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PilotLPTM [1.2K]

2 years ago

15

Suppose future astronauts are living on a Martian base when a major solar flare occurs. How long will it be before the charged p

articles ejected from the Sun during the flare reach them?

1 answer:

aev [14]2 years ago

8 0

Answer:

755.7 seconds

Explanation:

Distance between Sun and Mars = 226.71 * 10^9 m

Speed of light = 3 * 10^8 m/s

Speed = distance/time

Time = distance/speed

Time = (226.71 * 10^9)/(3 * 10^8)

Time = 755.7 s

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A 12-volt battery causes 0.60 ampere to flow through a circuit that contains a lamp and a resistor connected in parallel. The la

san4es73 [151]

The answer to this question is A

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1 year ago

Which of the following planets helped astronomers locate another planet?

worty [1.4K]

Answer: Pluto,Mercury

Explanation:

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2 years ago

The human eye can respond to as little as 10^−18 J of light energy. For a wavelength at the peak of visual sensitivity, 550 nm,

Bumek [7]

<span>The key equation is going to come from Mr Planck: E=h \nu

Where h is Plancks constant; and ν is the frequency. This equation gives you the energy per photon at a given frequency. Alas, you're given wavelength, but that's easy enough to convert to frequency given the following equation:

c= lambda / nu

where c is the speed of light; λ (lambda) is the wavelength; and ν is again frequency. As soon as you know the energy of a photon with a wavelength of 550nm, you should know how many photons you would require to accumulate 10^-18J. Be careful with your units.</span>

7 0

2 years ago

Read 2 more answers

A 1.5 m cylinder of radius 1.1 cm is made of a complicated mixture materials. Its resistivity depends on the distance x from the

Elis [28]

Answer:

a) $R = 171$ μΩ

b) $E = 1.7 *10^{-4} V/m$

c) $R_{2} = 1.16 *10^{-4}$ Ω

here * stand for multiplication

Explanation:

length of cylinder = 1.5 m

radius of cylinder = 1.1 cm

resistivity depends on the distance x from the left

$p(x)=a+bx^2$ ............(i)

using equation

$R = \frac{pl}{a}$

let dR is the resistance of thickness dx

$dR =\frac{p(x)dx}{a}$

where p(x) is resistivity l is length

a is area

$\int\limits^R_0 {dR} =\frac{1}{\pi r^2} \int\limits^L_0 {(a+bx^2)} \, dx \\$ .........................(2)

after integration

$R = \frac{[aL+\frac{bL^3}{3}] }{\pi r^2}$ ...............(3)

it is given $p(0) = a = 2.25 * 10 ^{-8}$ Ωm

$p(L) = a + b(L)^2$ = $8.5 * 10 ^{-8}$ Ωm

$8.5 * 10 ^{-8} = 2.25 * 10^{-8}+b(1.5)^2\\$

(here * stand for multiplication )

on solving we get

$b = 2.78* 10^{-8}$ Ωm

put each value of a and b and r value in equation 3rd we get

$R = \frac{[aL+\frac{bL^3}{3}] }{\pi r^2}$

$R = 1.71 * 10^{-4}$ Ω

$R = 171$ μΩ

FOR (b)

for mid point x = L/2

E = p(x)L

for x = L/2

$p(L/2) = a+b(L/2)^2$

for given current I = 1.75 A

so electric field

$E = \frac{[a+b(L/2)^2]I }{\pi r^2}$

by substitute the values

we get;

$E = 1.7 *10^{-4} V/m$

(here * stand for multiplication )

c ).

75 cm means length will be half

that is x = L/2

integrate the second equation with upper limit L/2

Let resistance is $R_{1}$

so after integration we get

$R_{1} = \frac{[a(L/2) +(b/3)(L^3/8)]}{\pi r^2}$

substitute the value of a , b and L we get

$R_{1} = 5.47 * 10 ^{-5}$ Ω

for second half resistance

$R_{2} = R- R_{1}$

$R_{2} = 1.7 *10^{-4} -5.47 *10^{-5}$

$R_{2} = 1.16 *10^{-4}$ Ω

(here * stand for multiplication )

5 0

2 years ago

Two objects, X and Y, move toward one another and eventually collide. Object X has a mass of 2M and is moving at a speed of 2v0

Sergeu [11.5K]

Answer:

c. The force exerted by X on Y is F to the right, and the force exerted by Yon X is F to the left.

Explanation:

If we take both objects as one single system, during the collision, assuming no external forces acting, the only forces present are the one that object X exerts on object Y, and the force that object Y does on object X.

These two forces, according to Newton's 3rd Law, form an action-reaction pair, and consequently, are equal in magnitude, acting in opposite directions.

As the object X is moving to the right, the force that produces, (F), is in the same direction (on Y), while for object Y, moving to the left, the force that produces (F also in magnitude) is in the same direction (on X), so the right answer is c.

The effect of the forces is different, due to masses are different, according Newton's 2nd Law.

6 0

2 years ago