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3 years ago

During an eclipse the sun the earth and the moon act as a light source,an obstacle or screen=

Physics

1 answer:

Zigmanuir [339]3 years ago

6 0

Answer:

Un eclipse solar se produce cuando la luna se interpone en el camino de la luz del sol y proyecta su sombra en la Tierra. Eso significa que durante el día, la luna se mueve por delante del sol y se pone oscuro. ... Este eclipse total se produce aproximadamente cada año y medio en algún lugar de la Tierra.

Explanation:

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He does whatever a spider can. Spider-Man, who has a mass of 76 [kg], is clinging onto an inclined wall forming an inclination a

Aliun [14]

Answer:570.54 N

Explanation:

Given

mass of man=76 kg

$\theta =50^{\circ}$

As man is standing over inclined building therefore

its weight has two components i.e. sin and cos component

Force perpendicular to inclined wall

$F=mgcos\theta =76\times 9.8\times \sin 50$

F=570.54 N

4 0

3 years ago

An object is thrown with velocity v from the edge of a cliff above level ground. Neglect air resistance. In order for the object

musickatia [10]

Answer:

Explanation:

The range of a projectile is given by R = v²sin2θ/g.

For maximum range, sin2θ = 1 ⇒ 2θ = sin⁻¹(1) = 90°

2θ = 90°

θ = 90°/2 = 45°

So the maximum horizontal distance R is in the range 0 < θ < 45°, if θ is the angle above the horizontal.

4 0

3 years ago

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An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect

Dvinal [7]

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

The initial speed of the electron is $v_i = 3.2 *10^5 m/s$

The mass of electron is $m = 9.11*10^{-31}kg$

Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

Let $KE_i$ be the initial kinetic energy of the electron \

Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

Considering that energy is conserved,

The energy at the initial position of the electron = The energy at the final position of the electron

i.e

$KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial position of the electron and at distance d of the electron to the proton

Here $U_1 = 0$

So the equation becomes

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

$k$ is electrostatic constant with value $8.99*10^9 N \cdot m^2 /C^2$

i.e $q = 1.602 *10^{-19}C$

$v_d$ is the velocity at distance d from the proton = 2 $v_i$

So the equation becomes

$\frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}$

$\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}$

$3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}$

Making d the subject of the formula

$d = \frac{2k(q)^2}{3mv_i^2}$

$= \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}$

$=1.66*10^{-9}m$

6 0

3 years ago

Which of the following is a correct example of the current in a light bulb?

Goryan [66]

Answer:

60 Amperes

Explanation:

current is the main factor, so current is measured in Amperes

3 0

3 years ago

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A dropped ball accelerates as it falls. but a ball rolling on a horizontal surface has zero acceleration because _______.

nordsb [41]

During a fall, the ball experiences an external force which is acted upon by gravity. This gravitation force induces a constant gravitational acceleration on the object equivalent to 9.81 meters per square second. Hence, the dropped ball is accelerating.

On the other hand, the ball does not experience external force hence the acceleration is zero.

4 0

3 years ago

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