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2 years ago

3

Two rockets having the same acceleration start from rest, but the first rocket travels for twice as much time as the second one.

If rocket A goes for 310 km, how far will B go? If rocket A reaches a speed of 320 , what speed will rocket B reach?

1 answer:

Liula [17]2 years ago

3 0

Let ta be the time the first rocket travels and tb be the time the second rocket travels. ta = 2tb
We apply:
sa = ut + 1/2 at², with u = 0 since they start from rest.
310000 = 1/2a(ta)²
a = 620000 / 4(tb)²

sb = 1/2 a(tb)²
sb = 1/2 x 620000 / 4(tb)² x (tb)²
sb = 77.5 Km

We apply v = u + at, with u = 0
va = a(ta)
a = 320/2(tb)

vb = a(tb)
vb = (320/2tb)(tb)
vb = 160

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Answer:

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d. The momentum of the 3 kg trolley is -12 kg·m/s

e. The velocity of the 3 kg trolley = -4 m/s

Explanation:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0

c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s

d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s

e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley

∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s

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2 years ago

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Dovator [93]

Answer:

2188 years

Explanation:

given,

see level rising at the rate = 3.2 mm/yr

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time it will take rise 7 m = ?

we know,

1 m = 1000 mm

1 mm = 0.001 m

rate of increase of rate = 0.0032 m/year

time taken to rise = $\dfrac{7}{0.0032}\ years$

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FrozenT [24]

Answer:

f=15.5 Hz

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$A=\pi*r^2\\A= \pi*(5.9x10^-4m/2)^2=2.734*10^-7m^2$

$R=(1.68*10^-8)*(14.4m)/(2.734*10^-7m^2)= 0.884$ Ω

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Therefore, voltage drop inside generator =

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Actual EMF required is

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Note that this is an RMS value.

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$20.29v=(60)*(0.650T)*(0.06m)^2$ *ω

ω $=144.5\frac{rad}{s}$

f=ω/(2π)=

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2 years ago

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jasenka [17]

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We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

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So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

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We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

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2 years ago

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guapka [62]

Answer:

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