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VikaD [51]
2 years ago
3

Radiometric dating of a magnetic anomaly stripe of rock that is 225 km away from the mid-ocean ridge axis gives an age of 9 mill

ion years. Assuming a constant rate, seafloor spreading in this area occurs at a rate of __________.
A. 5 cm per year
B. 1,012.5 km per year
C. 20,000 cm per year
D. 50 km per year
Physics
1 answer:
VashaNatasha [74]2 years ago
7 0

Answer:

The seafloor spreading in this area occurs at a rate of 5 cm per year.

Explanation:

It is given that, the radiometric dating of a magnetic anomaly stripe of rock that is 225 km away from the mid-ocean ridge axis gives an age of 9 million years.

We need to find the rate at which the seafloor is spreading in this area. The total distance will be, d = 225 km × 2 = 450 km

Time, t = 9 million years

The rate at which the seafloor is spreading in this area is given by :

v=\dfrac{d}{t}

v=\dfrac{4.5\times 10^7\ cm}{9\times 10^6\ year}

v=5\ cm/year

So, the seafloor spreading in this area occurs at a rate of 5 cm per year. Hence, this is the required solution.

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How can controlling the way light bends and reflects be used to help people?
Vinil7 [7]

too much sun is dangerous for humans and can cause cancer so it's important that light is reflected for example a pool reflects water back to space that is why water sometimes is cold because it reflects light

4 0
2 years ago
A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved
MariettaO [177]

Answer:

4.17 m/s

Explanation:

To solve this problem, let's start by analyzing the vertical motion of the pea.

The initial vertical velocity of the pea is

u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s

Now we can solve the problem by applying the suvat equation:

v_y^2-u_y^2=2as

where

v_y is the vertical velocity when the pea hits the ceiling

a=g=-9.8 m/s^2 is the acceleration of gravity

s = 1.90 is the distance from the ceiling

Solving for v_y,

v_y = \sqrt{u_y^2+2as}=\sqrt{(6.90)^2+2(-9.8)(1.90)}=3.22 m/s

Instead, the horizontal velocity remains constant during the whole motion, and it is given by

v_x = u cos \theta = (7.39)(cos 69.0^{\circ})=2.65 m/s

Therefore, the speed of the pea when it hits the ceiling is

v=\sqrt{v_x^2+v_y^2}=\sqrt{2.65^2+3.22^2}=4.17 m/s

5 0
2 years ago
A 1150 kg car is on a 8.70° hill. using x-y axis tilted down the plane, what is the x-component of the weight?
Fed [463]
I assume the x-y axis are tilted such that the x-axis is parallel to the surface of the hill while the y-axis is perpendicular to it.

In this case, the x-component of the weight is given by:
W_x =mg \sin \theta
where
m is the mass of the car
g is the acceleration of gravity
\theta is the angle of the hill

Substituting numbers into the formula, we find
W_x=(1150 kg)(9.81 m/s^2)(\sin 8.70^{\circ})=1706 N
6 0
2 years ago
A 2.0-m-tall man is 5.0 m from the converging lens of a camera. His image appears on a detector that is 50 mm behind the lens. H
ladessa [460]

Answer:

20 cm

Explanation:

We can solve the problem by using the magnification equation:

M=\frac{h_i}{h_o}=-\frac{q}{p}

where

h_i is the size of the image

h_o = 2.0 m is the height of the real object (the man)

q=50 mm =0.050 m is the distance of the image from the lens

p = 5.0 m is the distance of the object (the man) from the lens

Solving the formula for h_i, we find

h_i = -\frac{q}{p}h_o=-\frac{0.050 m}{5.0 m}(2.0 m)=-0.02 m = -20 cm

And the negative sign means the image is inverted.

6 0
2 years ago
To practice Problem-Solving Strategy 29.1: Faraday's Law. A metal detector uses a changing magnetic field to detect metallic obj
ExtremeBDS [4]

Answer:

1.138\times 10^{-3}V

Explanation:

Apply Faraday's Newmann Lenz law to determine the induced emf in the loop:

\epsilon=\frac{d\phi}{dt}

where:

d\Phi-variation of the magnetic flux

dt-is the variation of time

#The magnetic flux through the coil is expressed as:

\Phi=NBA \ Cos \theta

Where:

N- number of circular loops

A-is the Area of each loop(A=\pi r^2=\pi \times 5^2=78.5398)

B-is the magnetic strength of the field.

\theta=15\textdegree- is the angle between the direction of the magnetic field and the normal to the area of the coil.

\epsilon=-\frac{d(78.5398\times 10^{-3}NB \ Cos \theta)}{dt}\\\\=-(78.5398\times 10^{-3}N\ Cos \theta)}{\frac{dB}{dt}

\frac{dB}{dt}-=0.0250T/s is given as rate at which the magnetic field increases.

#Substitute in the emf equation:

=-(78.5398\times 10^{-3} m^2 \times 6\ Cos 15 \textdegree)\times 0.0250T/s\\\\=1.138\times 10^{-3}V

Hence, the induced emf is 1.138\times 10^{-3}V

4 0
2 years ago
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