Answer:
Net force = 129.4, Force as multiple of weight of her hand = 18.84
Explanation:
Given Data:
Total body weight = 56.0 kg ;
no. of turns = 2.5/second ;
hand to hand distance = 1.5m ;
weight of hand = 1.25% of body weight ;
Solution:
mass of hand = 
*56 = 0.7kg ;
radius = d/2 = 1.5/2 = 0.75m ;
Now we need to find velocity, as we know that velocity can be calculated by dividing distance by time
v = d/t = 
= 11.775 m/s or 12 m/s;
a.
The formula to calculate force is given as
F = mv²/r = (0.7*11.775²)/0.75 = 129.4 N
b.
To calculate force as multiple of weight on her hand, we need to calculate the gravitational force W on her hand first.
W = gm = 9.81 * 0.7 = 6.867 N
Now the wieght on her hand can be represented by
= 129.4 / 6.867 = 18.84
Answer:
A. 420 J
Explanation:
Given the following data;
Mass = 30.45 g
Specific heat capacity = 4.18 J/g °C.
Temperature = 3.3°C
To find the quantity of heat;
Heat capacity is given by the formula;
Where;
Q represents the heat capacity or quantity of heat.
m represents the mass of an object.
c represents the specific heat capacity of water.
t represents the temperature.
Substituting into the equation, we have;
Q = 420.03 ≈ 420 Joules
Answer:
29.4m/s
Explanation:
Given parameters:
Time = 3s
Unknown:
Average velocity = ?
Solution:
To solve this problem, we use the expression below:
v = u + gt
v is the average velocity
u is the initial velocity = 0m/s
g is the acceleration due to gravity = 9.8m/s²
t is the time
So;
v = 0 + (9.8 x 3) = 29.4m/s
Answer:
I_weight = M L²
this value is much larger and with it it is easier to restore balance.I
Explanation:
When man walks a tightrope, he carries a linear velocity, this velocity is related to the angular velocity by
v = w r
For man to maintain equilibrium needs the total moment to be zero
∑τ = I α
S τ = 0
The forces on the home are the weight of the masses, the weight of the man and the support on the rope, the latter two are zero taque the distance to the center of rotation is zero.
Therefore the moment of the masses and the open is the one that must be zero.
If the man carries only the bar, we could approximate it by two open one on each side of the axis of rotation formed by the free of the rope
I = ⅓ m L² / 4
As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.
If, in addition to the opening, each of them carries a specific weight, the moment of inertia of this weight is
I_weight = M L²
this value is much larger and with it it is easier to restore balance.
Nitrogen oxides play a critical role in photochemical smog. They give the smog its yellowish-brown hue. Indoor residential appliances like gas stoves and gas or wood heaters can be significant emitters of nitrogen oxides in poorly ventilated environments.
- Nitrogen dioxide (NO₂), ozone (O₃), peroxyacetyl nitrate (PAN), and chemical compounds with the -CHO group are the main harmful elements of photochemical smog (aldehydes). If present in high enough amounts, PAN and aldehydes can harm plants and irritate the eyes.
- The greatest sources of emissions are power plants, heavy construction equipment driven by diesel, other moveable engines, and industrial boilers. Cars, trucks, and buses are next in line.
Therefore , on conclusion i.e. two gases with molecules consisting of nitrogen and oxygen atoms are nitric oxide (NO) and nitrogen dioxide (NO₂). These nitrogen oxides play a part in the development of smog and acid rain, adding to the issue of air pollution.
To know more about photochemical smog
brainly.com/question/15635778
#SPJ1
