the information in scientific notation. The moon is approximately 3. 63 × 108 meters from Earth. What is
Usimov [2.4K]

Answer:

d

Explanation:

7 0
2 years ago
She sights two sailboats going due east from the tower. The angles of depression to the two boats are 42o and 29o. If the observ
Reika [66]

Answer:

The boats are  934.65 feet apart

Explanation:

Given:

The angles of depression to the two boats are 42 degrees and 29 degrees

Height of the observation deck i =  1,353 feet

To Find:

How far apart are the boats (y )= ?

Solution:

<em><u>Step 1 : Finding the value of x(Refer the figure attached)</u></em>

We can use the tangent ratio to find the x value

tan(42^{\circ}) = \frac{1353}{x}

x = \frac{1353}{tan(42^{\circ}) }

x = 590.47 feet

<em><u>Step 2 : Finding the value of  z (Refer the figure attached)</u></em>

tan(29^{\circ}) = \frac{1353}{z }

z  = \frac{1353}{tan(29^{\circ})}

z = 1525.12  feet

<em><u>Step 3 : Finding the value of  y (Refer the figure attached</u></em>)

y =  z -x

y = 1525.12 - 590.47

y = 934.65 feet

Thus the two boats are 934.65 feet apart

3 0
3 years ago
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