To solve letter a:
d1 = 85t1 = 16 km,
85t1 = 16,
t1 = 16 / 85 = 0.1882 h = 11.29 min.
d2 = 115t2 = 16 km,
115t2 = 16,
t2 = 16 / 115 = 0.139 h = 8.35 min.
t1 - t2 = 11.29 - 8.35 = 2.94 min.
Car #2 arrives 2.94 minutes sooner.
To solve letter b:
15 min = 1/4 h = 0.25 h.
d1 = d2,
115t = 85(t + 0.25),
115t = 85t + 21.25,
115t - 85t = 21.25,
30t = 21.25,
t = 21.25 / 30 = 0.71 h,
d = 115 * 0.71 = 81.65 km.
Answer: opening of the nicotinic acetylcholine receptor channels.
Explanation:
Neuromuscular junction is a special junction formed between a motor neurone and a muscle fibre. The junction is fortified with nerves and receptors that helps in the transmission of signals from the motor neurone to the muscle fibre in order to bring about the desired voluntary movements through muscular contraction.
Nicotinic acetylcholine receptor are activated through the binding of acetylcholine at the neuromuscular junction. This action leads to influx of sodium ions to accomplish endplate potential.
Answer:
Explanation:
To solve this problem we use the formula for accelerated motion:
We will take the initial position as our reference (
) and the downward direction as positive. Since the rock departs from rest we have:
Which means our acceleration would be:
Using our values:
Answer:
F n = 0.2 N
Explanation:
given,
you are exerting force of 10 N on the ball.
mass of the ball = 1 kg
acceleration due to gravity = 9.8 m/s²
normal force on the ball = ?
normal force is force exerted by the object to counteract the force from other object.
normal force acting on the ball will be
F n = F - mg
F n = 10 - 1 × 9.8
F n = 10 -9.8
F n = 0.2 N
Hence, normal force acting on the ball is equal to 0.2 N
First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.
