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Mashcka [7]
3 years ago
15

Study the image. Point 1 is beneath water moving toward land. Point 2 is next to an erupting vocano. Point 3 is beneath Earth ap

ostrophe s crust moving toward the water. Point 4 is the same layer as Point 1 moving down below Point 3. At which point is subduction occurring? 1 2 3 4
Physics
2 answers:
juin [17]3 years ago
7 0

Answer:

I believe the answer is D) 4

Explanation:

Subduction is the sideways and downward movement of the edge of a plate of the earth's crust into the mantle beneath another plate.

svet-max [94.6K]3 years ago
5 0

Answer:

c

Explanation:

mart me as brailist

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A base is a substance that's the chemical opposite of an acid.<br> True or false
anygoal [31]

True

Explanation:

A base is a substance that is often used as the chemical opposite of an acid.

Both behaves in opposite way to one another.

They can be said to complementary or conjugate chemicals.

  • According to Bronsted-Lowry  theory, an acid is a proton donor and a base is a proton acceptor.
  • The lewis theory states that an acid is an electron pair acceptor while a base is an electron pair donor.
  • Acids turn blue litmus paper red and bases turns red litmus paper blue.

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6 0
4 years ago
Cart 1 has an initial velocity and hits cart 2 which is stationary. after a perfectly inelastic collision, the combined carts ar
Tomtit [17]

Option(a)  the mass of cart 2 is twice that of the mass of cart 1 is the right answer.

The mass of cart 2 is twice that of the mass of cart 1  is correct about the mass of cart 2.

Let's demonstrate the issue using variables:

Let,

m1=mass of cart 1

m2=mass of cart 2

v1 = velocity of cart 1 before collision

v2 = velocity of cart 2 before collision

v' = velocity of the carts after collision

Using the conservation of momentum for perfectly inelastic collisions:

m1v1 + m2v2 = (m1 + m2)v'

v2 = 0 because it is stationary

v' = 1/3*v1

m1v1 = (m1+m2)(1/3)(v1)

m1 = 1/3*m1 + 1/3*m2

1/3*m2 = m1 - 1/3*m1

1/3*m2 = 2/3*m1

m2 = 2m1

From this we can conclude that the mass of cart 2 is twice that of the mass of cart 1.

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4 0
2 years ago
Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s west
Aleks04 [339]

Explanation:

Given that,

Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.

Taking eastward as positive direction, we have:

v_B=+5\ m/sis the velocity of Bill with respect to Amy (which is stationary)

v_c=15\ m/s is the velocity of Carlos with respect to Amy.

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

v_B=+5\ m/s

Therefore, Carlos velocity in Bill's reference frame will be

v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s

So, the magnitude is 20 m/s and the direction is westward (negative sign).

7 0
3 years ago
an ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causing the block to accelerate uni
stiv31 [10]
The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.

Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg

F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)

F = 83 N
3 0
3 years ago
2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or
krek1111 [17]

a) The work done by the tree is -1.44\cdot 10^5 J

b) The amount of force applied is 2880 N

Explanation:

a)

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J

The work is negative because it is done in the direction opposite to the direction of motion of the car.

b)

The work done by the tree on the car can also be rewritten as

W=Fd

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

W=-1.44\cdot 10^5 J is the work done

d=50.0 cm = 0.50 m is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

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3 0
4 years ago
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