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m_a_m_a [10]
3 years ago
15

The Hawaiian hot spot sits below the Pacific plate. As the plate moves over the hot spot, a chain of volcanoes is formed. The Ea

st Molokai Volcano is ~1.76 million years old, and about 178 km away from the current location of the Hawaiian hot spot. Assuming that the hot spot is in a fixed location, how fast (at what rate) is the Pacific plate moving?
Physics
1 answer:
Viktor [21]3 years ago
6 0

Answer:

3.2048179721\times 10^{-9}\ m/s

Explanation:

Assuming that the pacific plate moved 178 km in 1.76 million years.

s = Distance = 178 km

t = Time taken = 1.76 million years

Speed is given by

v=\dfrac{s}{t}\\\Rightarrow v=\dfrac{178000}{1.76\times 10^6\times 365.25\times 24\times 3600}\\\Rightarrow v=3.2048179721\times 10^{-9}\ m/s

The speed of the plate is 3.2048179721\times 10^{-9}\ m/s

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How much energy is required to raise the temperature of 50.0 grams of water 10.0 degree C? (Explain yourself answer in joules!)
maria [59]

The amount of energy needed is 2093 J

Explanation:

The amount of energy needed to increase the temperature of a substance by \Delta T is given by the equation

Q=mC\Delta T

where

m is the mass of the substance

C is its specific heat capacity

\Delta T is the increase in temperature

For the water in this problem, we have

m = 50.0 g = 0.050 kg

C=4186 J/g^{\circ}C (specific heat capacity of water)

\Delta T=10.0^{\circ}C

Therefore, the amount of energy needed is

Q=(0.050)(4186)(10)=2093 J

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

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3 years ago
A rope exerts a 280 N force while pulling an 80 Kg skier upward along a hill inclined at 12o. The rope pulls parallel to the hil
user100 [1]

Answer:

The speed of the skier after moving 100 m up the slope are of V= 25.23 m/s.

Explanation:

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g= 9,8 m/s²

N= m*g*sin(α)

N= 163 Newtons

Fr= μ * N

Fr= 24.45 Newtons

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a= (280N - 24.5N) / 80kg

a= 3.19 m/s²

d= a * t² / 2

t=√(2*d/a)

t= 7.91 sec

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V= 3.19 m/s² * 7.91 s

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Answer:

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A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia
Musya8 [376]

Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

When the tiger reaches the ground, y = 0:

0 = 6.00 m - 1/2 · 9.81 m/s² · t²

2 · (-6.00 m) / -9.81 m/s² = t²

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We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:

x = x0 + v0 · t

At t = 1.11 s, x = 2.00 m

x0 = 0

2.00 m = v0 · 1.11 s

2.00 m / 1.11 s = v0

v0 = 1.80 m/s

The initial speed of the tiger is 1.80 m/s

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beks73 [17]

Answer:

An image is formed on the retina with light rays converging most at the cornea and upon entering and exiting the lens. Rays from the top and bottom of the object are traced and produce an inverted real image on the retina. The distance to the object is drawn smaller than scale

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