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SVETLANKA909090 [29]
3 years ago
11

If we have a new moon today, when we will have the next full moon?.

Physics
1 answer:
Sav [38]3 years ago
6 0

29.5 days

It takes 27 days, 7 hours, and 43 minutes for our Moon to complete one full orbit around Earth. This is called the sidereal month, and is measured by our Moon's position relative to distant “fixed” stars. However, it takes our Moon about 29.5 days to complete one cycle of phases (from new Moon to new Moon).

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Two identical trucks have mass 5100 kg when empty, and the maximum permissible load for each is 8000 kg. the first truck, carryi
Oksanka [162]
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg. The key here is the conservation of momentum. For the first truck, the momentum is 0(5100 + 4300) The second truck has a starting momentum of 60(5100 + x) And finally, after the collision, the momentum of the whole system is 42(5100 + 4300 + 5100 + x) So let's set the equations for before and after the collision equal to each other. 0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x) And solve for x, first by adding the constant terms 0(5100 + 4300) + 60(5100 + x) = 42(14500 + x) Getting rid of the zero term 60(5100 + x) = 42(14500 + x) Distribute the 60 and the 42. 60*5100 + 60x = 42*14500 + 42x 306000 + 60x = 609000 + 42x Subtract 42x from both sides 306000 + 18x = 609000 Subtract 306000 from both sides 18x = 303000 And divide both sides by 18 x = 16833.33 So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums. 60(5100 + 16833.33) = 60(21933.33) = 1316000 42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000 They match. The 2nd truck was definitely over loaded.</span>
6 0
4 years ago
In young’s double slit experiment, the measured fringe width is 0.5 mm for a Sodium light of 589 nm at a distance of 1.5 m. A br
Levart [38]

Answer:

(A).  The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

Explanation:

Given that,

Fringe width d = 0.5 mm

Wavelength = 589 nm

Distance of screen and slit D = 1.5 m

Distance of bright fringe y = 1 cm

(A) We need to calculate the order of the bright fringe

Using formula of wavelength

\lambda=\dfrac{dy}{mD}

m=\dfrac{d y}{\lambda D}

Put the value into the formula

m=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{589\times10^{-9}\times1.5}

m=5.65 = 6

(B). We need to calculate the width of the bright fringe

Using formula of width of fringe

\beta=\dfrac{yd}{D}

Put the value in to the formula

\beta=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{1.5}

\beta=3.33\times10^{-6}\ m

\beta=3.33\ \mu m

Hence, (A).  The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

3 0
3 years ago
A point charge is placed 100 m from a 6 uC charge generating an electric field. What is the
kvv77 [185]

The strength of the electric field is 5 N/C

Explanation:

The magnitude of the electric field produced by a single-point charge is given by:

E=\frac{kQ}{r^2}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

Q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

Q=6 \mu C = 6\cdot 10^{-6}C is the charge producing the field

r = 100 m is the distance from the charge at  which we want to calculate the field

Substituting into the equation, we find the s trength of the electric field:

E=\frac{(8.99\cdot 10^9)(6\cdot 10^{-6})}{(100)^2}=5.4 N/C \sim 5 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago
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Answer:you in connections too?

Explanation:

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3 years ago
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