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4 years ago

several months after a forest is cut down, a nearby river becomes much muddier than ever before. Infer how the river changed

1 answer:

4 0

Well this is just my opinion but I think because all the plants were gone they cant stop mud from rolling into it. Also when it rains the trees wont suck up the water, that will make it over flow and release into the land or dirt on the outsides of it. Mixing those two gives mud and it can potentially flow back in the river in largeish masses causing the river to mix with the mud makinging it more muddy. Hope this helps :D

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A circus performer on a pair of stilts exerts a pressure of 32 kPa on the ground. If the performer stands on one stilt, what pre

Agata [3.3K]

Answer:

64 kPa

Explanation:

The pressure exerted by a force on a surface is given by

$p=\frac{F}{A}$

where

p is the pressure

F is the force

A is the area on which the force is exerted

In this problem, let's call:

F = the weight of the performer, which is the force

A = the area of 1 stilt

At the beginning, the performer is standing on both stilts, so the area on which he exerts pressure is 2A. So the pressure is

$p=\frac{F}{2A}=32 kPa$ (1)

Later, he stands on one stilt only. The force exerted is still the same (his weight), however, the area is now reduced to A; therefore, the new pressure is

$p'=\frac{F}{A} = 2(\frac{F}{2A})=2p$

which is twice the value calculated in (1); so, the new pressure is

$p'=2(32)=64 kPa$

3 0

4 years ago

imagine that a new planet is discovered with two moons of equal mass, moon A and moon B. The mass of the new planet is greater t

sdas [7]

The gravitational pull on moon A will be lower since it is further away from the new planet.

3 0

4 years ago

Read 2 more answers

At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r

Oksana_A [137]

Answer:

For proton:

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

For electron:

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s

Explanation:

For a proton we have:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

$W = \Delta Vq$ (1)

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C

V: is the potential

Also, the work is equal to:

$W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}$ (2)

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

$v_{a}$ : is the velocity in the point a

$v_{b}$ : is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

$\Delta Vq = \frac{1}{2}mv_{a}^{2}$

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}$

$v_{a} = 2.9 \cdot 10^{4} m/s$

For an electron we have:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

$\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}$

Since $v_{a}$ = 0 (starts from rest) and $m_{e}$ = 9.1x10⁻³¹ kg (electron's mass), we have:

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}$

$v_{b} = 1.2 \cdot 10^{6} m/s$

I hope it helps you!

6 0

4 years ago

Calculate the wavelength and frequency at which the intensity of the radiation is a maximum for a blackbody at 298 K. You will n

Juliette [100K]

Answer:

Wavelength, $\lambda=9.72\times 10^{-6}\ m$

Frequency, $f=3.08\times 10^{13}\ Hz$

Explanation:

We need to find the intensity of the radiation is a maximum for a black body at 298 K. It can be calculated using Wein's displacement law. It is given by :

$\lambda T=2.898\times 10^{-3}\ m-K$

$\lambda=\dfrac{2.898\times 10^{-3}}{T}$

Here, T = 298 K

$\lambda=\dfrac{2.898\times 10^{-3}}{298}$

$\lambda=9.72\times 10^{-6}\ m$

If f is the frequency of black body radiation. It is given by :

$f=\dfrac{c}{\lambda}$

$f=\dfrac{3\times 10^8}{9.72\times 10^{-6}}$

$f=3.08\times 10^{13}\ Hz$

Hence, this is the required solution.

6 0

3 years ago

Tommy runs his juicer every morning. The juicer uses 90 w of power & the current supplied in 4.5 A. How many volts are neces

Airida [17]

Answer:

20 Volts

Explanation:

Ohm's Law and Power Equations

Ohm's law equation (formula): V = I × R and the power law equation

(formula): P = I × V. P = power, I or J = Latin: influare, international ampere, or intensity and R = resistance. V = voltage, electric potential difference Δ V or E = electromotive force (emf = voltage).

Voltage or volts E or V: volts V

Resistivity or resistance R: ohms Ω

Wattage or power P: watts W

Amperage or current I: amperes, amps A

6 0

3 years ago