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3 years ago

5

Consider the following scenario: You live near the shoreline and hear a report that an earthquake hit the ocean floor near your

location. Your town may experience
A. a tsunami.
B. snow.
C. rain.
D. hail.

2 answers:

Harlamova29_29 [7]3 years ago

5 0

A. a tsunami
(if the earthquake is hitting the ocean, the water will get effected)

worty [1.4K]3 years ago

5 0

A. If an earthquake hits the ocean floor it causes waves that are known as tsunamis.

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A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the block, measured in meters per second.

$\Sigma F$ - Horizontal net force, measured in newtons.

$\Delta t$ - Impact time, measured in seconds.

Now we clear the final speed in (2):

$v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $m = 14\,kg$ , $\Sigma F = 98\,N$ and $\Delta t = 1.40\,s$ , then final speed of the ice block is:

$v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}$

$v_{f} = 9.8\,\frac{m}{s}$

The final speed of the block of ice is 9.8 meters per second.

6 0

2 years ago

Mr Ndlovu works in centurion and works at Rosebank.It takes him 13minutes to get to work and he needs to be there at 08h15

VARVARA [1.3K]

Answer:

8:02 no less

Explanation:

8:15 - 13mins = 8:02

6 0

2 years ago

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 2.5 x 10-15 m before

Tasya [4]

Explanation:

It is given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, $d=2.5\times 10^{-15}\ m$

Speed of light, $c=3\times 10^{8}\ m\s$

Let t is the time interval required for the strong interaction to occur. The speed is given by :

$c=\dfrac{d}{t}$

$t=\dfrac{d}{c}$

$t=\dfrac{2.5\times 10^{-15}\ m}{3\times 10^{8}\ m/s}$

$t=8.33\times 10^{-24}\ s$

So, the time interval required for the strong interaction to occur is $8.33\times 10^{-24}\ s$ . Hence, this is the required solution.

8 0

2 years ago

Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the

IRISSAK [1]

Answer:

a₁ = 0.63 m/s² (East)

a₂ = -1.18 m/s² (West)

Explanation:

m₁ = 95 Kg

m₂ = 51 Kg

F = 60 N

a₁ = ?

a₂ = ?

To get the acceleration (magnitude and direction) of the man we apply

∑Fx = m*a (⇒)

F = m₁*a₁ ⇒ 60 N = 95 Kg*a₁

⇒ a₁ = (60N / 95Kg) = 0.63 m/s² (⇒) East

To get the acceleration (magnitude and direction) of the woman we apply

∑Fx = m*a (⇒)

F = -m₂*a₂ ⇒ 60 N = -51 Kg*a₂

⇒ a₂ = (60N / 51Kg) = -1.18 m/s² (West)

For every case we apply Newton’s 3 d Law

8 0

2 years ago

Which pair of equations can describe the path of a particle moving with an acceleration that is perpendicular to the velocity of

antoniya [11.8K]

Answer:

The question clearly describes the circular motion.

The circular motion equation is

$a_{radial} = \frac{v^2}{r}$

The path of the particle is circular.

Explanation:

In circular motion, the radial acceleration is always towards the center and constant in magnitude. Furthermore, the velocity of the circular motion is always tangential to the circle, that is it is always perpendicular to the radius, hence the acceleration.

6 0

2 years ago