Answer:
The value of F= - 830 N
Since the force is negative, it implies direction of the force applied was due south.
Explanation:
Given data:
Mass = 1000-kg
Distance, d = 240 m
Initial velocity, v1 = 20.0 m/s
Final velocity, v2 = 0 (since the car came to rest after brake was applied)
v2²= v1² + 2ad (using one of the equation of motion)
0= 20² + (2 x a x 240)
0= 400 + 480 a
a = - 400/480
a = - 0.83 m/s²
Then, imputing the value of a into
F = ma
F = 1000 kg x ( - 0.83 m/s²)
F= - 830 N
The car was driving toward the north, and since the force is negative, it implies direction of the force applied was due south.
Answer:
θ=108rad
t =10.29seconds
α=-8.17rad/s²
Explanation:
Given that
At t=0, Wo=24rad/sec
Constant angular acceleration =30rad/s²
At t=2, θ=432rad as it try to stop because the circuit break
Angular motion
W=Wo+αt
θ=Wot+1/2αt²
W²=Wo²+2αθ
We need to find θ between 0sec to 2sec when the wheel stop
a. θ=Wot+1/2αt²
θ=24×2+1/2×30×2²
θ=48+60
θ=108rad.
b. W=Wo+αt
W=24+30×2
W=84rad/s
This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.
Wo=84rad/sec
W=0rad/s, because the wheel stop at θ=432rad
Using W²=Wo²+2αθ
0²=84²+2×α×432
-84²=864α
α=-8.17rad/s²
It is negative because it is decelerating
Now, time taken for the wheel to stop
W=Wo+αt
0=84-8.17t
-84=-8.17t
Then t =10.29seconds.
a. θ=108rad
b. t =10.29seconds
c. α=-8.17rad/s²
Answer:
energy is stored is 2.2 × 10⁻¹³ J
Explanation:
The capacitance of the cell is given with the expression
C = (KE₀A) / d
k is the dielectric constant, A is the area of the cell, d is the thickness of the cell.
Now given that; the diameter is 50,
Area A = 4πR²
A = 4π × ( 25 × 10⁻⁶ m)²
A = 7850×10⁻¹² m²
our capacitance C = (KE₀A) / d
C = [9 ( 8.85 × 10⁻¹² C²/N.m² × 7850×10⁻¹² m² )] / 7×10⁻⁹ m
C = 8.93 × 10⁻¹¹ F
Now Energy stored
E = 1/2 × CV²
E = 1/2 × (8.93 × 10⁻¹¹ F) × ( 70 × 10⁻³ V)²
E = 2.2 × 10⁻¹³ J
Therefore energy is stored is 2.2 × 10⁻¹³ J
Answer:
a. E = 122.4 N/C
b. E = 58.2 N/C
c. E = 0
Explanation:
The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.
In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.
A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.
B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.
C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.
<span>Height = y
T = 5.8 sec
tf = time to fall
ts = time for sound to travel distance y
T = tf + ts = √[2y/g] + y/330
or
(T - y/330)² = 2y/g
a quadratic in y.
</span><span>y = 141.4 m
</span>hope this helps
