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3 years ago

Pls tell me Question: 1 way to keep people safe from hurricane.

Physics

2 answers:

melamori03 [73]3 years ago

8 0

Go in creative mode and fly

Fittoniya [83]3 years ago

6 0

Answer:

Hurricanes move fast so what you want to do is get into a basement or somthing underground so it will be harder for the hurricane to hit you

Explanation:

thanks for the coins

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Which of the following is the best definition of ethics?

Licemer1 [7]

Answer:

B) laws that require scientists to research certain things

4 0

3 years ago

Read 2 more answers

Describe what is happening to the speed during the period (I). 0s - 10s __________________________________________________ (II).

aleksley [76]

Answer:

- There was a constant acceleration at 0 to 10s

- There was a zero acceleration at 10 to 25s

- There was a constant deceleration at 25 to 30s

Explanation:

<em>See attachment for complete question.</em>

Solving (a): What happens at 0s to 10s

There was a constant acceleration and this is proven below.

At time 0, velocity = 15

At time 10, velocity = 30

This is represented as:

$(t_1,v_1) = (0,15)$

$(t_2,v_2) = (10,30)$

Acceleration (A) is the rate of change of velocity against time.

So:

$A = \frac{v_2 - v_1}{t_2-t_1}$

$A = \frac{30-15}{10 - 0}$

$A = \frac{15}{10}$

$A = 1.5$

<em>Since the acceleration is positive, then it shows a constant acceleration.</em>

Solving (b): What happens at 10s to 25s

There was a zero acceleration and this is because the velocity do not change.

See proof below

At time 10, velocity = 30

At time 25, velocity = 30

This is represented as:

$(t_1,v_1) = (10,30)$

$(t_2,v_2) = (25,30)$

Acceleration (A) is the rate of change of velocity against time.

So:

$A = \frac{v_2 - v_1}{t_2-t_1}$

$A = \frac{30-30}{25 - 10}$

$A = \frac{0}{15}$

$A = 0$

Solving (c): What happens at 25s to 30s

There was a constant deceleration and this is proven below.

At time 25, velocity = 30

At time 30, velocity = 0

This is represented as:

$(t_1,v_1) = (25,30)$

$(t_2,v_2) = (30,0)$

Acceleration (A) is the rate of change of velocity against time.

So:

$A = \frac{v_2 - v_1}{t_2-t_1}$

$A = \frac{0-30}{30-25}$

$A = \frac{-30}{5}$

$A = -6$

<em>Since the acceleration is negative, then it shows a constant deceleration</em>

4 0

3 years ago

Satellite C revolves around Earth 10 times a day. What is the radius of its orbit, measured from Earth's center? Assume that Ear

frez [133]

The radius of its orbit, measured from Earth's center, will be 1.44 × 10⁷ mm.

<h3>What is Newton's law of gravitation?</h3>

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

Given data in problem is;

The mass of Earth is, $\rm m_E = 5.98 \times 10^{24} \ kg$

Gravitational constant, G =6.674 × 10⁻¹¹ N m₂/kg²

The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

$\rm F_g = \frac{Gm_sm_e}{r^2}$

The centripetal force due to rotation of the satellite;

$\rm F_c = \frac{m_s v^2}{r}$

The centripetal and the gravitational force are equal;

$\rm F_g = F_c \\\\ \frac{Gm_sm_e}{r^2} = \frac{m_s v^2}{r} \\\\ r = G \frac{m_E }{v^2 } \\\\ r = G \frac{m_E }{(r \omega )^2 } \\\\ r = \sqrt[3]{\frac{Gm_E}{\omega^2}} \\\\ r = \sqrt[3]{\frac{6.67 \times 10^{-11}(5.98 \times 10^{24})}{(3.63 \times 10^{-4})^2}} \\\\ r = 1.44 \times 10^7 \ mm$

Hence, the radius of its orbit measured from Earth's center will be 1.44 × 10⁷ mm.

To learn more about Newton's law of gravitation, refer to the link.

brainly.com/question/9699135.

#SPJ1

8 0

2 years ago

The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This sp

Delvig [45]

Answer:

Explanation:

Here is the full question and answer,

The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.

When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.

Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.

Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?

Answer

A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.

Calculate the speed of the archer fish.

The time of the flight of spitted water is,

$t = \frac{{2v\sin \theta }}{g}$

Substitute $9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for g and $60^\circ$ for $\theta$ in above equation.

$t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\$

Spitted water will travel $0.80{\rm{ m}}$ horizontally.

Displacement of water in this time period is

$x = vt\cos \theta$

Substitute $\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}$ for $t\rm 60^\circ[tex] for [tex]\theta$ and $0.80{\rm{ m}}$ for x in above equation.

$\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.

Calculate the height of insect above the surface.

Vertical component of the velocity is,

${v_v} = v\sin \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for v and $60^\circ$ for $\theta$ in above equation.

$\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

Horizontal component of the velocity is,

${v_h} = v\cos \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for v and $60^\circ$ for $\theta$ in above equation.

$\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

When horizontal $({0.60\;{\rm{m}}}$ distance away from the fish.

The time of flight for distance (d) is ,

$t = \frac{d}{{{v_h}}}$

Substitute $0.60\;{\rm{m}}$ for d and $1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for ${v_h}$ in equation $t = \frac{d}{{{v_h}}}$

$\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\$

Distance of the insect above the surface is,

$s = {v_v}t + \frac{1}{2}g{t^2}$

Substitute $2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for ${v_v}$ and $0.3987{\rm{ s}}$ for t and $- 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for g in above equation.

$\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\$

7 0

3 years ago

Explain how earths lithosphere and asthenosphere work together

Aleonysh [2.5K]

Answer:

The lithosphere can affect the atmosphere when tectonic plates move and cause an eruption, where magma below spews up as lava above.

Explanation:

The lithosphere is broken into giant plates that fit around the globe like puzzle pieces. These puzzle pieces move a little bit each year as they slide on top of a somewhat fluid part of the mantle called the asthenosphere.

3 0

3 years ago