Answer:
A. 4,9 m/s2
B. 2,0 m/s2
C. 120 N
Explanation:
In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:
If the package is barely lifted, that means that T=m_2*g; then:
Solving the equation for a_mín, we have:
Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:
For the monkey:
For the package:
The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:
For the package:
We have two unknowns and two equations, so we can proceed. We can match both tensions and have:
Solving a, we have
We can then replace this value of a in one for the sums of force and find the tension T:
The car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.
<h3>Average velocity of the car</h3>
The average velocity of the car is calculated as follows;
x(t) = a + bt + ct2
v = dx/dt
v(t) = b + 2ct
v(0) = -10.1 m/s + 2(1.1)(0) = -10.1 m/s
v(10) = -10.1 + 2(1.1)(10) = 11.9 m/s
<h3>Average velocity</h3>
V = ¹/₂[v(0) + v(10)]
V = ¹/₂ (-10.1 + 11.9 )
V = 0.9 m/s
Thus, the car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.
Learn more about velocity here: brainly.com/question/4931057
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The moment of inertia of a point mass about an arbitrary point is given by:
I = mr²
I is the moment of inertia
m is the mass
r is the distance between the arbitrary point and the point mass
The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.
The total moment of inertia of the system is the sum of the moments of each mass, i.e.
I = ∑mr²
The moment of inertia of each of the two inner masses is
I = m(ℓ/2)² = mℓ²/4
The moment of inertia of each of the two outer masses is
I = m(3ℓ/2)² = 9mℓ²/4
The total moment of inertia of the system is
I = 2[mℓ²/4]+2[9mℓ²/4]
I = mℓ²/2+9mℓ²/2
I = 10mℓ²/2
I = 5mℓ²
Answer:
0.0025116weber/m²
Explanation:
Magnetic field density (B) is the ratio of the magnetic flux (¶) through the loop to its cross sectional area (A).
Mathematically;
B = ¶/A
¶ = BA
Given B = 0.23Tesla which is the magnitude of the magnetic field
Dimension of the rectangular loop = 7.8 cm by 14 cm
Area of the rectangular loop perpendicular to the field B = 7.8cm×14cm
= 109.2cm²
Converting this value to m²
Area of the loop = 109.2 × 10^-4
Area of the loop = 0.01092m²
Magneto flux = 0.23×0.01092
Magnetic flux = 0.0025116weber/m²
Answer:
Time taken by A and B is 1.2 hr.
Explanation:
Given that
Time taken by tank when all(A+B+C) are open = 1 hr
Time taken by tank when A+C are open = 1.5 hr
Time taken by tank when B+C are open = 2 hr
If we treat as filling of tank is a work then
Work = time x rate
Lets take work is 1 unit
1 = 1(1/a+1/b+1/c) ---------1
1 = 1.5(1/a+1/c) ----------2
1 = 2(1/b+1/c) --------3
From equation 1 and 3
1=1(1/a+1/2)
a=2
Form equation 2
1 = 1.5(1/2+1/c)
c=6
From equation 3
1 = 2(1/b+1/6)
b=3
So time taken by
A is alone to fill tank is 2 hr
B is alone to fill tank is 3 hr
C is alone to fill tank is 6 hr
So
Time taken by A and B is 1.2 hr.