All categories

4 years ago

How long does a radar signal take to travel from earth to venus and back when venus is brightest?

Physics

1 answer:

Sergio [31]4 years ago

5 0

Given data :

Velocity of radar signal, v = Velocity of light = 3 x 10^8 m/s

Greatest distance between Venus and Earth d1 = 0.47 AU = 0.47 AU (a.5 x 10^11 m / 1 AU) = 7.1x10^10 m

Closest distance between Venus and Earth, d2 = 0.28 AU = 0.28 AU (1.5 x 10^11 m / 1 AU)= 4.2 x 10^10 m

The time take by radar signal to travel from Earth to Venus and back when Venus is brightest is calculated as

T1 = 2d1/v = 2(7.a x 10^10 m / 3 x10^8 m/s) = 473 s

The time taken by radar signal to travel from Earth to Venus and back when Venus is closest is calculated as

You might be interested in

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

3 years ago

A jet with mass m = 90000.0 kg jet accelerates down the runway for takeoff at 1.6 m/s2.

leva [86]

This question requires the use of the equation of motion:
v = u + at [v is final velocity (0), u is initial velocity (24), a is acceleration, t is time (13)]
to calculate the acceleration. This can then be multiplied by the mass of the plane to obtain the net force via:
F = ma (F is force, m is mass, a is acceleration)
First, we calculate the acceleration:
0 = 24 + 13(a)
a = -24/13 m/s^2
The force is then:
F = 90000 * (-24/13)
F = -1.66*10^5 Newtons
The negative sign indicates that the force and acceleration are in the opposite direction as the velocity (since we took velocity to be positive)

6 0

4 years ago

The speed v of a sound wave traveling in a medium that has bulk modulus b and mass density ρ (mass divided by the volume) is v=b

PilotLPTM [1.2K]

As it is given that Bulk modulus and density related to velocity of sound

$v = \sqrt{\frac{B}{\rho}}$

by rearranging the equation we can say

$B = \rho * v^2$

now we need to find the SI unit of Bulk modulus here

we can find it by plug in the units of density and speed here

$B = \frac{kg}{m^3} * (\frac{m}{s})^2$

so SI unit will be

$B = \frac{kg}{m* s^2}$

SO above is the SI unit of bulk Modulus

3 0

3 years ago

a) A unit of time sometimes used in microscopic physics is the shake. One shake equals 10–8 s. Are there more shakes in a second

lawyer [7]

Answer:

a) Yes, there are $10^8$ shakes in a second ( $1 s \frac{1 shake}{10^{-8}s}=10^8 shake$ ) in a year there are 31,536,000 seconds... that is 3.1536x $10^7$ ( $1year\frac{365 day}{1 year} \frac{24 h}{1 day} \frac{60 min}{1 h}\frac{60s}{1min}=3.1536 *10^7$ )

b) Note that the defined universe day will have 60*60*24=86400 universe seconds if the seconds are defined as normal seconds.

Also one universe day (or 86400 universe seconds) is equivalent to 1010 years. Now using a rule of three we can know the seconds that humans have existed.

$106 years * \frac{86,400 universe-seconds}{1,010 years}=9,067.728 s$

That is about 2 and half hours.

3 0

4 years ago

An isloated point charge produce an electric field with magnitude E at a point 2 m away. At a point 1 m from the charge magnitud

Nina [5.8K]

Answer:

the correct answer is C, E’= 4E

Explanation:

In this exercise you are asked to calculate the electric field at a given point

E = $k \frac{q}{r^2}$

indicates that the field is E for r = 2m

E = $\frac{ k q}{4}$ (1)

the field is requested for a distance r = 1 m

E ’= k \frac{q}{r'^2}

E ’= k q / 1

from equation 1

4E = k q

we substitute

E’= 4E

so the correct answer is C

8 0

3 years ago