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Sloan [31]
3 years ago
14

In comparison with other ocean basins, major sedimentary features such as continental rises and abyssal plains are relatively ra

re in the Pacific. The primary reason for their scarcity in the Pacific is that __________.
Physics
1 answer:
Maksim231197 [3]3 years ago
3 0

Answer:

Why are continental rises and abyssal plains relatively rare in the Pacific? This is because the extensive system of trenches along the active margins of the Pacific, trap much of the sediments flowing off the continents, preventing them from building the broad, flat abyssal plains typical of the Atlantic ocean basins.

You might be interested in
A block is given a very brief push up a 20.0° frictionless incline to give it an initial speed of 12.0 m/s. (a) how far along th
shtirl [24]
The incline is frictionless, this means we can use the conservation of energy: the initial kinetic energy of the block
K= \frac{1}{2}mv^2
is converted into gravitational potential energy
U=mgh
where h is the height reached by the block as it stops. By equalizing the two formulas, we get
\frac{1}{2} mv^2=mgh
h= \frac{v^2}{2g}= \frac{(12.0 m/s)^2}{2(9.81 m/s^2)} =7.3 m

However, this is the maximum height reached by the block. The distance along the surface of the plane is given by:
d= \frac{h}{\sin 20^{\circ}}= \frac{7.3 m}{\sin 20^{\circ}}=21.3 m
4 0
4 years ago
A particle with charge −5 µC is located on
Nataly [62]

Answer:

36.25 N

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=9\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the magnitude of the two charges

r is the separation between the two charges

Moreover:

- The force is repulsive if the two  charges have same sign

- The force is attractive if the two charges have opposite sign

In this problem, we have 3 charges:

q_1=-5\mu C = -5\cdot 10^{-6}C is the charge located at x=+10 cm = +0.10 m

q_2=+6\mu C=+6\cdot 10^{-6}C is the charge located at x=-8 cm =-0.08 m

q_3=+2\mu C=+2\cdot 10^{-6}C is the charge located at x=-2 cm=-0.02 m

The force between charge 1 and charge 3 is:

F_{13}=\frac{kq_1 q_3}{(x_1-x_3)^2}=\frac{(9\cdot 10^9)(5\cdot 10^{-6})(2\cdot 10^{-6})}{(0.10-(-0.02))^2}=6.25 N

And since the two charges have opposite sign, the force is attractive, so the force on charge 3 is to the right (towards charge 1).

The force between charge 2 and charge 3 is:

F_{23}=\frac{kq_2 q_3}{(x_2-x_3)^2}=\frac{(9\cdot 10^9)(6\cdot 10^{-6})(2\cdot 10^{-6})}{(-0.08-(-0.02))^2}=30.0 N

And since the two charges have same sign, the force is repulsive, so the force on charge 3 is to the right (away from charge 2).

So the two forces on charge 3 have same direction (to the right), so the net force is the sum of the two forces:

F=F_{13}+F_{23}=6.25+30.0=36.25 N

8 0
4 years ago
In pulling two identical carry-on bags through the airport, Mr. Myers and his 13 year old grandson, Vincent, do the same amount
Novay_Z [31]

Answer:

Mr Myers and his son use the same force to pull the bags between the gates

Explanation:

The work done by Mr. Myers in pulling the carryon bags = The work done by his 13 year old grandson in pulling the identical bag

Let F₁ represent the force used by Mr Myers, and let F₂ represent the force F₂ used by his grandson

Let d represent the distance through the gate

Therefore, given that Work done, W = Force, F × Distance, we have;

The work done by Mr Myers between the gates, W₁ = F₁ × d

The work done by his grandson between the gates, W₂ = F₂ × d

Where, the work done by both Mr Myers and his grandson are equal, we have;

W₁ = W₂ and therefore, F₁ × d = F₂ × d, which gives;

F₁ = F₂, the force used by both Mr Myers and his son between the gates are equal.

5 0
3 years ago
A quantum-mechanical system initially in its ground level absorbs a photon and ends up in the first excited state. The system th
Ludmilka [50]

Answer:

(b) a hydrogen atom

Explanation:

Wavelength is inversely proportional to energy. So we have to find the system for which the second photon has smaller energy than the first one.

For the harmonic oscillator, the energy level spacing remains the same as the quantum number increases.

For the hydrogen atom, the energy level spacing decreases as the quantum number increases.

For the particle in a box, the energy level spacing increases as the quantum number increases.

5 0
3 years ago
You are a world-famous physicist-lawyer defending a client who has been charged with murder. It is alleged
natta225 [31]

Answer:

It was not fired from the client's gun because the chair slid only 3 centimeters . If it had been fired from the client's gun the chair would slid 25.82 centimeters.

Explanation:

According to the law of conservation of momentum the momentum of the system before collision must be equal to the momentum of the system after the collision.

M1u1= m2u2

Let M1 = mass of the chair = 20kg

     m2= mass of the bullet= 10g= 0.001kg

      u1= velocity of the chair before collision = zero m/s

      u2 =  velocity of the bullet before collision = zero m/s

         v1= velocity of the chair after collision = ?  m/s  

        v2 =  velocity of the bullet after collision = 450 m/s

After collision their velocities change from u1 to v1 and u2 to v2 so

M1v1= m2v2

v1= m2v2/M1

v1= 0.01 *450/ 20=  0.225 m/s

Now according to the law of conservation of energy the energy of the system before collision must be equal to the energy of the system after the collision.

The energy of the chair after the bullet is hit is

KE of the chair + KE of the bullet=  1/2 (M)(v1)²+ 1/2 m(v2)²=

1/2 ( 20) (0.225 )² + 1/2 (0.01) (450)²

= 0.50625 + 1012.5=  1013.00625 Joules

Frictional force = Coefficient of kinetic force of wood on wood ( M+m) g

                           = 0.2* ( 20.01) 9.8=   39.2196 N

Work done by friction = frictional force * distance

If law of conservation of energy is applied  the KE  must be equal to the work done

KE = W

W= f*d

KE= F*d

d = KE/f= 1013.00625/ 39.2196= 25.82 cm

The chair did not move 25.82 cm .

It only moved 3 centimeter.

Hence the bullet fired was not from the client's gun.

7 0
3 years ago
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