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777dan777 [17]
3 years ago
11

A 128 kbs point to point link is set up between Earth and a rover on Mars. Use the distance between Earth and Mars as 55Gm and t

hat data travels over the link at the speed of light, 3x108 m/s. A camera on the rover takes pictures of its surroundings and sends the pictures to Earth. The pictures are 5 Mb in size. How long would it take a picture to reach the Earth?
Physics
1 answer:
saul85 [17]3 years ago
4 0

Answer:

Explanation:

Time Taken = Transmit + Propagation Delay

Propagation Delay = Distance/Speed of Light

= \frac{55\times 10^9}{3\times 10^8}\\\\= 183.334 seconds

Transmit = Size/Bandwidth

=\frac{5\times 10^6}{128\times 10^3}\\\\= 39.062 seconds

So, Total time = 183.334+39.062 = 222.396 seconds.

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Difference between discrete and continuous charge distribution?
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Discrete systems are those systems in which are made up of finite component particles a which are non-homogeneously arranged such that no smooth variation exists. It is such that all constituent particles have properties which vary randomly. They are direct opposite to continuous systems, which are smooth arrangement of particles which cannot be individually taken into consideration.
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5 0
3 years ago
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What is an equilibrant?
Stolb23 [73]
I’m pretty sure it is an object with a net force of zero. All forces are balanced and EQUAL
4 0
3 years ago
A4) A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.3
Klio2033 [76]

Answer:

\theta=30^{\circ}

Explanation:

It is given that,

Length of the wire, L = 0.6 m

Current flowing inside the wire, I = 2 A

Uniform magnetic field, B = 0.3 T

Force experienced by the wire in the magnetic field, F = 0.18 N

To find,

The angle made by the wire with the magnetic field.

Solve,

We know that the magnetic force acting on the wire inside the magnetic field is given by :

F=ILB\ sin\theta

sin\theta=\dfrac{F}{ILB}

sin\theta=\dfrac{0.18}{2\times 0.6\times 0.3}

\theta=30^{\circ}

Therefore, the wire makes an angle of 30 degrees with respect to magnetic field.

5 0
3 years ago
The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m
algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

a)

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

E=PE+KE=const.

The potential energy is given by

PE=\frac{1}{2}kx^2

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

E=KE_{max}=\frac{1}{2}mv_{max}^2 (1)

where

m is the mass of the block

v_{max}=1.25 m/s is the maximum speed

We also know that the total energy is

E=0.18 J

Re-arranging eq.(1), we can find the mass:

m=\frac{2E}{v_{max}^2}=\frac{2(0.18)}{(1.25)^2}=0.23 kg

b)

The maximum speed in a spring-mass system is also given by

v_{max} =\sqrt{\frac{k}{m}} A

where

k is the spring constant

m is the mass

A is the amplitude

Here we have:

v_{max}=1.25 m/s is the maximum speed

m = 0.23 kg is the mass

A = 14.0 cm = 0.14 m is the amplitude

Solving for k, we find the spring constant

k=\frac{v_{max}^2}{A^2}m=\frac{1.25^2}{0.14^2}(0.23)=18.3 N/m

c)

The frequency in a spring-mass system is given by

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k is the spring constant

m is the mass

In this problem, we have:

k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz

d)

We can solve this part by using the law of conservation of energy; in fact, we have

E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2

Where v is the speed of the system when the displacement is equal to x.

We know that the total energy of the system is

E = 0.18 J

Also we know that

k = 18.3 N/m is the spring constant

m = 0.23 kg is the mass

Substituting

x = 7.00 cm = 0.07 m

We can solve the equation to find the corresponding speed v:

v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s

#LearnwithBrainly

3 0
3 years ago
A hot–air balloon is moving at a speed of 10.0 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–d
Ad libitum [116K]

Answer:

Option (D) is correct.

Explanation:

The balloon lands horizontally at a distance of 420 m from a point where it as released.

Velocity of air balloon along +X axis =10 m/s

velocity of ball=4 m/s along + X axis

the velocity of balloon gets added to the velocity of ball. So the resultant velocity of the balloon=10+4 = 14 m/s

time taken= 30 s

The distance traveled is given by d= v t

d= 14 (30)

d= 420 m

Thus the balloon lands horizontally at a distance of 420 m from a point where it as released.

6 0
3 years ago
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