Answer:
The normal force the seat exerted on the driver is 125 N.
Explanation:
Given;
mass of the car, m = 2000 kg
speed of the car, u = 100 km/h = 27.78 m/s
radius of curvature of the hill, r = 100 m
mass of the driver, = 60 kg
The centripetal force of the driver at top of the hill is given as;
where;
Fc is the centripetal force
is downward force due to weight of the driver
is upward or normal force on the drive
Therefore, the normal force the seat exerted on the driver is 125 N.
Given:
10^10 electrons per second
To justify that coulomb is a very large unit for practical use, we need to convert the quantity of electron given to Coulombs:
From literature,
1 Coulomb is equivalent to 6.242×10^18 electrons<span>.
So,
= 10^10 electrons * (1 coulomb/</span><span>6.242×10^18</span> electrons) / second
<span>= 1.602 x 10^-9 coulumbs
This value is too small to be used in an actual setting.
</span><span>
</span>
Answer:
.a = 849.05 m / s²
Explanation
The centripetal acceleration is
a = v² / r
Linear and angular velocity are related
v = w r
Angular velocity and frequency are related by
w = 2π f
Let's replace
a = w² r
a = 4π² f² r
Let's reduce to the SI system
f = 2.30 rev / s (2π rad / 1 rev) = 14.45 rad / s
.r = 10.3 cm = 0.103 m
Let's calculate
a = 4π² 14.45² 0.103
.a = 849.05 m / s²
the more pressure put on the string, the more frequency and higher pitch.
