Answer:
v = (78.0 i ^ - 70.27 j ^) m/s, v = 105 m / s
, θ = 318º
Explanation:
We have a projectile launch problem, let's start by calculating the time it takes to get through the canyon
y = t - ½ gt2
As the motorcyclist comes out horizontally, the speed he has is the horizontal speed (vox) and the initial vertical speed is zero (I go = 0)
y = 0 - ½ g t2
t = √ 2y / g
t = √ (2 252 /9.8)
t = 7.17 s
Let's calculate the vertical speed for this time
= - gt
= 0 - gt
= - 9.8 7.17
= - 70.27 m / s
We can give the result in two ways
First:
v = (78.0 i ^ - 70.27 j ^) m / s
Second:
using the Pythagorean theorem and trigonometry
v² = vₓ² + ²
v = √ [(78.0)² + (-70.42)²] = √ (11042.98)
v = 105 m / s
tan θ₁ = y / vₓ
tan θ₁ = -70.42 / 78.0
θ₁ = 42º
If we measure this angle from the positive direction of the x-axis counterclockwise
θ = 360 - θ₁
θ = 360 - 42
θ = 318º
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
Answer:
ML2T-2
Explanation:
Work= Force × displacement
=>
work= [MLT^-2] [L]
W= [ML2T-2]
Answer:
last answer
Explanation:
gripping it with the blade facing downward is the most efficient and safe way to use an exacto knife
Complete question:
Consider the hypothetical reaction 4A + 2B → C + 3D
Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?
Answer:
the final concentration of A is 0.992 M.
Explanation:
Given;
time of reaction, t = 4.0 s
rate of change of the concentration of B = -0.0760 M/s
initial concentration of A = 1.600 M
⇒Determine the rate of change of the concentration of A.
From the given reaction: 4A + 2B → C + 3D
2 moles of B ---------------> 4 moles of A
-0.0760 M/s of B -----------> x
⇒Determine the change in concentration of A after 4s;
ΔA = -0.152 M/s x 4s
ΔA = -0.608 M
⇒ Determine the final concentration of A after 4s
A = A₀ + ΔA
A = 1.6 M + (-0.608 M)
A = 1.6 M - 0.608 M
A = 0.992 M
Therefore, the final concentration of A is 0.992 M.