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2 years ago

A total solar eclipse is visible from

Physics

2 answers:

leonid [27]2 years ago

6 0

Answer:

b. only within the moon’s umbra.

Explanation:

A total solar eclipse occurs when the moon blocks the sunlight falling on the Earth. It occurs at the bright side of the Earth when Earth passes through the umbra of the moon.

When sun, moon and Earth align in a straight line, the disk of the moon is able to block the disk of the sun causing the sunlight to get blocked. Shadow of moon falls on the Earth. A total eclipse occurs when the sun is complete blocked and there is total darkness. It is visible from those areas which pass through the moon's umbra.

True [87]2 years ago

5 0

Hey! Hope that this helps. I'm fairly certain the answer is B, because I know that it's not visible from all over the earth, so that eliminates A, but doing some research I found out that an observer in the umbra experiences a total eclipse.

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To distinguish between properties of the two major types of supernovae: massive star supernovae and white dwarf supernovae. all

const2013 [10]

<span>In my opinion, I myself believe that there are only two supernovae. The first is the white dwarf. It makes sense because if something is too big for its size, it will "explode". Just like a basketball with too much air. Massive star supernovae is like something has reached it's full potential and cannot get any bigger or better.</span>

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2 years ago

If a child starts from rest at point A and lands in the water at point B, a horizontal distance L = 2.52 m from the base of the

tamaranim1 [39]

Answer:

The height of the water slide is 0.878 m

Explanation:

Given that,

Distance = 2.52 m

Suppose Children slide down a friction less water slide that ends at a height of 1.80 m above the pool.

We need to calculate the time

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value in the equation

$1.80=0+\dfrac{1}{2}\times9.8\times t^2$

$t^2=\dfrac{1.80\times2}{9.8}$

$t=\sqrt{\dfrac{1.80\times2}{9.8}}$

$t=0.606\ sec$

We need to calculate the velocity

Using formula of velocity

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

Put the value into the formula

$v=\dfrac{2.52}{0.606}$

$v=4.15\ m/s$

We need to calculate height

Using conservation of energy

$\dfrac{1}{2}mv^2=mgh$

$h=\dfrac{v^2}{2g}$

Put the value into the formula

$h=\dfrac{4.15^2}{2\times9.8}$

$h=0.878\ m$

Hence, The height of the water slide is 0.878 m.

4 0

2 years ago

A 3.9 kg ball traveling towards a soccer player at a velocity of -3.5 m/s rebounds off the soccer player's foot at a velocity of

Tresset [83]

Answer: 2.92 s

Explanation:

Given

Mass of ball is $m=3.9\ kg$

The initial velocity of the ball is $u=-3.5\ m/s$

Velocity after the rebound is $v=15.9\ m/s$

Force during the contact is $F=25.9\ N$

We know, change in momentum is Impulse

$\Rightarrow F\cdot \Delta t=m(\Delta v)$

$\Rightarrow 25.9\cdot \Delta t=3.9(15.9-(-3.5))\\\\\Rightarrow \Delta t=\dfrac{3.9\times 19.4}{25.9}=2.92\ s$

Thus, the force is applied for 2.92 s

4 0

2 years ago

video for A door 1 m wide, of mass 15 kg, is hinged at one side so that it can rotate without friction about a vertical axis. It

natka813 [3]

Answer:

$\omega_f = 0.4\ rad/s$

Explanation:

given,

width of door dimension = 1 m

mass of door = 15 Kg

mass of bullet = 10 g = 0.001 Kg

speed of bullet = 400 m/s

$I_{total} =I_{door} + I_{bullet}$

$I_{total} =\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2$

a) from conservation of angular momentum

$L_i = L_f$

$mv\dfrac{W}{2} = I_{total}\omega_f$

$mv\dfrac{W}{2}= (\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f$

$\omega_f= \dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}$

$\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}$

$\omega_f= \dfrac{\dfrac{0.01\times 400}{2}}{\dfrac{15\times 1 }{3}+(\dfrac{0.01\times 1}{4})}$

$\omega_f = 0.4\ rad/s$

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2 years ago

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Answer:

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Explanation:

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