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3 years ago

12

Which of these gets so bright as to be seen in daylight at times? mars saturn jupiter mercury venus?

2 answers:

fenix001 [56]3 years ago

8 0

While these could all be seen with a telescope during daytime, Venus<span> is the only one that could be viewed with the naked eye in daylight.</span>

tamaranim1 [39]3 years ago

5 0

Venus sometimes does that.

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4) For the situation pictured below, F1 = 20.0 N east and F2 = 30.0 N west ,

maksim [4K]

Answer:

10.0 N West

Explanation:

Imagine East is positive and West is negative. Essentially now we have 2 forces pulling opposite each other on the same plane, so we can add them up.

F1=20N

F2=-30N

F1+F2=Net Force

20+(-30)=-10

Therefore the Net Force is 10N West.

8 0

3 years ago

On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the

Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

$\frac{F_E}{F_G}=\frac{qE}{mg}$ (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

$\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764$

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

$F_E-F_G=ma\\\\a=\frac{qE-mg}{m}$

you replace the values of all variables:

$a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}$

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0

3 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago

If the Sun suddenly went dark, we would not know it until its light stopped arriving on Earth. How long would that be, in second

Gre4nikov [31]

Answer: 500 s

Explanation:

Speed $v$ is defined as a relation between the distance $d$ and time $t$ :

$v=\frac{d}{t}$

Where:

$v=3(10)^{8}m/s$ is the speed of light in vacuum

$d=1.5(10)^{11}m$ is the distance between the Earth and Sun

$t$ is the time it takes to the light to travel the distance $d$

Isolating $t$ :

$t=\frac{d}{v}$

$t=\frac{1.5(10)^{11}m}{3(10)^{8}m/s}$

Finally:

$t=500 s$

5 0

3 years ago

How do i find the weight of the suitcase<br>pls help asap

Citrus2011 [14]

F=W=mg
12N(given)= m*9.8
m= 1.22 kg

3 0

3 years ago