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3 years ago

What conditions are required for a lunar eclipse?

1 answer:

6 0

For a total lunar eclipse, the sun, earth and the moon must be in align with earth in between the sun and the moon.
Hope this helps.

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A cart on a horizontal, linear track has a fan attached to it. The cart is positioned at one end of the track, and the fan is tu

kompoz [17]

Answer:

$F = 0.0566 \ N$

$t = 6.147 \ s$

Explanation:

From the question we are told that

The distance travel in 4.22 s is $s = 1.44 \ m$

The mass of the cart plus the fan is $m = 350 \ g = 0.35 \ kg$

Generally from kinematic equation we have that

$s = ut + \frac{1}{2} * a * t^2$

Here u is the initial velocity with value $u = 0 \ m/s$

$1.44= 0 * t + \frac{1}{2} * a * 4.22^2$

=> $a = 0.1617 \ m/s^2$

Generally the net force is

$F = m * a$

=> $F = 0.35 * 0.1617$

=> $F = 0.0566 \ N$

Gnerally the new mass of the cart plus the fan is $M = 656 \ g = 0.656 \ kg$

The distance considered is $s_1 = 1.63 \ m$

Generally the new acceleration of the cart is mathematically represented as

$F = M * a_1$

=> $a_1 = \frac{F}{M}$

=> $a_1 = \frac{0.0566}{0.656}$

=> $a_1 = 0.08628 \ m/s^2$

Gnerally from kinematic equation we have

$s = ut + \frac{1}{2} * a_1 * t ^2$

Here u is the initial velocity and the value is zero because it started from rest

=> $1.63 = 0 * t + \frac{1}{2} * 0.08628* t ^2$

=> $t = 6.147 \ s$

6 0

3 years ago

A car starting at rest accelerates to a speed of 6miles/hr in 5 sec what is the average acceleration

Luden [163]

Answer:

1.2 mi/hr/sec

Explanation:

<em>a = Vf - Vi / t (formula)</em>

-----------------------------------

6mi/hr - 0 mi/hr 6 mi/hr

a = ____________ = _______ = 1.2 mi/hr/sec

5 sec 5 sec

5 0

3 years ago

Read 2 more answers

30 points, brainiest, high school physics problem, please help with explanations.

Ymorist [56]

Answer:

I think the answer is C too

4 0

3 years ago

A cylindrical bucket, open at the top, is 28.0 cm high and 11.0 cm in diameter. A circular hole with a cross-sectional area 1.55

svlad2 [7]

Answer:

so height is 0.1283 m

Explanation:

given data

height = 28 cm

diameter = 11 cm

cross-sectional area = 1.55 cm2

water flow rate = 2.46×10^−4 m3/s

to find out

How high will the water in the bucket rise

solution

we know that here

potential energy = kinetic energy

mgh = 1/2 mv²

multiply both sides by the 2 and we get

2mgh=mv²

solve it we get

√(2gh) = v ....................1

h = v²/2g ...............2

and

flow rate = A V

2.46×10^−4 = V 1.55×10^−4

V = 1.5870 m/s

so from 2

h = v²/2g

h = 1.5870²/ 2(9.81)

h = 0.1283 m

so height is 0.1283 m

6 0

4 years ago

What form of energy is released into the atmosphere by the earth's surface

Crank

Answer:

Thermal Energy (Heat)

8 0

3 years ago

Read 2 more answers