The acceleration of the box up the ramp is 9.65 m/s².
<h3>
What is the magnitude of acceleration of the box?</h3>
The magnitude of the acceleration of the box is calculated by applying Newton's second law of motion as shown below;
F(net) = ma
where;
- m is the mass of the box
- a is the acceleration of the box
The net force on the box is calculated as follows;
F(net) = F - Ff
F(net) = F - μmgcosθ
where;
- θ is the inclination of the plane
- μ is coefficient of friction
F(net) = 170 - (0.3 x 15 x 9.8 x cos55)
F(net) = 144.7
The acceleration of the box is calculated as;
a = F(net) / m
a = (144.7) / (15)
a = 9.65 m/s²
Thus, the acceleration of the box up the ramp is 9.65 m/s².
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Answer:
l= 3.002 cm
Explanation:
Given that
n= 70 turns
B= 1.2 T
θ= 15°
I= 1.5 A
τ = 0.0294 N⋅m
Lets take length of sides is l.
We know that
τ = n I A B sin θ
Area of square ,A= l²
Now by putting the value
τ = n I A B sin θ
0.0294 = 70 x 1.5 x l² x 1.2 x sin 15°
l² = 0.000901 m²
l² = 9.01 cm²
l= 3.002 cm
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