Answer:
Current, I = 0.0011 A
Explanation:
It is given that,
Diameter of rod, d = 2.56 cm
Radius of rod, r = 1.28 cm = 0.0128 m
The resistivity of the pure silicon, 
Length of rod, l = 20 cm = 0.2 m
Voltage, 
The resistivity of the rod is given by :
R = 893692.30 ohms
Current flowing in the rod is calculated using Ohm's law as :
V = I R
I = 0.0011 A
So, the current flowing in the rod is 0.0011 A. Hence, this is the required solution.
Answer: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4 - Chemical Equation Balancer
Equation is already balanced.
Explanation: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4
<span>B) 0.6 N
I suspect you have a minor error in your question. Claiming a coefficient of static friction of 0.30N is nonsensical. Putting the Newton there is incorrect. The figure of 0.25 for the coefficient of kinetic friction looks OK. So with that correction in mind, let's solve the problem.
The coefficient of static friction is the multiplier to apply to the normal force in order to start the object moving. And the coefficient of kinetic friction (which is usually smaller than the coefficient of static friction) is the multiplied to the normal force in order to keep the object moving. You've been given a normal force of 2N, so you need to multiply the coefficient of static friction by that in order to get the amount of force it takes to start the shoe moving. So:
0.30 * 2N = 0.6N
And if you look at your options, you'll see that option "B" matches exactly.</span>
Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s
Answer:
U² = 142.86 N
U¹ = 357.14 N
Explanation:
Taking summation of the moment about point A, we get the following equilibrium equation: (taking clockwise direction as positive)
where,
W = weight of boy = 500 N
U² = reaction ay B = ?
Therefore,
<u>U² = 142.86 N</u>
Now, taking summation of forces on the plank. Taking upward direction as positive, for equilibrium position:
<u>U¹ = 357.14 N</u>
