Hi there!
(a)
Recall that:
W = Work (J)
F = Force (N)
d = Displacement (m)
Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.
To the nearest multiple of ten:
(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.
Thus:
(c)
Similarly, the normal force is perpendicular to the displacement, so:
(d)
Recall that the force of kinetic friction is given by:
Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
In multiples of ten:
(e)
Simply add up the above values of work to find the net work.
Nearest multiple of ten:
(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
Nearest multiple of ten:
Atoms, Protons, Neutrons, and Electrons
Answer:
The mass of the man is 71 kg
Explanation:
Given;
kinetic energy of the man, K.E = 887.5 J
velocity of the man, v = 5 m/s
The mass of the man is calculated as follows;
K.E = ¹/₂mv²
where;
m is the mass of the man
2K.E = mv²
m = 2K.E / v²
m = (2 x 887.5) / (5)²
m = 71 kg
Therefore, the mass of the man is 71 kg
The gravitational force would get stronger because the farther the two masses are separated the more gravitational force will be used to pull them together the closer they are the less gravitational pull is used to pull them together
Answer:
160 years.
Explanation:
From the question given above, the following data were obtained:
Initial count rate (Cᵢ) = 400 count/min
Half-life (t½) = 40 years
Final count rate (Cբ) = 25 count/min
Time (t) =?
Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:
Initial count rate (Cᵢ) = 400 count/min
Final count rate (Cբ) = 25 count/min
Number of half-lives (n) =?
Cբ = 1/2ⁿ × Cᵢ
25 = 1/2ⁿ × 400
Cross multiply
25 × 2ⁿ = 400
Divide both side by 25
2ⁿ = 400/25
2ⁿ = 16
Express 16 in index form with 2 as the base
2ⁿ = 2⁴
n = 4
Thus, 4 half-lives has elapsed.
Finally, we shall determine the time taken for the radioactive material to decay to the rate of 25 counts per minute. This can be obtained as follow:
Half-life (t½) = 40 years
Number of half-lives (n) = 4
Time (t) =?
n = t / t½
4 = t / 40
Cross multiply
t = 4 × 40
t = 160 years.
Thus, it will take 160 years for the radioactive material to decay to the rate of 25 counts per minute.
