Answer:
a) 5m/s2
b) 4 sec
c) 6 sec
d) 90 m
e) Answer in the file attached as it is a graph
Explanation:
This question can be solved using equations of motion. The two equations are:
2(a)(s) = v² – u²
v = u + (a)(t)
a = acceleration
s = distance
v = final velocity
u = initial velocity
t = time
a) Given final velocity = 50m/s and initial velocity = 30 m/s for 160m journey
Using 2(a)(s) = v² - u²
2(a)(160) = 50² – 30²
320 (a) = 2500 - 900
a = 1600/320
a = 5m/s²
b) The acceleration remains constant throughout so we can use it in this part as well.
Using v = u + (a)(t)
50 = 30 + (5)(t)
t = 4 sec
c) The acceleration remains constant throughout so we can use it in this part as well. Now the initial velocity will be 0m/s and final will be 30m/s
Using v = u + (a)(t)
30 = 0 + 5(t)
t = 6 sec
d) The acceleration remains constant throughout so we can use it in this part as well. Now the initial velocity will be 0m/s and final will be 30m/s
Using 2(a)(s) = v² - u²
2(5) (s) = 30² – 0²
10(s) = 900
s = 90 m
e) Graphs are attached as image
70 Years for the oil to last
Answer:
<h2>0.069 N, in the X direction</h2>
Explanation:
According to Flemming's left hand rule, it sates that if the first three fingers of the left hand are held mutually at right angles to one another, the fore finger will point in the direction of magnetic field, the middle finger will point in direction of current, while the thumb will point to the direction of force.
Mathematically the law is stated as
F= BIL
given data
Magnetic field B= 0.43T
Current I= 4.9 A
length of conductor L= 3.3cm to meter , 3.3/100= 0.033 m
Applying the formula the force is calculated as
F= 0.43*4.9* 0.033= 0.069 N
According to Flemming's rule the direction of all parameters are mutually perpendicular to one another, then the Force is in the X direction
Answer:
Explanation:
Bridge length A B = 36.5 m
weight W = 2.56 x 10⁵ N acting at middle point that is at 18.25 m from either end.
weight of truck W₁ = 5.25 x 10⁴ N standing at 10.2 m from A .
Let force at A be F₁ and at B be F₂ .
Balancing total upward and downward force
F₁ + F₂ = ( 2.56 + .525 ) x 10⁵ = 3.085 x 10⁵ N
Taking torque of all the forces , for rotational balancing
2.56 x 10⁵ x 18.25 + .525 x 10⁵ x 10.2 = F₂ x 36.5
(46.72 + 5.355 )x 10⁵ = F₂ x 36.5
F₂ = 1.42 x 10⁵ N
F₁ = 1.665 x 10⁵ N