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2 years ago

Why do you think spiders are disliked

Physics

2 answers:

Rama09 [41]2 years ago

7 0

Answer:

<h2>why do you think spiders are disliked?</h2>

The presence of body hair seems to be significantly associated with both fear and disgust of spiders (Davey, 1991). We suggest that the rationale for this perception (fear of hairiness) is that body hair (or fur) when standing up in many mammals occur when the animal is threatened.

Explanation:

<h3>Always remember:</h3><h3>beluga always love answer your question by following these steps:</h3><h3>Follow beluga</h3><h3>ask me</h3><h3>Brainliest me first ok</h3><h3 /><h3>take note please: everyone i answer question I got happy:)</h3><h3 /><h3>please don't delete Your question beacause Brainliest Marks & points</h3><h3>will be lose</h3>

SCORPION-xisa [38]2 years ago

4 0

Answer:

it is called arachnophobia

Explanation:

most reason are they way they walk and jump and the people who know they have 4 eyes

Hope This Helped i Like Spiders (^^vv^^) (spider smiley face)

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The peak of the trajectory occurs at time t1. This is the point where the ball reaches its maximum height ymax. At the peak the

Digiron [165]

<span>Part B
What are the values of the intial velocity vector components v0,x and v0,y (both in m/s) as well as the acceleration vector components a0,x and a0,y (both in m/s2)? Here the subscript 0 means "at time t0."
15.0, 26.0, 0, -9.80

</span><span>Part C
What are the values of the velocity vector components v1,x and v1,y (both in m/s) as well as the acceleration vector components a1,x and a1,y (both in m/s2)? Here the subscript 1 means that these are all at time t1.
15, 26, 0, -9.81</span><span>

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3 years ago

A _ increases it decreases voltage in a power line

Levart [38]

A transformer increases and decreases voltage.

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3 years ago

A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and

cricket20 [7]

Answer:

The frequency $f$ = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = $\frac{d}{2}$ = $\frac{44}{2}$ = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = $\frac{d}{2}$ = $\frac{3.0}{2}$ = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = $\frac{d}{2}$ = $\frac{2.0}{2}$ = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

$P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2$

$pgy_1=\frac{1}{2}pv^2_2 +pgy_2$

$v_2=\sqrt{2g(y_1-y_2)}$

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = $\sqrt{2*9.8*(0.35-0)}$

v₂ = $\sqrt {6.86}$

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

$v_3 = \frac{v_2r_2^2}{v_3^2}$

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = $(2.62 m/s)*$ $(\frac{1.5mm^2}{1.0mm^2} )$

v₃ = 0.0589 m/s

∴ velocity of the fluid in the cylinder = 0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

$f=\frac{v_s}{4(h-v_3t)}$

where;

$v_s$ = velocity of sound

h = height of the fluid

v₃ = velocity of the fluid in the cylinder

$f=\frac{343}{4(0.40-(0.0589)(0.4)}$

$f= \frac{343}{0.6576}$

$f$ = 521.59 Hz

∴ The frequency $f$ = 521.59 Hz

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

$\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})$

$\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)$

$\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}$

where;

$v_s$ (velocity of sound) = 343 m/s

v₃ (velocity of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

$\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}$

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s when the cylinder has been filling for 4.0 s.

8 0

3 years ago

blank refers to the method of spreading fertilizer evenly over the entire field by hand it is done at the blank stage

Andrej [43]

Answer:

Broadcasting is the method, not sure about the stage it is done in

Explanation:

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3 years ago

If a statement is true, select true. if it's false, select false. <br>

Elena L [17]

3 is false 2 is true and the rest true

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3 years ago

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