I can't seem to figure out the angle between T1 and T2. So suppose, it is 10º; then T2 makes an angle of 35º w/r/t horizontal, and T1 makes an angle of 45º.
Sum the moments about the base of the crane; Σ M = 0. 0 = T2*cos35*L*cos40 + T1*cos45*L*cos40 - T2*sin35*L*sin40 - T1*sin45*L*sin40 - W*(L/2)*sin40 - T1*L*sin40 → length L cancels where W = 18 kN
0 = 0.259*T2 - 43kN T2 = 166 kN
Answer:
0.56 atm
Explanation:
First of all, we need to find the number of moles of the gas.
We know that
m = 1.00 g is the mass of the gas
is the molar mass of the carbon dioxide
So, the number of moles of the gas is

Now we can find the pressure of the gas by using the ideal gas equation:

where
p is the pressure
is the volume
n = 0.023 mol is the number of moles
is the gas constant
is the temperature of the gas
Solving the equation for p, we find

And since we have

the pressure in atmospheres is

Answer:
The peak current carried by the axon is 5.85 x 10⁻⁸ A
Explanation:
Given;
distance of the field from the axon, r = 1.3 mm
peak magnetic field strength, B = 9 x 10⁻¹² T
To determine the peak current carried by the axon, apply the following equation;

where;
B is the peak magnetic field
r is the distance of the magnetic field from axon
μ is permeability of free space = 4π x 10⁻⁷
I is the peak current
Re-arrange the equation and solve for "I"

Therefore, the peak current carried by the axon is 5.85 x 10⁻⁸ A
Answer:
A) F = - 8.5 10² N, B) I = 21 N s
Explanation:
A) We can solve this problem using the relationship of momentum and momentum
I = Δp
in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction
v₀ = 8.50 m / s
v_f = -8.50 m / s
F t = m v_f -m v₀
F =
let's calculate
F =
F = - 8.5 10² N
B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations
v² = v₀² - 2g (y- y₀)
as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0
v =
calculate
v =
v = 14 m / s
to calculate the momentum we use
I = Δp
I = m v_f - mv₀
when it hits the ground its speed drops to zero
we substitute
I = 1.50 (0-14)
I = -21 N s
the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is
I = 21 N s