Answer: Yep! I made a yt vid on them so look this up:
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Answer:
mparing this value with the weight of the vehicle we see W₂> 1.6 10⁴ N
therefore the woman can lift the car
Explanation:
For this exercise on Pascal's principle, we use that the pressure in every incompressible liquid is the same
P₁ = P₂
the press is defined by
P = F / A
for the woman's side
the area of a circle is
A = π D₁² / 4
P₁ = W (4 /π d₁²)
for the car side
A₂ =π d₂²
P₂ = W₂ (4 /π d₂²)
we substitute in the first equation
W (4 /π d₁²) = W₂ 4 /π d₂²)
W / d₁² = W₂ / D₂²
W₂ = (d₂ / d₁)² W
the weight of the woman is
W = mg
we calculate
W₂ = (24/3)₂ 50 9.8
W₂ = 3,136 10⁴ N
when comparing this value with the weight of the vehicle we see
W₂> 1.6 10⁴ N
therefore the woman can lift the car
The correct answer is: Option (3) 9.8 N/kg
Explanation:
According to Newton's Law of Gravitation:
--- (1)
Where G = Gravitational constant = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²
m = Mass of the body = 2 kg
M = Mass of the Earth = 5.972 × 10²⁴ kg
R = Distance of the object from the center of the Earth = Radius of the Earth + Object's distance from the surface of the Earth = (6371 * 10³) + 3.0 = 6371003 m
Plug in the values in (1):
(1)=>
Now that we have force strength at the location, we can use:
Force = mass * gravitational-field-strength
Plug in the values:
19.63 = 2.0 * gravitational-field-strength
gravitational-field-strength = 19.63/2 = 9.82 N/kg
Hence the correct answer is Option (3) 9.8 N/kg
Answer:
here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location
Explanation:
As we know that gravitational field is defined as the force experienced by the satellite per unit of mass
so we will have
now in order to find the acceleration of the satellite we know by Newton's II law
so we will have
so here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location
Answer:
≈ 6.68 m/s
Explanation:
A suitable formula is ...
vf^2 -vi^2 = 2ad
where vi and vf are the initial and final velocities, a is the acceleration, and d is the distance covered.
We note that if the initial launch direction is upward, the velocity of the ball when it comes back to its initial position is the same speed, but in the downward direction. Hence the problem is no different than if the ball were initially launched downward.
Then ...
vf = √(2ad +vi^2) = √(2·9.8 m/s^2·1.0 m+(5 m/s)^2) = √44.6 m/s
vf ≈ 6.68 m/s
The ball hits the ground with a speed of about 6.68 meters per second.
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We assume the launch direction is either up or down.