1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
vova2212 [387]
4 years ago
13

Light travels at 3 × 108 m/s, and it takes about 8 min for light from the sun to travel to Earth. Based on this, the order of ma

gnitude of the distance from the sun to Earth is g
Physics
1 answer:
N76 [4]4 years ago
5 0

Answer:

The order of magnitude of the distance from the sun to Earth is 10⁸ km.

Explanation:

The order of magnitude of the distance from the sun to Earth can be calculated as follows:

c = \frac{x}{t}

Where:

c: is the speed of light = 3x10⁸ m/s

t: is the time = 8 min

Hence, the distance is:

x = c*t = 3 \cdot 10^{8} m/s*8 min*\frac{60 s}{1 min} = 1.44 \cdot 10^{11} m = 1.44 \cdot 10^{8} km

Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.

I hope it helps you!

You might be interested in
A water pump draws about 9 A when connected to 240 V. What is the cost (with electrical energy at 13 cents per kWh) of running t
Fiesta28 [93]

Answer:

4.49 dollars

Explanation:

i = 9 A, V = 240 V, t = 16 h

Energy = V x i x t = 240 x 9 x 16 = 34560 W h = 34.56 kWh

The cost of 1 kWh is 13 cents.

Cost of 34.56 kWh = 13 x 34.56 = 449.28 cents = 449.28 / 100 = 4.49 dollars

8 0
4 years ago
Equation of motion, for a simplicity assume that the cart begins to moves at t=0 and reaches a speed of 0.44m/s at t=1.8 sec. Wr
nignag [31]

Answer:

cudkldllfkfklldldlflfkfkjkkfkfkfllflfkfkkkfllf

6 0
3 years ago
Is it possible to have a charge of 5 x 10-20 C? Why?
ruslelena [56]

1) No

2) Yes

3) No

4) Equal and opposite

5) 32400 N

6) Repulsive

7) The electric force is 2.3\cdot 10^{39} times bigger than the gravitational force

Explanation:

1)

In nature, the minimum possible charge that an object can have is the charge of the electron, which is called fundamental charge:

e=1.6\cdot 10^{-19}C

Electrons are indivisible particles (they cannot be separated), this means that an object can have at least the charge equal to the charge of one electron (in fact, it cannot have a charge less than e, because it would meant that the object has a "fractional number" of electrons).

In this problem, the object has a charge of

Q=5\cdot 10^{-20}C

If we compare this value to e, we notice that Q, so no object can have a charge of Q.

2)

As we said in part 1), an object should have an integer number of electrons in order to be charged.

This means that the charge of an object must be an integer multiple of the fundamental charge, so we can write it as:

Q=ne

where

Q is the charge of the object

n is an integer multiple

e is the fundamental charge

Here we have

Q=2.4\cdot 10^{-18}C

Substituting the value of e, we find n:

n=\frac{Q}{e}=\frac{2.4\cdot 10^{-18}}{1.6\cdot 10^{-19}}=15

n is integer, so this value of the charge is possible.

3)

We now do the same procedure for the new object in this part, which has a charge of

Q=2.0\cdot 10^{-19}C

Again, the charge on this object can be written as

Q=ne

where

n is the number of electrons in the object

Using the value of the fundamental charge,

e=1.6\cdot 10^{-19}C

We find:

n=\frac{Q}{e}=\frac{2.0\cdot 10^{-19}}{1.6\cdot 10^{-19}}=1.25

n is not integer, so this value of charge is not possible, since an object cannot have a fractional number of electrons.

4)

To solve this part, we use Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (Action force), then object B exerts an equal and opposite force on object A (reaction force)".

In this problem, we have two objects:

- A charge Q

- A charge 5Q

Charge Q exerts an electric force on charge 5Q, and we can call this action force. At the same time, charge 5Q exerts an electric force on charge Q (reaction force), and according to Newton's 3rd law, the two forces are equal and opposite.

5)

The magnitude of the electric force between two single-point charges is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the two charges

In this problem we have:

q_1=+4.5\cdot 10^{-6}C is charge 1

q_2=+7.2\cdot 10^{-6}C is charge 2

r = 0.30 cm = 0.003 m is the separation

So, the electric force  between the two charges is

F=(9\cdot 10^9)\frac{(4.5\cdot 10^{-6})(7.2\cdot 10^{-6})}{(0.003)^2}=32400 N

6)

The electric force between two charged objects has direction as follows:

- If the two objects have charges of opposite signs (+ and -), the force between them is attractive

- If the two objects have charges of same sign (++ or --), the force between them is repulsive

In this problem, the two charges are:

q_1=+4.5\cdot 10^{-6}C is charge 1

q_2=+7.2\cdot 10^{-6}C is charge 2

We see that the two charges have same sign: therefore, the force between them is repulsive.

7)

The electric force between the proton and the electron in the atom can be written as

F_E=k\frac{q_1 q_2}{r^2}

where

q_1 = q_2 = e = 1.6\cdot 10^{-19}C is the magnitude of the charge of the proton and of the electron

r=5.3\cdot 10^{-11} m is the separation between them

So the force can be rewritten as

F_E=\frac{ke^2}{r^2}

The gravitational force between the proton and the electron can be written as

F_G=G\frac{m_p m_e}{r^2}

where

G is the gravitational constant

m_p = 1.67\cdot 10^{-27}kg is the proton mass

m_e=9.11\cdot 10^{-27}kg is the electron mass

Comparing the 2 forces,

\frac{F_E}{F_G}=\frac{ke^2}{Gm_p m_e}=\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})^2}{(6.67\cdot 10^{-11})(1.67\cdot 10^{-27})(9.11\cdot 10^{-31})}=2.3\cdot 10^{39}

8 0
3 years ago
A Ferris wheel is a vertical, circular amusement ride with radius 9.5 m. Riders sit on seats that swivel to remain horizontal. T
Kruka [31]

Answer:

928.34 N

Explanation:

Radius=r=9.5 m

Time period of revolution=T=7.6 s

Mass of rider=m=57 kg

We have to find the rate of change of the rider's momentum at the bottom of the ride

According to Newton's second law of motion

F_{net}=ma_c

N-mg=mr\omega^2

N=mg+mr(\frac{2\pi}{T})^2

Where \pi=3.14,g=9.8 m/s^2

Using the formula

N=57\times 9.8+57\times 9.5(\frac{2\times 3.14}{7.6})^2

N=928.34 N

Hence, the rate of change of the rider's momentum=928.34 N

4 0
3 years ago
Sea level is currently rising at 3.3 mm/yr, and scientists predict that global warming could cause a rise in sea level of 7 m if
olga2289 [7]

Answer:

A. 2,121 years

Explanation:

This question can be looked at as a speed and distance question where 3.3 mm/yr is the speed and 7 m is the distance. So as the formula normally goes;

time = distance/speed

       = \frac{7 m}{3.3mm/yr} =\frac{7m}{0.0033m/yr} = 2121 years

Please note that it was important to change the rate of rising of the sea from mm/year to m/year because the distance was in metres

5 0
3 years ago
Other questions:
  • A green croquet ball of mass 0.50 kg is rolling at +12 m/s. It collides with a blue croquet ball that also
    11·1 answer
  • A skydiver falls toward the ground at a constant velocity. Which statement best applies Newton’s laws of motion to explain the s
    6·2 answers
  • What types of bonds do you think the good conductors of electricity have?
    11·1 answer
  • Which two objects repel each other?
    13·1 answer
  • A bridge is made with segments of concrete 50 m long. If the linear expansion coefficient is 12  10–6 (°C)–1 , how much spacing
    5·1 answer
  • What height would a 4 kg book need to be to have a potential energy of
    12·1 answer
  • What is the potential difference when the current in a circuit is 5mA and resistance is 30 Ohms
    13·1 answer
  • A bromine atom has an atomic number of 35 and an atomic mass of 80. What is the structure of this atom?
    12·2 answers
  • Which is not poor conductor?
    12·1 answer
  • A crash test dummy with a mass of 55kg is placed into a car traveling at 40m/s. During the first trial the dummy does not wear a
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!