Answer:
Explanation:
Hello.
In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:
The distance is clearly 1743 km and the time is:
Thus, the velocity turns out:
Which is a typical velocity for a plane to allow it be stable when flying.
Best regards.
The correct formula to use is: F = G [M1*M2] /r^2
Where,
G = the force of gravity
M1 = the mass of the first object [the mass of the astronaut]
M2 = the mass of the second object [the mass of the planet]
r = the distance between the two objects in metre
F = 6.67 * 10^-11 [66.5 * 8.43 * 10^23] / [4.40 * 10^6]^2
F = 193N.<span />
Answer:
8.9m/s
Explanation:
Final velocity = initial velocity + acceleration *time
Vf=vi+at
Vf=1.4 m/s
Vi=?
A=-2.5m/s^2
T=3
1.4m/s=vi+(-2.5m/s^2)(3s)
Vi=8.9m/s
The speed of the pin after the elastic collision is 9 m/s east.
<h3>
Final speed of the pin</h3>
The final speed of the pin is calculated by applying the principle of conservation of linear momentum as follows;
m1u1 + mu2 = m1v1 + m2v2
where;
- m is the mass of the objects
- u is the initial speed of the objects
- v is the final speed of the objects
4(1.4) + 0.4(0) = 4(0.5) + 0.4v2
5.6 = 2 + 0.4v2
5.6 - 2 = 0.4v2
3.6 = 0.4v2
v2 = 3.6/0.4
v2 = 9 m/s
Thus, The speed of the pin after the elastic collision is 9 m/s east.
Learn more about linear momentum here: brainly.com/question/7538238
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Answer:
E = 9.4 10⁶ N / C
, The field goes from the inner cylinder to the outside
Explanation:
The best way to work this problem is with Gauss's law
Ф = E. dA = qint / ε₀
We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.
The flow on the faces is zero, since the field goes in the radial direction of the cylinders.
The area of the cylinder is the length of the circle along the length of the cable
dA = 2π dr L
A = 2π r L
They indicate that the distance at which we must calculate the field is
r = 5 R₁
r = 5 1.3
r = 6.5 mm
The radius of the outer shell is
r₂ = 10 R₁
r₂ = 10 1.3
r₂ = 13 mm
r₂ > r
When comparing these two values we see that the field must be calculated between the two housings.
Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is
λ = q / L
Qint = λ L
Let's replace
E 2π r L = λ L /ε₀
E = 1 / 2piε₀ λ / r
Let's calculate
E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3
E = 9.4 10⁶ N / C
The field goes from the inner cylinder to the outside
