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3 years ago

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Which astronomer spent 20 years plotting the positions of the planets

1 answer:

Harlamova29_29 [7]3 years ago

3 0

Tycho Brahe is the answer... i already investigate this :)

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A 2 microcoulomb charge is placed at a distance of 0.25 m away from a 3.6 microcoulomb charge. Describe the type of electrostati

EleoNora [17]

Answer: 1.04N

Explanation:

Given

q1 = 2*10^-6C

q2 = 3.6*10^-6C

r = 0.25m

k = 9*10^9

Magnitude of electrostatic force can be calculated by using coulomb's law. Coulomb's law states that, "the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them."

F =(kq1q2) / r²

F = (9*10^9 * 2*10^-6 * 3.6*10^-6) / 0.25²

F = 0.0648/0.0625

F = 1.04N

The type of electrostatic force between the charges is the repulsive force

7 0

3 years ago

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Why is a camera lens round but the pictures come out square

sergey [27]

It is round or say spherical to widen the range of photography and it is designed to focus the image as the ray are coming from infinity so !!

I am not an expert of camera but ya it is the main theme !!

6 0

3 years ago

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A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target

stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

$\theta=35^{\circ}$

a.Let $v_0$ be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

$v_x=v_0cos\theta=v_0cos35$

$v_y=v_0sin\theta=v_0sin35$

$x=v_0cos\theta\times t=v_0cos35\times t$

$t=\frac{30}{v_0cos35}$

$h=v_yt-\frac{1}{2}gt^2$

Substitute the values

$1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2$

$1.8=30tan35-\frac{6574.6}{v^2_0}$

$\frac{6574.6}{v^2_0}=21-1.8=19.2$

$v^2_0=\frac{6574.6}{19.2}$

$v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s$

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

$t=\frac{30}{18.5cos35}$

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0

3 years ago

Which letter represents the wavelength of

Morgarella [4.7K]

The letter D represents the wavelength

4 0

3 years ago

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A 4.0 cm × 4.2 cm rectangle lies in the xy-plane. You may want to review (Pages 664 - 668) . Part A What is the electric flux th

padilas [110]

Answer:

(A). The flux is 0.336 N.m²/C

(B). The flux is zero.

Explanation:

Given that,

Length = 4.2 cm

Width = 4.0 cm

Electric field $E=(150 i-200 k)\ N/C$

Area vector is perpendicular to xy plane

(A). We need to calculate the flux

Using formula of flux

$\phi=E\cdot A$

Where, E = electric field

A = area

Put the value into the formula

$\phi=(150 i-200 k)\times(4.2\times10^{-2}\times4.0\times10^{-2})k$

$\phi=-200\times4.2\times10^{-2}\times4.0\times10^{-2}$

$\phi=-0.336\ N.m^2/C$

(B). Given electric field

$E=(150i-200j)\ N/C$

We need to calculate the flux

Using formula of flux

$\phi=E\cdot A$

Put the value into the formula

$\phi=(150 i-200 j)\times(4.2\times10^{-2}\times4.0\times10^{-2})k$

Here, The component of k is not given

So, the flux is

$\phi=0$

Hence, (A). The flux is -0.336 N.m²/C

(B). The flux is zero.

7 0

3 years ago

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