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
sukhopar [10]
3 years ago
6

"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and

the Sun."
Physics
1 answer:
spayn [35]3 years ago
8 0

Answer:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

You might be interested in
A block is 10cm long, 5cm wide and 2cm high and weighs 100g. What is the volume of the block? What is the density?
Ivahew [28]

Answer:

1gm/cm^3

Explanation:

its the answer

5 0
3 years ago
The downsprue leading into the runner of a certain mold has a length of 175 mm. The cross-sectional area at the base of the spru
adoni [48]

Answer:

(a) Velocity at bottom is 1.85 m/s

(b) Volume flow rate is 7.4 x 10⁻⁴ m³/s.

(c) The time required to fill the mold is 1.35 s.

Explanation:

(a)

Applying Bernoulli's Equation on both ends of the down sprue, with the assumptions that every point is at atmospheric pressure and the liquid metal at the pouring basin is at zero velocity. The equation then becomes:

V = √2gh

where,

V = velocity at bottom of down sprue

h = height of down sprue = 175 mm = 0.175 m

V = √2(9.8 m/s²)(0.175 m)

<u>V = 1.85 m/s</u>

<u></u>

(b)

The volume flow rate is given as:

Volume Flow Rate = (V)(A)

where,

V = velocity at bottom = 1.85 m/s

A = Area of bottom = 400 mm² = 0.0004 m²

Therefore,

Volume Flow Rate = (1.85 m/s)(0.0004 m²)

<u>Volume Flow Rate = 7.4 x 10⁻⁴ m³/s = 740 cm³/s</u>

(c)

The time required to fill the cavity is given as:

Volume Flow Rate = V/t

where,

V = Volume of mold Cavity = 0.001 m³

t = time required to fill the cavity = ?

Therefore,

t = V/Volume Flow Rate

t = 0.001 m³/7.4 x 10⁻⁴ m³/s

<u>t = 1.35 s</u>

<u></u>

5 0
3 years ago
A particle of mass 10 g and charge 72 μC moves through a uniform magnetic field, in a region where the free-fall acceleration is
sineoko [7]

Answer:

-0.07163\hat k\ T or 0.07163 T into the page

Explanation:

m = Mass of particle = 10 g

a = Acceleration due to gravity = -9.8j m/s²

v = Velocity of particle = 19i km/s

q = Charge of particle = 72 μC

B = Magnetic field

Here the magnetic and gravitational forces on the particle are applied in the opposite direction so,

F_b=F_g

F_b=qvBsin\theta\\\Rightarrow F_b=qvBsin90\\\Rightarrow F_b=72\times 10^{-6}\times 19000B

F_g=ma\\\Rightarrow F_g=0.01\times -9.8

72\times 10^{-6}\times 19000B=0.01\times -9.8\\\Rightarrow B=\frac{0.01\times -9.8}{72\times 10^{-6}\times 19000}\\\Rightarrow B=-0.07163\hat k\ T

The magnetic field is 0.07163 T into the page

5 0
3 years ago
If the radius of the atom is the distance from point A to D, where is the MOST likely location of the LEAST concentration of mas
lana [24]

Answer: B

Explanation:

3 0
3 years ago
*WILL MARK BRAINLIEST*
Arte-miy333 [17]

Answer:

1.52

Explanation:

5 0
3 years ago
Read 2 more answers
Other questions:
  • How could you increase the potential energy of the roller coaster shown below?
    10·2 answers
  • 17. The number of waves that pass a particular point in a body in a unit of time is called the _______ of the waves. A. rarefact
    14·1 answer
  • Give one example in which both physical and chemical changes takes places ​
    6·2 answers
  • Which most likely could cause acceleration? Check all that apply.
    7·1 answer
  • A solid disk of mass m1 = 9.3 kg and radius R = 0.22 m is rotating with a constant angular velocity of ω = 31 rad/s. A thin rect
    10·1 answer
  • A ball is thrown vertically downwards at speed vo from height h. Draw velocity vs. time &amp; acceleration vs. time graphs. In t
    10·1 answer
  • You are working as an expert witness for an attorney who is suing a shipping company. The company operates ships that carry crud
    8·1 answer
  • One type of BB gun uses a spring-driven plunger to blow the BB from its barrel.
    11·1 answer
  • A softball is hit into the outfield during a game. The initial velocity is 33.52 m/s, and the ball is hit at an angle of 27°. Ho
    10·1 answer
  • If two objects, like the eggs in the video, experience the same change in momentum but over time periods of
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!