Answer:
= 4.3 × 10 ⁻¹⁴ m
Explanation:
The alpha particle will be deflected when its kinetic energy is equal to the potential energy
Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C
Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C
Kinetic energy of the alpha particle = 5.28 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV)
= 8.459 × 10⁻¹³
k electrostatic force constant = 9 × 10⁹ N.m²/c²
Kinetic energy = potential energy = k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus
r = ( 9 × 10⁹ N.m²/c² × 3.2 × 10⁻¹⁹ C × 1.264 × 10⁻¹⁷C) / 8.459 × 10⁻¹³
= 4.3 × 10 ⁻¹⁴ m
Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
- Δx=760 m
- V₀=87 m/s
- t=13.6 s
- a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!
With that we can plug the values in to get:
Answer:
5 years worth of work (aka all of the homework i currently have)
The west constituent of their sequence needs to cancel out 58 mph crosswind. Subsequently a northwest direction is a 45-degree angle up to even with the destination. That is the third point out of the triangle and the right angle is at the destination. The top side is the west constituent of their flight the vertical side is their resultant travel and the hypotenuse is their definite distance flown. Since the 58 mph crosswind was negated by flying northwest, the distance from the beginning to the destination must be the same distance as the west component of their travel. The hypotenuse is square root of twice the side since it has 2 identical sides.
c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02
Alternative solution:
c = sqrt (2) * 58 = 1.414 * 58 = 82.02
Therefore, they have to fly 82.02 mph
Answer:
E) 800 km/h
Explanation:
The computation of the average vector velocity module of the plane, at that time is shown below:
The displacement vector is
d^2 = d1^2 + d2^2
where,
d1 = northeast displacement
d2 = southeast displacement
Now
d^2 = (120)^2 + (160)^2
= 14400 + 2560
= 40000
= √40000
d = 200 km
Now the average velocity is
V = ΔS ÷ Δt
= 200 ÷ 1 ÷ 4
= 200 × 4
= 800 km/h
