The gravitational force the sun experiences from the earth is 3.48×10²²N, which is exactly the same as the force the sun experiences from the earth.
- Gravity is a force that develops as a result of the attraction between mass-containing objects. The mass of the object has a direct relationship to the strength of this attraction. r equals the separation of two objects.
F = G (M₁M₂)/r²
Where, F the gravitational force
G=6.67×10⁻¹¹Nm²kg⁻² gravitational constant
M₁=5.98×10²⁴kg mass of earth
M₂= 1.99×10³⁰ kg the mass of the sun
r =15×10¹⁰ m is the distance between sun and earth
Putting all the values in above equation,
F = 6.67×10⁻¹¹Nm²kg⁻²(5.98×10²⁴kg 1.99×10³⁰ kg)/15×10¹⁰ m
On solving the above equation we get,
F = 3.48×10²²N
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Choice-a is a very rubbery, imprecise, ambiguous, slippery statement. But it's probably less wrong than any of the other choices on the list.
Answer:
The lines slope up from the red ball to the corresponding blue ball.
The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of
<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N
Then the work done by this force on the crate as it slides down the ramp is
<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J
The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so
<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2
Solve for <em>v</em> :
<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s
Answer:
t = (ti)ln(Ai/At)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Explanation:
Let Ai represent the initial amount and At represent the final amount of beryllium-11 remaining after time t
At = Ai/2^n ..... 1
Where n is the number of half-life that have passed.
n = t/half-life
Half life = 14
n = t/14
At = Ai/2^(t/14)
From equation 1.
2^n = Ai/At
Taking the natural logarithm of both sides;
nln(2) = ln(Ai/At)
n = ln(Ai/At)/ln(2)
Since n = t/14
t/14 = ln(Ai/At)/ln(2)
t = 14ln(Ai/At)/ln(2)
Ai = 800
At = 50
t = 14ln(800/50)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Let half life = ti
t = (ti)ln(Ai/At)/ln(2)
