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2 years ago

I need help with question two

Physics

1 answer:

zaharov [31]2 years ago

6 0

I can't see numbers here, so here are all the answers:
1) the frequency is c/λ = 3e8/556e-9 = 5.39e14Hz
2) light travels at <span>299,792,458 m/s. So in nanoseconds it's 0.299792458m. This is about 1/3 of a meter which is about one foot.
3) length is L = ct = (</span>299,792,458 m/s)(6e-15) = 1.799e-6m or 1.799μm

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Spinning satellites are particularly stable. suppose that the astronauts launching a particular satellite decide to increase its

Crank

Supposing velocity is speed and direction, and momentum is mass*velocity, if the velocity increases by a factor of 5, then so should the momentum regardless of the whole space thing.

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3 years ago

According to the law of conservation of energy, the total amount of energy in the universe does not change, but remains constant

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2 years ago

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A rectangular pane of glass is 91.1 cm wide and 155.9 cm long, and its area is equal to the length multiplied by the width. Usin

Kamila [148]

911.1×155.9 = 14,202.49
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2 years ago

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Two satellites are in circular orbits around a planet that has radius 9.00×106 m. One satellite has mass 68.0 kg, orbital radius

olasank [31]

Answer:

v₂ = 5131.42 m/s

Explanation:

given,

radius of the planet = r₁ = 9.00×10⁶ m

mass of the satellite = m₁ = 68 Kg

orbital radius = r₁ = 8 x 10⁷ m

orbital speed = v₁ = 4800 m/s

mass of second satellite = m₂ = 84.0 kg

orbital radius = r₂ = 7.00×10⁷ m

orbital speed of second satellite = v₂ = ?

using orbital speed of satellite

$v = \sqrt{\dfrac{GM}{r}}$

so,

$v \alpha \dfrac{1}{\sqrt{r}}$

now,

$\dfrac{v_1}{v_2} =\sqrt{\dfrac{r_2}{r_1}}$

$\dfrac{v_2}{4800} =\sqrt{\dfrac{8\times 10^7}{7 \times 10^7}}$

v₂ = 5131.42 m/s

The orbital speed of second satellite is equal to v₂ = 5131.42 m/s

7 0

2 years ago

A 2.0-kg block is held at rest against a spring of constant 2700 N/m, compressing it 5.5 cm. The block and the spring are locate

KonstantinChe [14]

Answer:

a) 4.1 J

b) -14 J

c) 4.8 m/s

Explanation:

The energy stored in the spring is given by:

$U_e=\frac{1}{2}*K*x^2\\\\U_e=\frac{1}{2}*2700N/m*(5.5*10^{-2}m)^2\\\\U_e=4.1J$

The mechanical energy loss is because of the work done by the friction force.

The friction force (only presented on the inclined surface) is given by:

$F_f=\µ*F_N$

$F_N=2.0kg*9.8m/s^2*cos(35)\\F_N=16N\\F_f=0.29*16N=4.6N$

We need to calculate the length of the ramp in order to calculate the work, the length of the ramp is the hypotenuse:

$sin(\theta)=\frac{OC}{h}\\h=\frac{OC}{sin(\theta)}\\\\h=\frac{1.7m}{sin(35)}\\\\h=3.0m$

So the work done by the friction force is:

$W_f=F_f*d*cos(\alpha)\\W_f=4.6N*3.0*cos(180)\\W_f=-14J$

the angle is 180 degrees because the force is opposite to the motion.

In order the know the final velocity we need to apply the Energy Conservation Theorem:

$K_1+U_{g1}+U_{e}+W_f=K_2+U_{g2}\\\\0+m*g*h_1+4.1J-14J=\frac{1}{2}*m*v^2+m*g*h_2\\\\2.0kg*9.8m/s^2*1.7m+4.1J-14J=\frac{1}{2}*2.0kg*v^2+2.0kg*9.8m/s^2*(0)\\\\33J+4.1J-14J=v^2\\\\v=\sqrt{23.1}\\\\v=4.8m/s$

5 0

2 years ago