(a) The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 10 m/s.
(b) The velocity and acceleration at any time t is v = (4ti + j) m/s and a = a = 4i m/s²
(c) The average acceleration in the time interval given in part (a) is 3.98 m/s².
<h3>Position of the particle</h3>
x = at²i + btj
x = 2t²i + tj
<h3>Average velocity, at t₁=2sec and t₂=3sec</h3>
Δv = Δx/Δt
x(2) = 2(2)²i + 2j
x(2) = 8i + 2j
|x(2)| = √(8² + 2²) = 8.246
x(3) = 2(3)²i + 3j
x(3) = 18i + 3j
|x(3)| = √(18² + 3²) = 18.248
Δv = (18.248 - 8.246)/(3 - 2)
Δv = 10 m/s
<h3>Velocity and acceleration at any time, t</h3>
v = dx/dt
v = (4ti + j) m/s
a = dv/dt
a = 4i m/s²
<h3>Average acceleration</h3>
v(2) = 4(2)i + j
v(2) = 8i + j
|v(2)| = 8.06 m/s
v(3) = 4(3)i + j
v(3) = 12i + j
|v(3)| = 12.04 m/s
a = (12.04 - 8.06)/(3 - 2)
a = 3.98 m/s²
Learn more about average acceleration here: brainly.com/question/104491
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This looks complicated, but it's actually not too tough.
The formula for the gravitational force between two objects is
Force = G (one mass) (other mass) / (distance²) .
The question GAVE us all of those numbers except the distance.
All we have to do is pluggum in, massage it around, and find
the distance.
Force = 4.18 x 10¹⁵ N
G = 6.673 x 10⁻¹¹ N·m²/kg²
One mass = 6.58 x 10²³ kg
Other mass = 9.3 x 10¹⁵ kg .
The only tricky thing about this is gonna be the arithmetic ...
keeping all the exponents straight.
Take the formula for the gravitational force and plug in
everything we know:
Force = (G) · (one mass) · (other mass) / (distance²)
4.18x10¹⁵N = (6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg) / (distance²).
Multiply each side by (distance²):
(distance²)·(4.18x10¹⁵N) = (6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg)
Divide each side by (4.18 x 10¹⁵ N) :
(distance²)=(6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg) / (4.18x10¹⁵N)
That's the end of the Physics and Algebra. The only thing left is Arithmetic.
We have to simplify that whole ugly thing on the right side of the equation,
and then take the square root of each side.
When I crunch down the right side of that equation, I get
(distance²) = 9.769 x 10¹³ m²
and when I take the square root of each side, I get
distance = 9.884 x 10⁶ meters . **
You should check my Arithmetic. **
(Pause occasionally to let your calculator cool off.)
BY THE WAY ...
That "distance" in the equation for gravitational force is the distance
between the CENTERS of the two objects.
This doesn't make much difference for Phobos, because Phobos isn't
much bigger than a big sweet potato. But it does make a difference for
Mars.
The 'distance' we find with all of this nonsense is NOT the distance
between Phobos and the surface of Mars. It's the distance between
Phobos and the CENTER of Mars, so it includes the planet's radius.
** Consulting online resources between Floogle and Flickerpedia,
I found that the orbital distance of Phobos from Mars varies between
9,234 km and 9,517 km. Add the planet's radius to these, and I'm
beginning to feel confidence in the results of my back-of-the-napkin
calculation. But you should still check my Arithmetic.
Answer:
1. they both act on an object in free fall
Explanation:
2. both help determine how fast the object will accelerate
Answer:
p = 1.16 10⁻¹⁴ C m and ΔU = 2.7 10 -11 J
Explanation:
The dipole moment of a dipole is the product of charges by distance
p = 2 a q
With 2a the distance between the charges and the magnitude of the charges
p = 1.7 10⁻⁹ 6.8 10⁻⁶
p = 1.16 10⁻¹⁴ C m
The potential energie dipole is described by the expression
U = - p E cos θ
Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line
Orientation parallel to the field
θ = 0º
U = 1.16 10⁻¹⁴ 1160 cos 0
U1 = 1.35 10⁻¹¹ J
Antiparallel orientation
θ = 180º
cos 180 = -1
U2 = -1.35 10⁻¹¹ J
The difference in energy between these two configurations is the subtraction of the energies
ΔU = | U1 -U2 |
ΔU = 1.35 10-11 - (-1.35 10-11)
ΔU = 2.7 10 -11 J
