Answer:
the time needed for her to close the door is 1.36 s.
Explanation:
given information:
Force, F = 220 N
width, r = 1.40 m
weight, W = 790 N
height, h = 3.00 m
angle, θ = 90° = π/2
to find the times needed to close the door we can use the following equation
θ = ω₀t + 1/2 αt²
where
θ = angle
ω = angular velocity
α = angular acceleration
t = time
in this case, the angular velocity is zero. thus,
θ = 1/2 αt²
now, we can find the angular speed by using the torque formula
τ = I α
where
τ = torque
I = Inertia
we know that
τ = F r
and
I = 1/3 mr²
so,
τ = I α
F r = 1/3 mr² α
α = 3 F/mr
= 3 F/(w/g)r
= 3 (220)/(790/9.8) 1.4
= 5.85 rad/s²
θ = 1/2 αt²
π/2 = 1/2 5.85 t²
t = 1.36 s
if spring constant is doubled, the mass on spring will be doubled as well. according to this formula, F=ke
k stands for spring constant and e stands for the length extended
During a climb UP the mountain, gravity does NO work on the climber.
Actually, it's more correct to say that gravity does NEGATIVE work
on him. The climber has to DO the positive work to haul himself up.
Work = (mass) x (gravity) x (height) .
For the guy in this problem:
Work = (67 kg) x (9.8 m/s²) x (3,500 meters)
= 2,298,100 joules.
If he eats no candy bars on the way, and completely depends on
his stored body fat for the energy, then he'll burn off
(2,298,100 joules) / (3.8 x 10⁷ joules/kg)
= 0.06 kg of fat.
That's only about 2.1 ounces. We KNOW he'll lose more weight than that,
climbing 11,000 feet. That's because climbing is pretty inefficient.
In addition to the potential energy you have to give your body weight,
you also have to expend energy breathing, digesting, metabolizing,
and sweating.
Answer: 2.37N
Explanation:
According to coulombs law which states that the force of attraction (F) between two charges (q1 and q2) is directly proportional to the product of their charges and inversely proportional to the square of the distance (r) between them. Mathematically,
F = kq1q2/r²
For the first two charges that are sitting 1.5 m apart with a force of 3 N between them, we have
3 = kq1q2/1.5²
3 = kq1q2/2.25
Kq1q2= 6.75... (1)
If the charges are now moved farther apart 2.25 m and one of the charges is increased by a factor of 4. The formula becomes
F2 = k(4q1)q2/2.25² (q1 has been increased by factor of 4)
k(4q1)q2 = 5.06F2 ... (2)
Dividing 2 by 1 we have
k(4q1)q2/kq1q2 = 5.06F2/3
4 = 5.06F2/3
5.06F2 = 12
F2= 12/5.06
F2 = 2.37N
Therefore the magnitude of the new force between the two charges is 2.37N
Answer:
To find the value of the unknown weight, we previously placed the 3 known weights and made a graph of the force against displacement
When hanging the weight is known, we measure the displacement and from the graph we can find the value of the hanging masses
We can also use the equation and multiply the constant K by the displacement and this is the applied weight.
Explanation:
For this problem we will use the translational equilibrium relation
F –W = 0
F = W
W = mg
The spring elastic force is
F = - k x
We substitute
k x = m g
Where we see that the force of the spring is equal to the weight of the body.
To find the value of the unknown weight, we previously placed the 3 known weights and made a graph of the force against displacement
When hanging the weight is known, we measure the displacement and from the graph we can find the value of the hanging masses
We can also use the equation and multiply the constant K by the displacement and this is the applied weight.