Explanation:
The heart rate of the astronaut is 78.5 beats per minute, which means that the time between heart beats is 0.0127 min. This will be the time t measured by the moving observer. The time t' measured by the stationary Earth-based observer is given by

a) If the astronaut is moving at 0.480c, the time t' is


This means that time between his heart beats as measured by Earth-based observer is 0.0145 min, which is equivalent to 69.1 beats per minute.
b) At v = 0.940c, the time t' is


So at this speed, the astronaut's heart rate is 1/(0.0372 min) or 26.9 beats per minute.
The final speed of the orange is 7.35 m/s
Explanation:
The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration
towards the ground. So we can use the following suvat equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time elapsed
For the orange in this problem, we have
u = 0 (it is dropped from rest)
is the acceleration
Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:

Learn more about free fall:
brainly.com/question/1748290
brainly.com/question/11042118
brainly.com/question/2455974
brainly.com/question/2607086
#LearnwithBrainly
Answer:D) His Facial expressions when they spend time together
Explanation:
Eva's should carefully observe the normal verbal signs by her lover because facial expression gives the hint to listener or viewer that one is interested towards other.
Some common facial Expression are
- Smiling Eyes:While smiling is, of course, an important expression when we speak, attempting to keep a constant smile could make you look more like a trained sociopath than a happy, enthusiastic presenter. These enables other person to feel the good vibes out of you and left a positive impact of your presence.
- Nod on: It is important to express interest in the feedback of your audience. No one likes an identity-centered snob. One way to show that you're actually engaged in the comments and questions they pose is by nodding while you're listening. Moving your head a little wider will help to express your attention as well.
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