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Flura [38]
3 years ago
14

A satellite with mass 500 kg is placed in a circular orbit about Earth (Mass= 5.98 x 10^24 kg), radius = (6.4 x 10^6), a distanc

e of 1500 km above the surface. (a) what is the force gravity acting on satellite? (b) what is the satellite’s acceleration? (c) what is the satellite’s orbital speed?
Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
4 0

Explanation:

a) F = GmM / r²

F = (6.67×10⁻¹¹) (500) (5.98×10²⁴) / (6.4×10⁶ + 1.5×10⁶)²

F = 3200 N

b) F = ma

3200 = 500a

a = 6.4 m/s²

c) a = v² / r

640 = v² / (6.4×10⁶ + 1.5×10⁶)

v = 7100 m/s

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3 years ago
Two narrow slits 0.02 mm apart are illuminated by light from a CuAr laser (λ = 633 nm) onto a screen. a) If the first fringe is
Masja [62]

Answer:

a) 0.063m

b) 2.72°

c) 3151 fringes

d) 1.87*10^-6m

Explanation:

a) To find the screen distance you use the following formula:

y=\frac{m\lambda D}{d}\\\\D=\frac{dy}{m\lambda}

D: screen distance

d: distance between slits

m: order of the fringes

λ: wavelength

By replacing the values of the parameters you obtain:

D=\frac{(0.02*10^{-3}m)(0.2*10^{-2}m)}{(1)(633*10^{-9}m)}=0.063m

b) The condition for dark fringes is given by:

\lambda(m+\frac{1}{2})=dsin\theta

for the first dark fringe the angle is:

\theta=sin^{-1}(\frac{\lambda(m+\frac{1}{2})}{d})\\\\\theta=sin^{-1}(\frac{(633*10^{-9}m)(1+\frac{1}{2})}{0.02*10^{-3}m})=2.72\°

c) the visible number of fringes is given by:

N=1+2\frac{D}{d}=1+\frac{0.063m}{0.02*10^{-3}}=3151 \ fringes

d) the wavelength of a laser in which its first order fringe coincides with the third one of the CuAr laser is:

y=\frac{(3)(633*10^{-9}m)(0.063m)}{0.02*10^{-3}m}=5.98*10^{-3}m\approx0.59cm\\\\\lambda'=\frac{dy}{mD}=\frac{(0.02*10^{-3}m)(0.59*10^{-2}m)}{(1)(0.063m)}=1.87*10^{-6}m

4 0
3 years ago
5. A 1052 kg truck, starting from rest, reaches a speed of 20.0 m/s in 6.20 s.
BaLLatris [955]

Answer:

a. 3,392.7 N

b. 3,392.7 N

Explanation:

We are given the following information;

  • Mass of the truck as 1052 kg
  • initial speed as 0 m/s
  • Final speed as 20.0 m/s
  • Time taken as 6.20 s

#a. We are required to calculate the acceleration;

We need to know the formula of getting acceleration;

a = (v-u)/t

Where v is the final velocity, u is the initial velocity

Therefore;

a = (20 m/s - 0 m/s)/6.20s

= 3.225 m/s²

Thus, the average acceleration of the truck is 3.225 m/s²

#b. We are required to calculate the net force on the truck

We need to know that;

According to the second Newton's law of motion, F=ma

Where F is the net force, m is the mass and a is the acceleration.

Therefore;

Net force, F = mass × Acceleration

                    = 1052 kg × 3.225 m/s²

                    = 3,392.7 N

Thus, the net force on the truck is 3,392.7 N

4 0
4 years ago
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