Answer:
6.69 m/s
4.483 m
1.42s
Explanation:
Given that:
Initial Velocity, u = 0
Final velocity, v =?
Acceleration, a = 35m/s²
1.) using the relation :
v² = u² + 2as
v² = 0 + 2(35) * 64*10^-2m
v² = 70 * 0.64
v = sqrt(44.8)
v = 6.693
v = 6.69 m/s
B.) height from the ground, h0 = 2.2
How high ball went , h:
Using :
v² = u² + 2as
Upward motion, g = - ve
0 = 6.69² + 2(-9.8)*(h - 2.2)
0= 6.69² - 19.6(h - 2.2)
44.7561 + 43.12 - 19.6h = 0
19.6h = 44.7561 - 43.12
h = 87.8761 / 19.6
h = 4.483 m
C.)
vt - 0.5gt² = h - h0
6.69t - 0.5(9.8)t²
6.69t - 4.9t² = 1.83 - 2.2
-4.9t² + 6.69t + 0.37 = 0
Using the quadratic equation solver :
Taking the positive root:
1.4185 = 1.42s
Answer:
a)Change in the speed = 1.41 m/s
b)The final speed will be 3.11 m/s
Explanation:
Given that
Acceleration ,a= 4.7 x 10⁻³ m/s²
a)
We know that
v= u + a t
v=final speed ,u=initial speed
t= time ,a= acceleration
Change in the speed
v- u = a t
t= 5 min = 5 x 60 s = 300 s
v- u = 4.7 x 10⁻³ x 5 x 60 m/s
v-u = 1.41 m/s
Change in the speed = 1.41 m/s
b)
Given that
u= 1.7 m/s
v-u = 1.41 m/s
v= 1.7 + 1.41 m/s
v=3.11 m/s
The final speed will be 3.11 m/s
Explanation:
Below is an attachment containing the solution.
<h2>
Answer:</h2>
4.2 C
<h2>
Explanation:</h2>
The charge (Q) moving is the product of the current(I) flowing through the torso of the person and the time taken (t) for the flow.
i.e
Q = I x t
Where;
I = current = 14.0A
t = time taken = 0.0300s
Substituting the values of I and t into the equation above gives
Q = 14.0 x 0.0300
Q = 4.2 C
Therefore quantity of charge moving is 4.2 C
