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velikii [3]
3 years ago
15

What is Obama's last name? PLZ PLZ PLZ THIS IS 500% OF MY GRADE PLS HELP PLS ​

Physics
2 answers:
kherson [118]3 years ago
5 0

Answer:

Explanation:

                              B                      A                   R                    A                 K

                            O                   B                     A                     M                  A

kotegsom [21]3 years ago
4 0

Answer:

Barack Hussein Obama II

Explanation:

I think

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A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.
marishachu [46]

Answer:

a) the magnitude of the force is

F= Q(\frac{kqs}{r^3}) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b)  the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = \frac{kq}{r^{2} }

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = \frac{kp}{r^{3} } , where p = q × s

E(r) = \frac{kqs}{r^{3} } where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q(\frac{kqs}{r^3})

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall  F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

8 0
3 years ago
A 0.140-kg baseball is pitched horizontally at 33.6 m/s. When a player hits the ball, it moves at the
Mars2501 [29]

Answer:

F = 1176 N

Explanation:

Given that,

Mass of a baseball, m = 0.140-kg

Initial speed of the baseball, u = 33.6 m/s

Final speed of the baseball, v = -33.6 m/s (in opposite direction)

The time of contact of the bat and the ball, t = 0.008 s

We need to find the  average force the bat exerts on the ball. The force acting on the ball is given by :

F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.14(-33.6-33.6)}{0.008}\\F= -1176\ N

So, the average force the bat exerts on the ball is 1176 N.

7 0
3 years ago
A delivery drone uniformly slowed down from 45 m/s to 23 m/s in a straight line over a 30-minute duration within a 16-kilometer
Neporo4naja [7]
I think it is letter A
7 0
3 years ago
An object with mass 110 kg moved in outer space. When it was at location < 13, -18, -2 > its speed was 19.5 m/s. A single
Serhud [2]

Answer:

v = 21.4m/s

Explanation:

Given r1 = < 13, -18, -2 > m,

F = < 250, 390, -220 > N, r2 = < 19, -23, -5 > m

Where r1 and r2 are position vectors in space and F is the constant force vector in space.

r2 - r1 = < 19–13, -23–(-18), -5–(-2) > m

Δr = < 6, -5, -3 > m

F = m×a

a = F/m

m = 110kg (scalar quantity)

a = < 250/110, 390/110, -220/110 >

a = < 2.27, 3.55, -2 > m/s²

Since the force is constant, the acceleration is also constant. Therefore the equations of constant acceleration motion apply here.

V² = u² +2aΔr

The magnitude of the acceleration is calculated as follows

a = √(2.27² + 3.55² +(-2)²)

a = √(21.7554)

a = 4.66m/s²

Magnitude of Δr,

Δr = √(6² + (-5)² +(-3)²) = √(70) = 8.37m

Having calculated the magnitude of the acceleration and displacement, we can now calculate the final velocity.

u = initial velocity = 19.5m/s

v² = 19.5² + 2×4.66×8.37

v² = 458.25

v = √(458.25)

v = 21.4m/s

6 0
3 years ago
This problem explores the behavior of charge on realistic (i.e. non-ideal) insulators. We take as an example a long insulating r
frez [133]

Answer:

A) weakly attracted

B) negative charge on end A with B remaining almost neutral

C) strongly repelled

Explanation:

A)

  • As the negatively charged ball is brought close to the end A of the road, positive charges on this end try to move away from there, leaving a small negative charge in this end, as they are not free to move as in a conductor.
  • As a consequence of this, the end A of the road will be weakly attracted by the negatively charged ball.

B)

  • After several contacts between the ball and the end A of the rod, an appreciable quantity of negative charge has been deposited in the end A, repelling positive charges towards the end B.
  • However, as the rod is an insulator, positive charges can't travel massively to the end B, so it remains almost neutral.

C)

  • After repeated contacts that have deposited a significative negative charge on the end A (which can't distribute evenly as in a conductor), when the ball (negatively charged) is brought close to this end of the rod, as both bodies have the same type of charge, they will be strongly repelled each other.
3 0
4 years ago
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