Answer: The launch speed is 43m/s
4.7secs after launch speed is 4m/s
Explanation: To solve this we use the first equation of motion but in this case our acceleration would be -10m/s² since we are going upwards against gravity(launch).
Vf = Vi - a*t
Where Vf is the final velocity after launch, Vi is the initial velocity at launch, t is time in secs then a is acceleration
a. From the question
t = 2.3secs
Vf = 20m/s
a = -10m/s²
Substituting into the above equation we have that,
20= Vi - 10* 2.3
20 = Vi - 23
Vi = 20+23
Vi = 43 m/s
Which is the speed at launch.
b. The magnitude of speed (Vf) 4.7 sec after launch is calculated as follows using same procedure but here Vi is 43m/s as calculated
Vf = 43 - 10*4.7
Vf = 43 - 47
Vf = -4m/s
But since we are asked to find the magnitude we neglect the negative sign.
Answer:
ωf = 13 rad/s
Explanation:
- The angular acceleration, by definition, is just the rate of change of the angular velocity with respect to time, as follows:
- α = Δω/Δt = (ωf-ω₀) / (tfi-t₀)
- Choosing t₀ = 0, and rearranging terms, we have
where ω₀ = 5 rad/s, t = 4 s, α = 2 rad/s2
- Replacing these values in (1) and solving for ωf, we get:
- The wheel's angular velocity after 4s is 13 rad/s.
Friction or vibration is your answer
Answer:
λ = 6.97 *10⁻⁶ C/m
Explanation:
Conceptual Analysis:
The electric field at a distance r from a charge line of infinite length and constant charge per unit length is calculated as follows:
E= 2k*(λ/r) Formula (1)
Where:
E: electric field .( N/C)
k: Coulomb electric constant. (N*m²/C²)
λ: linear charge density. (C/m)
r : distance from the charge line to the surface where E calculates (m)
Known data
E= 6.6 x 10⁴ N/C
r = 1.9 m
k= 8.99 *10⁹ N*m²/C²
Problem development
We replace data in the formula (1):
E= 2k*(λ/r)
6.6*10⁴ = 2*(8.99 *10⁹)*(λ/1.9)
(6.6*10⁴)*(1.9) = 2*(8.99 *10⁹) *λ
(12.54* 10⁴) / (17.98*10⁹) = λ
λ = 6.97 *10⁻⁶ C/m