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LekaFEV [45]
3 years ago
11

Tectonic plates move ________. A. about a kilometer per year B. at different speeds C. about one yard per year D. at the same sp

eed all over the world E. about one meter per year
Physics
1 answer:
Alona [7]3 years ago
7 0

Answer:

B. at different speeds.

Explanation:

Tectonic plates are able to move because the Earth's lithosphere has greater mechanical strength than the underlying asthenosphere. Lateral density variations in the mantle result in convection; that is, the slow creeping motion of Earth's solid mantle. Plate movement is thought to be driven by a combination of the motion of the seafloor away from spreading ridges due to variations in topography (the ridge is a topographic high) and density changes in the crust (density increases as newly formed crust cools and moves away from the ridge. These average rates of plate separations can range widely. The Arctic Ridge has the slowest rate (less than 2.5 cm/yr), and the East Pacific Rise near Easter Island, in the South Pacific about 3,400 km west of Chile, has the fastest rate (more than 15 cm/yr).

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A car has a mass of 1520 kg. While traveling at 20 m⁄s, the driver applies the brakes to stop the car on a wet surface with a 0.
docker41 [41]

Answer:

(a)d₁ = 51.02 m

(b)d₂ =51.02m

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

Known data

m=1520 kg  : mass of the  car

μk= 0.4 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the car

We define the x-axis in the direction parallel to the movement of the  car and the y-axis in the direction perpendicular to it.

W: Weight of the block : In vertical direction  downward

FN : Normal force :  In vertical direction  upward

f : Friction force:  In horizontal direction  

Calculated of the W

W= m*g

W=  1520 kg* 9.8 m/s² = 14896 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay    ay = 0

FN - Wy = 0

FN = Wy

FN = 14896 N

Calculated of the f

f = μk* N= (0.4)* (14896 N )

f = 5958.4 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

- f = m*a

-5958.4 = (1520)*a

a  =  (-5958.4) /  ((1520)

a = -3.92 m/s²

(a) displacement of the car (d₁)

Because the car moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d₁ Formula (2)

Where:  

d:displacement  (m)

v₀: initial speed  (m/s)

vf: final speed   (m/s)

Data:

v₀ = 20 m⁄s

vf = 0

a = --3.92 m/s²

We replace data in the formula (2)  to calculate the distance along the ramp the block reaches before stopping (d₁)

vf²=v₀²+2*a*d ₁

0 = (20)²+2*(-3.92)*d ₁

2*(3.92)*d₁  = (20)²

d₁ = (20)² / (7.84)

d₁  = 51.02 m

(b)  Different car

m₂ = 1.5 *1520 kg

μk₂= 0.4

W₂= m*g

W₂=   (1.5) *1520 kg* 9.8 m/s² = (1.5)*14896 N  

FN₂=  (1.5)*14896 N  

f= 0.4* (1.5)*14896 N  

a = - f/m₂ = - 0.4* (1.5)*14896 N  /(1.5) *1520

a = -3.92   m/s²

vf²=v₀²+2*a*d₂

vf=0 , v₀=20 m⁄s , a = -3.92   m/s²

d₂ = d₁ = 51.02m

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How much kinetic energy does a 6kg cart have when moving at 4 m/s?
Arturiano [62]

Given:

mass = 6 kg

velocity = 4 m/s

To find:

Kinetic energy of the cart = ?

Formula used:

Kinetic energy = \frac{1}{2} m v^{2}

Where m = mass of the cart

v = velocity with which the cart is moving

Solution:

Kinetic energy of the moving cart is given by,

Kinetic energy = \frac{1}{2} m v^{2}

Where m = mass of the cart

v = velocity with which the cart is moving

Kinetic energy =\frac{1}{2} \times 6 \times \ 4 \times 4

Kinetic energy = 48 Joule

Thus, the kinetic energy of the moving cart is 48 Joule.

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