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Rufina [12.5K]
2 years ago
9

Assume that you stay on the Earth's surface. What is the ratio of the sun's gravitational force on you to the earth's gravitatio

nal force on you? Express your answer using two significant figures.
Physics
1 answer:
andrew11 [14]2 years ago
4 0

First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg. 

F = Gm₁m₂/d²

where

m₁ = 5.972×10²⁴ kg (mass of earth)

m₂ = 60 kg

d = 6,371,000 m (radius of Earth)

G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²

F = 589.18 N

Next, we find the gravitational force exerted by the Sun by replacing,

m₁ = 1.989 × 10³⁰ kg

Distance between centers of sun and earth = 149.6×10⁹ m

Thus,

d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m

Thus,

F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²

F = 0.356  N

Ratio = 0.356  N/589.18 N

Ratio = 6.04

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Kate is researching air pollution and finds some information on ozone. She knows that ozone is a good thing as part of the ozone
jek_recluse [69]

"At ground level, ozone contributes to smog" so it is also an air pollutant.

Option: A

<u>Explanation</u>:

ozone is naturally present in stratosphere and acts as shield against harmful ultraviolet radiations. But it acts a pollutant contributing to global warming when it is present in lower level atmosphere particularly troposphere. In this level it combines with primary pollutants that is "nitrogen oxides" and "volatile organic" compounds to form secondary pollutant which absorbs outgoing radiation and contributes in raising the temperature. It has harmful impacts on vegetation as well as human health.

7 0
2 years ago
The headlights of a car emit light of wavelength 400 nm and are separated by 1.2 m. The headlights are viewed by an observer who
densk [106]

Answer:

The most correct option is;

B. 10 km

Explanation:

L = \frac{y \times d}{1.22 \times  \lambda} = \frac{1.2 \times 0.004}{1.22 \times  400 \times 10^{-9}} = 9836.066 \ km

Where:

y = Distance between the two headlights

d = Aperture of observers eye

λ = Wavelength of light

L = Distance between the observer and the headlight

Therefore, from the above solution, the distance between the observer and the headlights is 9386.066 km which is approximately 10 km.

Also we have

sinθ = y/L = 1.22 (λ/d)  

= 1.22 \times \frac{400 \times 10^{-9}}{0.004}

sinθ = 1.22×10⁻⁴ rad

6 0
2 years ago
Read 2 more answers
A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s2 . How big is the frictional force on the block
solniwko [45]
We can first calculate the net force using the given information.

By Newton's second law, F(net) = ma:

F(net) = 25 * 4.3 = 107.5

We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):

f = F(net) - F(app) = 150 - 107.5 = 42.5 N

Now we can calculate the coefficient of friction, u, using the normal force, F(N):

f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
4 0
2 years ago
If it takes an airplane 15 minutes to go from 30 mph to 330 mph, what is its acceleration?
zzz [600]
Using the a=vf-vi divided by tf-ti:
A is acceleration
Vf is final velocity- 330
Vi is intial velocity-30
Tf is final time-15
Ti is initial time-0
A = 330-30 divided by 15-0
A = 300 divided by 15
A= 20 m/s^2 
Hope this helps
3 0
2 years ago
What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?
Stolb23 [73]

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

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

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

8 0
2 years ago
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