Answer:
James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down
Explanation:
Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards
so here we can say
Upwards force = downwards Force + weight of snow
while if we find the other force which is acting downwards
then for that force we can say that net torque must be balanced
so here we have

so here we have

so here we can say that upward force by which we push up is always more than the downwards force
Answer:
26.2 m/s
Explanation:
We can find the speed of the cannonball just by analyzing its vertical motion. In fact, the initial vertical velocity is
(1)
where u is the initial speed and
is the angle of projection.
We can therefore use the following suvat equation for the vertical motion of the ball:

where
is the vertical velocity at time t, and
is the acceleration of gravity. The time of flight is 3.78 s, so we know that the ball reaches its maximum height at half this time:

And at the maximum height, the vertical velocity is zero:

Substituting these values, we find the initial vertical velocity:

And using eq.(1) we now find the initial speed:

Answer:
Explanation:
Given that, .
R = 12 ohms
C = 500μf.
Time t =? When the charge reaches 99.99% of maximum
The charge on a RC circuit is given as
A discharging circuit
Q = Qo•exp(-t/RC)
Where RC is the time constant
τ = RC = 12 × 500 ×10^-6
τ = 0.006 sec
The maximum charge is Qo,
Therefore Q = 99.99% of Qo
Then, Q = 99.99/100 × Qo
Q = 0.9999Qo
So, substituting this into the equation above
Q = Qo•exp(-t/RC)
0.9999Qo = Qo•exp(-t / 0.006)
Divide both side by Qo
0.9999 = exp(-t / 0.006)
Take In of both sodes
In(0.9999) = In(exp(-t / 0.006))
-1 × 10^-4 = -t / 0.006
t = -1 × 10^-4 × - 0.006
t = 6 × 10^-7 second
So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge
Answer:
σ = 1.09 mm
Explanation:
Step 1: Identify the given parameters
rod diameter = 20 mm
stiffness constant (k) = 55 MN/m = 55X10⁶N/m
applied force (f) = 60 KN = 60 X 10³N
young modulus (E) = 200 Gpa = 200 X 10⁹pa
Step 2: calculate length of the rod, L
K = \frac{A*E}{L}K=
L
A∗E
L = \frac{A*E}{K}L=
K
A∗E
A=\frac{\pi d^{2}}{4}A=
4
πd
2
d = 20-mm = 0.02 m
A=\frac{\pi (0.02)^{2}}{4}A=
4
π(0.02)
2
A = 0.0003 m²
L = \frac{A*E}{K}L=
K
A∗E
L = \frac{(0.0003142)*(200X10^9)}{55X10^6}L=
55X10
6
(0.0003142)∗(200X10
9
)
L = 1.14 m
Step 3: calculate the displacement of the rod, σ
\sigma = \frac{F*L}{A*E}σ=
A∗E
F∗L
\sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}σ=
(0.0003142)∗(200X10
9
)
(60X10
3
)∗(1.14)
σ = 0.00109 m
σ = 1.09 mm
Therefore, the displacement at the end of A is 1.09 mm