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2 years ago

For each scenario below, choose the best graph. (a) Maria bikes from home to work.

Physics

1 answer:

BlackZzzverrR [31]2 years ago

5 0

A line graph that plots either speed or distance against time.

Really, the type of graph depends on what quantity you want the graph to show.

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What controls whether a lava flow is aa or pahoehoe

nevsk [136]

Explanation:

Basaltic lava

Basaltic lava generally takes two distinct forms known by the Hawaiian terms pahoehoe and aa. Pahoehoe has a smooth wavy surface that resembles twisted rope. It advances by extruding molten toes of lava beneath a thin, flexible crust. As it travels pahoehoe lava often changes to blocky flows called aa.

7 0

3 years ago

The infant's tendency to turn its head toward things that touch its cheek is known as the

katovenus [111]

Answer:

I think it is Rooting Reflex

4 0

3 years ago

Read 2 more answers

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

3 0

3 years ago

A 0.73-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar

Tasya [4]

Answer:

A) B = 5.4 10⁻⁵ T, B) the positive side of the bar is to the West

Explanation:

A) For this exercise we must use the expression of Faraday's law for a moving body

fem = $- \frac{d \phi }{dt}$

fem = $- \frac{d (B l y}{dt}= - B l v$ - d (B l y) / dt = - B lv

B = $- \frac{fem}{l \ v}$

we calculate

B = - 7.9 10⁻⁴ /(0.73 20)

B = 5.4 10⁻⁵ T

B) to determine which side of the bar is positive, we must use the right hand rule

the thumb points in the direction of the rod movement to the south, the magnetic field points in the horizontal direction and the rod is in the east-west direction.

Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West

4 0

3 years ago

Four equal masses m are so small they can be treated as points, and they are equallyspaced along a long, stiff mass less wire. T

gavmur [86]

The moment of inertia of a point mass about an arbitrary point is given by:

I = mr²

I is the moment of inertia

m is the mass

r is the distance between the arbitrary point and the point mass

The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.

The total moment of inertia of the system is the sum of the moments of each mass, i.e.

I = ∑mr²

The moment of inertia of each of the two inner masses is

I = m(ℓ/2)² = mℓ²/4

The moment of inertia of each of the two outer masses is

I = m(3ℓ/2)² = 9mℓ²/4

The total moment of inertia of the system is

I = 2[mℓ²/4]+2[9mℓ²/4]

I = mℓ²/2+9mℓ²/2

I = 10mℓ²/2

I = 5mℓ²

4 0

3 years ago