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Vedmedyk [2.9K]
3 years ago
8

1 The Earth's crust is made up of plates that can move. Which of the following topographic features could be directly formed by

the movement of the Earth's plates? I. midocean ridges II. mountains III. sand dunes IV. coral reefs A. I, II, and III only B. II, III, and IV only C. I and II only D. II and III only
Physics
1 answer:
olga_2 [115]3 years ago
7 0
Can you write this better and from my info of what i know i know that one way is the trenches in the ocean like marianas trench  <span />
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EXPLAIN HOW ENERGY IS TRANSMITTED THROUGH A MEDIUM
jek_recluse [69]
I say it helped then because TrueType had room
5 0
3 years ago
Read 2 more answers
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

3 0
3 years ago
Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?
shtirl [24]

Answer:

L=6.21m

Explanation:

For the simple pendulum problem we need to remember that:

\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0,

where \theta is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

\omega^{2}=\frac{g}{L},

where \omega is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

\omega=\frac{2\pi}{T},

where T is the oscillation period. Now, we can easily solve for L:

(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m

3 0
3 years ago
A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is
Maru [420]

Answer:

A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is the surface charge density of the sheet"?

Explanation:

Surface charge density is a measure of how much electric charge is accumulated over a surface. It can be calculated as the charge per unit area.

We will convert all parameters in SI units.

Charge = Q = -5.02nC

Q  = -5.02×10^{-9}C

As it is clear from question that Sheet is a square (All sides will be of equal length)

Area = A = (21.8×10^{-2}m) (21.8×10^{-2}m)  = 4.75×10^{-4}m²

A  = 4.75×10^{-4}m²

Surface charge density = Q/A

Surface charge density = (-5.02×10^{-9}C)/(4.75×10^{-4}m²)

Surface charge density = -1.057×10^{-5} Cm^{-2}

3 0
3 years ago
A cart travels 4.00 meters east and then4.00 meters north. determine the magnitude ofthe cart’s resultant displacement.
gogolik [260]
^
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Square root of (4^2 + 4^2) = 4*squareRoot(2)
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