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Vesna [10]
3 years ago
14

There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in

England as a red glow on and around the Moon. How long (in s) after the asteroid hit the Moon, which is 3.58 ✕ 105 km away, would the light first arrive on Earth?
Physics
1 answer:
soldi70 [24.7K]3 years ago
4 0

Answer:

1.28 s

Explanation:

The expression to find the distance traveled by light is,

d = ct

Here is distance is d,  v is the speed and t is the time.

Rearrange the expression for the time,

t = \frac{d}{c}

Substitute 3.58 x 10^{8} m for d and 3 .00 x 10^{8}  m/s  for c in the expression of time,

t = \frac{3.58 * 10^{8 } m  }{3.00 * 10^{8} m/s} = 1.28 s

Hence light will arrive at earth after.

1.28 s

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What volume of h2 gas (in l), measured at 756 mmhg and 90 ∘c, is required to synthesize 23.0 g ch3oh?
Naddika [18.5K]

First let us calculate for the moles of CH3OH formed:

moles CH3OH = 23 g / (32 g / mol) = 0.71875 mol

We see that there are 2 moles of H2 per mole of CH3OH, so:

moles H2 = 0.71875 mol * 2 = 1.4375 mol

 

Assuming ideal gas behaviour, we use the formula:

PV = nRT

V = nRT / P

V = 1.4375 mol * (62.36367 L mmHg / mol K) * (90 + 237.15 K) / 756 mm Hg

<span>V = 43.06 Liters</span>

6 0
3 years ago
Jose is loading his luggage into his car so that he can go to visit his grandmother. He lifts his suitcase up a 10 m staircase i
maxonik [38]

Answer:

The work done on the suitcase is, W = 600 J

Explanation:

Given,

The average force exerted by Jose on his suitcase,  F = 60 N

Jose carried the suitcase to a distance, S = 10 m

The work done on the suitcase is given by the relation

                           <em>W = F x S</em>

Substituting the given values in the above equation,

                            W = 60 N x 10 m

                                 = 600 J

Hence, the work done on the suitcase is, W = 600 J

3 0
3 years ago
The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =
luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = \frac{\textup{Distance}}{\textup{Velocity}}

or

Time = \frac{\textup{0.016}}{\textup{23}}

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = \frac{q\times8.20\times10^4}{1\times10^{-11}}

Now from the Newton's equation of motion

d=ut+\frac{1}{2}at^2

where,  

d is the distance

u is the initial speed  

a is the acceleration

t is the time

or

0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2

or

q = 9.98 × 10⁻⁹ C

4 0
3 years ago
(physical science) could someone please help me out with this lab? if i’m being honest i did the lab but i lost all of my work :
djverab [1.8K]

Explanation:

hdhhrhhrhehhshsujwuuwuwuwwh

6 0
2 years ago
PLEASE HELP FAST The object distance for a convex lens is 15.0 cm, and the image distance is 5.0 cm. The height of the object is
IgorLugansk [536]

Answer:

The image height is 3.0 cm

Explanation:

Given;

object distance, d_o = 15.0 cm

image distance, d_i = 5.0 cm

height of the object, h_o = 9.0 cm

height of the image, h_i = ?

Apply lens equation;

\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm

Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.

4 0
3 years ago
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