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Oliga [24]
4 years ago
7

Where can volcanoes form? A. Where there are cracks in the crust B. Along fault lines C. Where the crust is thin and can be rupt

ured D. All of the above
Physics
2 answers:
schepotkina [342]4 years ago
8 0
The answer is all of the above
jasenka [17]4 years ago
6 0
Hot spots, Divergent plate boundaries (such as rifts and mid-ocean ridges), and. Convergent plate boundaries (subduction zones)
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Which landscape will erode more quickly? Assume the same precipitation and average temperature in each landscape.
Artemon [7]

Answer:

  B.  Landscape B

Explanation:

Shale is fine sediment pressed together to form rock.

Sandstone is larger (sand-grain-sized) sediment cemented together to form rock.

Shale erodes faster, as evidenced by the second attachment. That attachment shows erosion of a rock face consisting of interbedded shale and sandstone. The shale has receded significantly, leaving the sandstone layers with space between them.

5 0
3 years ago
C
kkurt [141]

Which letter represents the position at which the basketball has the greatest potential energy? Explain. Point C. At this point, which is the highest point, all of the ball's energy is gravitational potential energy.

6 0
3 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
Find the net downward force on the tank's flat bottom, of area 1.60 m2 , exerted by the water and air inside the tank and the ai
Dovator [93]

Answer:

The net downward force on the tank is 1.85\times10^{5}\ N

Explanation:

Given that,

Area = 1.60 m²

Suppose the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 meters per second per second. The pressure at the surface of the water will be 150 K Pa , and the depth of the water will be 14.4 m . The pressure of the air in the building outside the tank will be 88.0 K Pa.

We need to calculate the net downward force on the tank

Using formula of formula

F=(P+\rho\times g\times h-P_{out})A

Where, P = pressure

g = gravity at mars

h = height

A = area

Put the value into the formula

F=(150\times10^3+1.00\times10^3\times3.71\times14.4-88.0\times10^{3})\times1.60

F=1.85\times10^{5}\ N

Hence, The net downward force on the tank is 1.85\times10^{5}\ N

8 0
4 years ago
A rope with a mass density of 1 kg/m has one end tied to a vertical support. You hold the other end so that the rope is horizont
PIT_PIT [208]

Answer:

v' = 2.83 m/s

Explanation:

Velocity of wave in stretched string is given by the formula

v = \sqrt{\frac{T}{\mu}}

here we know that

T = 4 N

also we know that linear mass density is given as

\mu = 1 kg/m

so we have

v = \sqrt{\frac{4}{1}} = 2 m/s

now the tension in the string is double

so the velocity is given as

v' = \sqrt{\frac{8}{1}} = 2\sqrt2 m/s

v' = 2.83 m/s

4 0
3 years ago
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