Answer:
F = 39.2 N (hand force) and N = 68.6 N (shoulder force)
Explanation:
In this exercise we must use the rotational and translational equilibrium conditions, we have several forces: the weight (W) of the pole applied at its geometric center, the load (w1) at one end, the shoulder support (N) 60 cm from the load and hand force (F) at the other end of the pole
Let's set the reference system at the fit point of the shoulder
∑ τ = 0
We will assume that the counterclockwise turns are positive
w₁ 0.60 + W 0.1 + F₁ 0 - F 0.4 = 0
all distances are measured from the support of the man (x₀ = 0.60 m)
F = (w₁ 0.60 + W 0.1) / 0.4
F = (m₁ 0.6 + m 0.1) g / 0.4
let's calculate
F = (2.6 0.6 + 0.4 0.1) 9.8 / 0.4
F = 39.2 N
this is the force that the hand must exert to keep the system in balance
We apply the translational equilibrium condition
-w₁ -W + N - F = 0
N = w₁ + W + F
N = (m₁ + m) g + F
let's calculate
N = (2.6 + 0.4) 9.8 + 39.2
N = 68.6 N
Which of the following is a characteristic of a base?
A. pH greater than 7
B. pH between 4 and 13
C. pH between 4 and 9
D. pH less than 7
Answer:
A. pH greater than 7
Explanation:
Because a base has a pH scale of 8 to 14
A 58-kg boy swings a baseball bat, which causes a 0.140-kg baseball to move toward 3rd base with a velocity of 38.0 m/s.
Calculate the kinetic energy of the baseball (rounding your answer to the integer).
Answer:
101 J
Explanation:
Answer:
[tex]\boxed {\boxed {\sf 101 \ Joules}}[/tex]
Explanation:
Kinetic energy can be found using the following formula:
[tex]KE=\frac{1}{2}mv^2[/tex]
The mass of the baseball is 0.140 kilograms and the velocity is 38.0 meters per second.
[tex]m= 0.140 \ kg \\v= 38.0 \ m/s[/tex]
Substitute the values into the formula.
[tex]KE=\frac{1}{2} [0.140 \ kg][(38.0 \ m/s)^2][/tex]
First, evaluate the exponent.
(38.0 m/s)²= 38.0 m/s * 38.0 m/s = 1444 m²/s²[tex]KE=\frac{1}{2}(0.140 \ kg)(1444 \ m^2/s^2)[/tex]
Multiply the two numbers in parentheses together.
[tex]KE=\frac{1}{2}(202.16 \ kg*m^2/s^2)[/tex]
Multiply the fraction by the number, or divide the number by 2.
[tex]KE=101.08 \ kg*m^2/s^2[/tex]
Round to the nearest whole number. The 0 in the tenth place tells us we can leave the number as is.
[tex]KE\approx 101 \ kg*m^2/s^2[/tex]
1 kg*m²/s² is equal to 1 Joule. Therefore, our answer of 101 kg*m²/s² is equal to 101 Joules (J).
[tex]KE\approx 101 \ J[/tex]
The kinetic energy of the baseball is about 101 Joules.
characteristics of radiation from the sun's energy would be
A. heat from a distance source, elcetromagnatic waves and energy that travels through space.
B. Direct contact of the energy source of sufaces
C. Movement of heat through fluids
WHICH ONE IS THE ANSWER HELPPP
Answer:
Answer is A
Explanation:
PLS HELP THIS IS DUE IN TWO HOURS
Answer:
C
Explanation:
A 3kg ball moving at 8 m/s strikes a 2kg ball at rest,if the collision is elastic,what is the speed of the lighter ball if the heavier ball moves at 2m/s in the opposite direction
Answer:
Speed of lighter ball is 4 m/s.
Explanation:
Applying the principle of conservation of linear momentum,
momentum before collision = momentum after collision.
[tex]m_{1}[/tex][tex]u_{1}[/tex] + [tex]m_{2}[/tex] [tex]u_{2}[/tex] = [tex]m_{1}[/tex][tex]v_{1}[/tex] - [tex]m_{2}[/tex][tex]v_{2}[/tex]
[tex]m_{1}[/tex] = 3 kg, [tex]u_{1}[/tex] = 8 m/s, [tex]m_{2}[/tex] = 2 kg, [tex]u_{2}[/tex] = 0 m/s ( since it is at rest), [tex]v_{1}[/tex] = 2 m/s, [tex]v_{2}[/tex] = ?
(3 x 8) + (2 x 0) = (8 x 2) - (2 x [tex]v_{2}[/tex])
24 + 0 = 16 - 2[tex]v_{2}[/tex]
2[tex]v_{2}[/tex] = 16 - 24
2[tex]v_{2}[/tex] = -8
[tex]v_{2}[/tex] = [tex]\frac{-8}{2}[/tex]
= -4 m/s
This implies that the light ball moves at the speed of 4 m/s in the opposite direction of the heavier ball after collision.
Which part of a DC generator keeps the electric current flowing in only one
direction?
A. The permanent magnets
B. The commutator
o
C. The power source
D. The wire coil
A truck initially traveling at a speed of 22 m/s increases at a constant rate of 2.4 m/s^2 for 3.2s. What is the total distance travelled by the truck during this?
Answer:
82.7 m
Explanation:
u= 22m/s
a= 2.4 m/s^2.
t= 3.2 secs
Therefore the distance travelled can be calculated as follows
S= ut + 1/2at^2
= 22 × 3.2 + 1/2 × 2.4 × 3.2^2
= 70.4 + 1/2×24.58
= 70.4 + 12.29
= 82.7 m
Hence the distance travelled by the truck is 82.7 m