Answer:
A) t = 29.3 s
B) V = 58.56 ft/s
C) a_net = 2.3 ft/s²
Explanation:
A) The formula for radial acceleration is given as;
a_c = V²/R
We are given;
Radius;R = 350 ft
So, a_c = V²/350
Where V is velocity
Tangential acceleration;a_t = 2 ft/s²
Formula for net acceleration is;
a_net = √((a_c)² + (a_t)²)
We are given a_net = 10 ft/s²
Thus;
10 = √(V²/350)² + (2²)
10² = V⁴/350² + 4
100 - 4 = V⁴/350²
96 × 350² = V⁴
V = 58.56 ft/s
Now, formula for angular velocity is;
ω = V/r
ω = 58.56/350
ω = 0.1673 rad/s
Angular acceleration is given by;
α = a_t/r
α = 2/350
α = 0.00571 rad/s²
Time needed will be gotten from the formula;
t = ω/α
t = 0.1673/0.00571
t = 29.3 s
B) we are told total acceleration is 10 ft/s², thus it's the same as velocity gotten earlier which is 58.56 ft/s
C) we are told that the speed is now 20 ft/s
Thus;
a_c = 20²/350
a_c = 1.1429 ft/s²
Since a_net = √((a_c)² + (a_t)²)
We are given a_t = 2 ft/s²
Thua;
a_net = √(1.1429² + 2²)
a_net = √5.30622041
a_net = 2.3 ft/s²
Answer:
dont know because I am a student lol
Answer:
q' = 0.5q
Hence, the charge stored is also halved
Explanation:
The capacitance of a capacitor is given by the following formula:
------- eqn(1)
where,
C = Capacitance of the capacitor
q = charge stored in the capacitor
V = potential across the capacitor
Now, if we make potential half of its initial value:
<u></u>
<u>q' = 0.5q </u>
<u>Hence, the charge stored is also halved </u>
Answer:
v=15.24 m/s
Explanation:
Given that
Mass ,m= 1600 kg
radius ,r= 28 m
Coefficient of friction ,μ = 0.83
The radial force on the car when it takes turn

The friction force on the car
Fr= μ m g
The condition for motion without losing traction
F= Fr

v²=μ r g

Now by putting the values
( take g=10m/s²)
v=15.24 m/s
The speed of the car will be 15.24 m/s
Answer:
Low Potential energy and High Kinetic energy
Explanation:
Hope this helps and have a good day! Apologies if it's wrong.<3