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2 years ago

How much power does the athlete produce if he has 28,056J of work over 45s?

Physics

1 answer:

Jlenok [28]2 years ago

5 0

Explanation:

Power = Work done ÷ Time

P = 28056 ÷ 45 Watt

P = 623.46 Watt

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A) Find the gravitational field strength of an asteroid with the mass of 3.2 * 10^3 kg and an average radius of 30 km when at a

MrMuchimi

a) $1.96\cdot 10^{-16} m/s^2$

The gravitational field strength near the surface of the asteroid is given by:

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the asteroid

R the radius of the asteroid

h is the distance from the surface

Substituting the data of the asteroid:

$M=3.2\cdot 10^3 kg$ is the mass

$R=30 km = 30000 m$ is the radius of the asteroid

$h=3 km = 3000 m$ is the distance from the surface

We find

$g=\frac{(6.67\cdot 10^{-11})(3.2\cdot 10^3)}{(30000+3000)^2}=1.96\cdot 10^{-16} m/s^2$

b) i) $5.53\cdot 10^9 s$

The acceleration of the astronaut popped out at 3 km from the surface is exactly that calculated at part a):

$g=1.96\cdot 10^{-16} m/s^2$

So, since its motion is at constant acceleration, we can find the time he takes to reach the surface using suvat equations:

$s=ut+\frac{1}{2}gt^2$

where

s = 3 km = 3000 m is his displacement to reach the surface

u = 0 is his initial velocity

t is the time

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(3000)}{1.96\cdot 10^{-16} m/s^2}}=5.53\cdot 10^9 s$

b) ii) $1.08\cdot 10^{-6} m/s$

Again, we can use another suvat equation:

$v=u+gt$

where

v is the final velocity

u is the initial velocity

g is the acceleration of gravity

t is the time

Since we have

u = 0

$t=5.53\cdot 10^9 s$

$g=1.96\cdot 10^{-16} m/s^2$

The velocity of the astronaut at the surface will be

$v=0+(1.96\cdot 10^{-16} m/s^2)(5.53\cdot 10^9)=1.08\cdot 10^{-6} m/s$

b) iii) 175 years

The duration of one year here is

$T=3.16\cdot 10^7 s$

And the time it takes for the astronaut to reach the surface of the asteroid is

$t=5.53\cdot 10^9 s$

Therefore, to find the number of years, we just need to divide the total time by the duration of one year:

$n=\frac{t}{T}=\frac{5.53\cdot 10^9 s}{3.16\cdot 10^7}=175$

So, the astronaut will take 175 years to reach the surface.

8 0

4 years ago

A beam of light strikes a mirror at angle of 23° to the normal. What is the angle between the incident and reflected beam?

My name is Ann [436]

Answer:

well, during reflection

angle of incident is equal to angle of reflection so if angles of incident is 23° , then angle of reflection is also equal to 23° .

so the total angle between the reflected and incident ray will be equal to angle of incidence plus angle of reflection

i.e. angle i + angle r = 23° + 23° = 46°

hence the required angle is 46°

i hope it helped .

4 0

3 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.

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