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2 years ago

03. A rocket and its payload have a total mass of 5.0 × 104 kg. How large is the force produced by the engine (the thrust) when

Physics

1 answer:

morpeh [17]2 years ago

3 0

Answer:

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ILL GIVE BRAINLYEST

lawyer [7]

Answer:

3rd picture straight line going up right

Explanation:

3rd picture

5 0

3 years ago

When the distance between two stars decreases by one-third, the force between them

pashok25 [27]

Answer:

the force will increase by a factor 2.25

Explanation:

The gravitational force between the two stars is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two stars

r is the distance between the stars

If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance

$r'=\frac{2}{3}r$

So the new force will be

$F'=G\frac{m_1 m_2}{(\frac{2}{3}r)^2}=\frac{9}{4} G\frac{m_1 m_2}{r^2}=2.25 F$

So, the force will be 2.25 times the previous value.

4 0

3 years ago

I will give Brainliest to whoever can answer!!!!

Alexus [3.1K]

Answer:

The nodes and anti nodes would reverse roles.

Explanation:

I believe it has to do with the path differences. If waves are in phase, then the path differences are such that the waves reach the screen with crests superimposing crests and troughs superimposing troughs. This happens when the periods of each wave are equal or the paths themselves differ by a whole number multiple of the wavelength (λ, 2λ, 3λ, ...).

Now make these waves out of phase. Then half of the waves will travel half a wavelength farther than the rest. So the path difference will be 0.5λ, 1.5λ, 2.5λ, ....

4 0

3 years ago

Heather and Jerry are standing on a bridge 49 mm above a river. Heather throws a rock straight down with a speed of 17 m/sm/s .

lara [203]

Answer:

3.467 s

Explanation:

given,

distance , d = 49 mm = 0.049 m

initial speed of the of the rock, v = 17 m/s

time taken by the Heather rock to reach water

using equation of motion

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

taking downward as negative

$-0.049 = -17 t -\dfrac{1}{2}\times 9.8\times t^2$

4.9 t² + 17 t - 0.049 = 0

now,

$t_1 = \dfrac{-(17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}$

t₁ = -3.47 s , 0.0028 s

rejecting negative values

t₁ = 0.0028 s

now, time taken by the ball of Jerry

using equation of motion

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

taking downward as negative

$-0.049 = 17 t -\dfrac{1}{2}\times 9.8\times t^2$

4.9 t² - 17 t - 0.049 = 0

now,

$t_2 = \dfrac{-(-17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}$

t₂ = 3.47 s ,-0.0028 s

rejecting negative values

t₂ = 3.47 s

now, time elapsed is = t₂ - t₁ = 3.47 - 0.0028 = 3.467 s

5 0

3 years ago

12. By convention (agreement of the scientific community for consistency)

Reptile [31]

Answer:

. always start on the north pole and terminate (end) on the South Pole

Explanation:

4 0

3 years ago