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2 years ago

Plants make energy. Some of it they use to preform the necessary functions of living. What do they do with the excess energy?

Physics

1 answer:

Maksim231197 [3]2 years ago

4 0

Answer:

They can use it for when they are dormant in the winter or to grow more sources for storing and creating energy, or they store the energy (this energy would be considered stored energy).

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A charged particle is moving perpendicular to a magnetic field in a circle with a radius r. An identical charge particle enters

emmasim [6.3K]

Answer:

<em>The second particle will move through the field with a radius greater that the radius of the first particle</em>

Explanation:

For a charged particle, the force on the particle is given as

$F = \frac{mv^{2} }{r}$

also recall that work is force times the distance traveled

work = F x d

so, the work on the particle = F x d,

where the distance traveled by the particle in one revolution = $2\pi r$

Work on a particle = 2πrF = $2\pi mv^{2}$

This work is proportional to the energy of the particle.

And the work is also proportional to the radius of travel of the particles.

Since the second particle has a bigger speed v, when compared to the speed of the first particle, then, the the second particle has more energy, and thus will move through the field with a radius greater that the radius of the first particle.

3 0

2 years ago

At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the hor

Irina18 [472]

Answer:

The charge flows in coulombs is

$dq=1.843x10^{-5}C$

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

$I=N*\frac{dF}{Rdt}$

$\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}$

$dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}$

$N=1300$ , $\beta=0.703$ , $A=\pi*r^2=\pi*0.10^2=0.01\pi m^2$ , $R=99.4+202=301.4$ Ω

$\alpha_f=14.6$ , $\alpha_i=165.4$

Replacing :

$dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}$

$dq=1.843x10^{-5}C$

5 0

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

2 years ago

A coal - fired power station produces 100MJ of electrical energy when it is supplied with 400MJ of energy from its fuel. The eff

Alexus [3.1K]

Answer:

100 divide by 400 times 100 percent

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2 years ago

A boy is going down a slide. As he reaches the bottom, friction causes him to slow down and stop. Which one of Newton's laws is

lara31 [8.8K]

I think it’s the second law

8 0

3 years ago