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Lesechka [4]
2 years ago
13

Seafloor spreading is a process that occurs along ___________1____________. At ___________2____________ boundary, two plates mov

e away from each other resulting to spreading of seafloor, thereby creating new ___________3____________ crust. Near the mid-oceanic ridge _________4__________ rocks are found and far from it are older.
Physics
1 answer:
sergejj [24]2 years ago
5 0

Answer:

where is the answer choices

Explanation:

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Determine the projection (magnitude and sign), or component, of vector v1 along the direction of vector v2. Your answer could be
professor190 [17]

Answer:

- 1.07 ft

Explanation:

V1 = (-5, 7, 2)

V2 = (3, 1, 2)

Projection of v1 along v2, we use the following formula

=\frac{\overrightarrow{V1}.\overrightarrow{V2}}{V2}

So, the dot product of V1 and V2 is = - 5 (3) + 7 (1) + 2 (2) = -15 + 7 + 4 = -4

The magnitude of vector V2 is given by

= \sqrt{3^{2}+1^{2}+2^{2}}=3.74

So, the projection of V1 along V2 = - 4 / 3.74 = - 1.07 ft

Thus, the projection of V1 along V2 is - 1.07 ft.

so we need to find the direction of v2

7 0
3 years ago
Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.
amm1812
-- Equations  #2  and  #6  are both the same equation,
and are both correct.

-- If you divide each side by  'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by  'frequency', you get Equation #3,
which is also correct. 
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.
7 0
3 years ago
Read 2 more answers
The equation v=F^aL^M^-c were shows the relationship between velocity of the waves tensile force in the string length, L and mas
Travka [436]
I literally looked everywhere for the answer, and I still found nothing. I hope you get it right. Sorry.
8 0
3 years ago
An athlete whirls a 7.00 kg hammer 1.8 m from the axis of rotation in a horizontal circle, as
iogann1982 [59]

Answer:

A-500 N

Explanation:

The computation of the tension in the chain is shown below

As we know that

F = ma

where

F denotes force

m denotes mass = 7

And, a denotes acceleration

Now for the acceleration we have to do the following calculations

The speed (v) of the hammer is

v = Angular speed × radius

where,

Angular seed = 2 × π ÷ Time Period

So, v = 2 × π × r ÷ P

v = 2 × 3.14 × 1.8 ÷ 1

= 11.304 m/s

Now

a = v^2 ÷ r

= 70.98912 m/s^2

Now the tension is  

T = F = m × a

= 7 × 70.98912

= 496.92384 N

= 500 N

5 0
2 years ago
An ancient club is found that contains 100 g of pure carbon and has an activity of 6.5 decays per second. Determine its age assu
never [62]

Answer:

The age of living tree is 11104 years.

Explanation:

Given that,

Mass of pure carbon = 100 g

Activity of this carbon is = 6.5 decays per second = 6.5 x60 decays/min =390 decays/m

We need to calculate the decay rate

R=\dfrac{-dN}{dt}=\lambda N=\dfrac{0.693}{t_{\frac{1}{2}}}N....(I)

Where, N = number of radio active atoms

t_{\frac{1}{2}}=half life

We need to calculate the number of radio active atoms

For N_{12_{c}}

N_{12_{c}}=\dfrac{N_{A}}{M}

Where, N_{A} =Avogadro number

N_{12_{c}}=\dfrac{6.02\times10^{23}}{12}

N_{12_{c}}=5.02\times10^{22}\ nuclie/g

For N_{c_{14}}

N_{c_{14}}=1.30\times10^{-12}N_{12_{c}}

N_{c_{14}}=1.30\times10^{-12}\times5.02\times10^{22}

N_{c_{14}}=6.526\times10^{10}\ nuclei/g

Put the value in the equation (I)

R=\dfrac{0.693\times6.526\times10^{10}\times60}{5700\times3.16\times10^{7}}

R=15.0650\ decay/min g

100 g carbon will decay with rate

R=100\times15.0650=1507\ decay/min

We need to calculate the total half lives

(\dfrac{1}{2})^{n}=\dfrac{390}{1507}

2^n=\dfrac{1507}{390}

2^n=3.86

n ln 2=ln 3.86

n=\dfrac{ln 3.86}{ln 2}

n =1.948

We need to calculate the age of living tree

Using formula of age

t=n\times t_{\frac{1}{2}}

t=1.948\times5700

t=11103.6 =11104\ years

Hence, The age of living tree is 11104 years.

5 0
3 years ago
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