Answer:
3.8 secs
Explanation:
Parameters given:
Acceleration due to gravity, g = 9.8 
Initial velocity, u = 11.76 m/s
Final velocity, v = 49 m/s
Using one of Newton's equations of linear motion, we have that:
where t = time of flight of arrow
The sign is positive because the arrow is moving downward, in the same direction as gravitational force.
Therefore:
The arrow was in flight for 3.8 secs
Answer:
Explanation:
Let the velocity be v
Total energy at the bottom
= rotational + linear kinetic energy
= 1/2 Iω² + 1/2 mv² ( I moment of inertia of shell = mr² )
= 1/2 mr²ω² + 1/2 mv² ( v = ω r )
= 1/2 mv² +1/2 mv²
= mv²
mv² = mgh ( conservation of energy )
v² = gh
v = √gh
= √9.8 x 1.8
= 4.2 m /s
Answer:
a = 8.06 m/s²
Explanation:
The acceleration of this car can be found using the first equation of motion:
where,
a = acceleration = ?
vf = final speed = 26.8 m/s
vi = initial speed = 0 m/s
t = time = 3.323 s
Therefore,
<u>a = 8.06 m/s²</u>
Answer:
5.95 m
Explanation:
Given that the biggest loop is 40.0 m high. Suppose the speed at the top is 10.8 m/s and the corresponding centripetal acceleration is 2g
For the car to stick to the loop without falling down, at the top of the ride, the centripetal force must be equal to the weight of the car. That is,
(MV^2) / r = mg
V^2/ r = centripetal acceleration which is equal to 2g
2 × 9.8 = 10.8^2 / r
r = 116.64 /19.6
r = 5.95 m
Answer:
18.1 × 10⁻⁶ A = 18.1 μA
Explanation:
The current I in the wire is I = ∫∫J(r)rdrdθ
Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.
So, dI = J(r)rdrdθ
dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²
Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m
I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m = 0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA
