5. How much does a suitcase weigh if it has a mass of 22.5 kg?
2 answers:
Answer:
Weight of the suitcase, W = 220.5 N
Explanation:
It is given that,
Mass of the suitcase, m = 22.5 kg
Let W is the weight of the suitcase. The weight of an object is equal to the force acting on it due to the gravity. It is given by the product of mass and acceleration due to gravity.Mathematically, it is given by :
W = mg
W = 220.5 N
So, the weight of the suitcase is 220.5 N. Hence, this is the required solution.
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Answer:
a1 = 3.56 m/s²
Explanation:
We are given;
Mass of book on horizontal surface; m1 = 3 kg
Mass of hanging book; m2 = 4 kg
Diameter of pulley; D = 0.15 m
Radius of pulley; r = D/2 = 0.15/2 = 0.075 m
Change in displacement; Δx = Δy = 1 m
Time; t = 0.75
I've drawn a free body diagram to depict this question.
Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;
ΣF_x = T1 = m1 × a1
a1 is acceleration and can be calculated from Newton's 2nd equation of motion.
s = ut + ½at²
our s is now Δx and a1 is a.
Thus;
Δx = ut + ½a1(t²)
u is initial velocity and equal to zero because the 3 kg book was at rest initially.
Thus, plugging in the relevant values;
1 = 0 + ½a1(0.75²)
Multiply through by 2;
2 = 0.75²a1
a1 = 2/0.75²
a1 = 3.56 m/s²
Answer:
(a) B = 2.85 × Tesla
(b) I = I = 0.285 A
Explanation:
a. The strength of magnetic field, B, in a solenoid is determined by;
r =
⇒ B =
Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field
B =
=
B = 2.85 × Tesla
b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;
B = μ I N/L
⇒ I = B ÷ μN/L
where B is the magnetic field, μ is the permeability of free space = 4.0 ×Tm/A, N/L is the number of turns per length.
I = B ÷ μN/L
=
I = 0.285 A