I would go with both A and B. But you might want to wait for a better answer cause I'm not too sure
Answer:
Explanation:
Calorie is a common term used to describe the amount of energy that can be derived from food products.
We quantify foods based on the calories of energy they possess. A high calorific food will yield more energy to the body and is often desired for intense physical activities.
Calorie is defined as the amount of heat energy needed to raise the temperature of 1g of a substance by 1°C.
Food calories slightly differs in definition and they imply the amount of heat needed to raise the temperature of 1kg of water by 1°C.
<h2>Answer </h2>
The kinetic energy is 8100 J.
<u>Explanation</u>
Mass is 50.0kg and velocity is 18 m/s, the kinetic energy is:
As we know the formula of kinetic energy which is K.E = ½ ( mv ^ 2 ),
mass = m = 50.0kg
velocity = v = 18 m / s,
by putting values in the formula,
K.E = ½ ( mv ^ 2 ),
K.E = ½ ( 50kg ) . ( 18 m / s ) ^ 2
K.E = ½ ( 50kg ) . ( 324 ),
=> K.E = 1/2 ( 16200 ),
=> K.E = 16200 / 2,
=> K.E = 8100J.
Hence, the kinetic energy ( K.E ) is 8100 joule ( J ).
Answer:
a) fr = 266.92 N, fy = 1300 N, b) μ = 0.36
Explanation:
a) This is a balancing act.
Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive
-w x - W x₂ + R y = 0 (1)
usemso trigonometry to find distances
cos 60.08 = x / 7.5
x = 7.5 cos 60.08
x = 3.74 m
fireman
cos 60.08 = x₂ / 4
x2 = 4 cos 60
x2 = 2 m
wall support
sin 60.08 = y / 15
y = 15 are 60.08
y = 13 m
we substitute in equation 1
R y = w x + W x2
R = (w x + W x2) / y
R = (500 3.74 +800 2) / 13
R = 266.92 N
now let's write the expressions for the translational equilibrium
X axis
R -fr = 0
R = fr
fr = 266.92 N
Y Axis
Fy - w-W = 0
fy = 500 + 800
fy = 1300 N
b) ask the friction coefficient
the firefighter's distance is
cos 60.08 = x₃ / 9.00
x₃ = 9 cos 60
x₃ = 5.28 m
from equation 1
R = (w x + W x₃) / y
R = 500 3.74 + 800 5.28) / 13
R = 468.769 N
we saw that
fr = R = 468.769
The expression for the friction force is
fr = μ N
in this case the normal is the ratio to pesos
N = Fy
N = 1300 N
μ N = fr
μ = fr / N
μ = 468,769 / 1300
μ = 0.36
