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Tatiana [17]
3 years ago
7

An archer fish launches a droplet of water from the surface of a small lake at an angle of 70° above the horizontal. He is aimin

g at a juicy spider sitting on a leaf 40 cm to the east and on a branch 23 cm above the water surface. The fish is trying to knock the spider into the water so that the fish can eat the spider. (a) What must the speed of the water droplet be for the the fish to be successful?
Physics
1 answer:
Norma-Jean [14]3 years ago
4 0

Answer:

a). v=2776 m/s

Explanation:

The speed of the water droplet for the fish be successful is

Taking the distance in axis 'x' and 'y'

x_{tx}=40cm\frac{1m}{100cm}=0.40m\\x_{ty}=23cm\frac{1m}{100cm}=0.23m

The time is the velocity in axis 'x' with the angle 70 so

t=\frac{0.40m}{v_{x}*cos(70)}

Now using the time in terms of velocity the motion in axis 'y' can find the velocity to be the fish successful

x_{yf}=x_{yo}+v_{o}*t+\frac{1}{2}*g*t^{2}\\0.23m=0.40m\frac{vo*sin(70)}{vo*cos(70)} +\frac{1}{2}*9.8*(\frac{0.40}{vo*cos(70)} )^{2}\\0.23m=0.40m*vo*tan(70)+4.9*(\frac{0.16m^{2} }{vo^{2} *cos(70)^{2} }) \\vo^{2}cos(70)^{2}=\frac{0.16m^{2} }{0.177}\\vo=\sqrt{\frac{0.9021}{cos(70)^{2}}} \\vo=2.776 \frac{m}{s}

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The mass of the block in the drawing is 10 kg. The coefficient of static friction between the block and the vertical wall is 0.5
antiseptic1488 [7]

Answer:

The minimum force to start the block moving up the wall = 49 N

Explanation:

Friction: This is the force that tend to oppose the motion of two bodies in contact. The S.I unit of frictional force is Newton (N)

The minimum force required to start the block moving up the wall = Frictional Force.

I.e F = Frictional force.

And, F = μR..........................Equation 1

Where μ = coefficient of static friction, R = Normal reaction.

But R = mg ( on a level surface).................. Equation 2

Where m = mass, g = acceleration due to gravity.

Given: m = 10 kg,

Constant: g = 9.8 m/s²

substituting these values into Equation 2

R = 10 × 9.8

R = 98 N.

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Substituting these values into equation 1

F = 98 × 0.5

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Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at se
adoni [48]

Answer:

(a)\ F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

(b)\ F_{No} = 771.125N

Explanation:

Given

d_D = 6000ft ---- Altitude of container in Denver

A = 0.0155m^2 -- Surface Area of the container lid

P_D = 79000Pa --- Air pressure in Denver

P_{No} = 100250Pa --- Air pressure in New Orleans

<em>See comment for complete question</em>

Solving (a): The expression for F_{No

Force is calculated as:

F = \triangle P * A

The force in New Orleans is:

F_{No} = \triangle P * A

Since the inside pressure is half the pressure at sea level, then:

\triangle P = P_{No} - \frac{P_{area}}{2}

Where

P_{area} = 101000Pa --- Standard Pressure

Recall that:

F_{No} = \triangle P * A

This gives:

F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

Solving (b): The value of F_{No

In (a), we have:

F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

Where

A = 0.0155m^2

P_{No} = 100250Pa

P_{area} = 101000Pa

So, we have:

F_{No} = [100250 - \frac{101000}{2}] * 0.0155

F_{No} = [100250 - 50500] * 0.0155

F_{No} = 49750* 0.0155

F_{No} = 771.125N

4 0
2 years ago
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