Ask question

Ask question

All categories

3 years ago

15

The voyager 1 spacecraft is currently 20.7 billion km from earth and heading out of our solar system. how long does it take radi

o messages from voyager 1 to reach us

1 answer:

sveta [45]3 years ago

8 0

Answer:

20.7 billion = 20.7E9 converting to units of 10

20.7E9 km = 20.7E12 m

t = S /v = 20.7E12 m / 3.0E8 m/s = 6.9E4 s = 69000 s

69000 s = 1150 min = 19.2 hrs

You might be interested in

Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.

vlabodo [156]

GAVNSTVVHUYCZ swffggff

4 0

3 years ago

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te

QveST [7]

Answer:

Please see attachment

Explanation:

Please see attachment

8 0

3 years ago

A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge

Kisachek [45]

Answer:

$\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

$\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r$

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

$\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

5 0

3 years ago

A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.

sladkih [1.3K]

Explanation:

Given that,

Charge 1, $q_1=6.75\ nC=6.75 \times 10^{-9}\ C$

Charge 2, $q_2=4.46\ nC=4.46\times 10^{-9}\ C$

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

$F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}$

$F=6.84\times 10^{-8}\ N$

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.

5 0

3 years ago

Can you cool a kitchen by leaving the refrigerator door open while the refrigerator is operating? a. No. Since the refrigerator

slava [35]

Answer:

no

Explanation:

because will the cool air will be leaving the fridge the motor in the bottom will be running and eventually the mixture of the cool and hot air will balance out

7 0

3 years ago

Read 2 more answers