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:
-0.0047 rad/s²
335.103 seconds
99.18 seconds
Explanation:
= Final angular velocity
= Initial angular velocity = 1.5 ra/s
= Angular acceleration
= Angle of rotation = 40 rev
t = Time taken
Equation of rotational motion

Acceleration while slowing down is -0.0047 rad/s²

Time taken to slow down is 335.103 seconds

Solving the equation

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103
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Answer:
Explanation:
Current in a wire is 120mA
I = 120mA = 120 × 10^-3 A
I = 0.12 A
If the voltage applied at across the wire is tripled
From ohms law
V=IR
R = V / I
Since R is constant
Then,
V / I = K
Then, we can say
V / I = V' / I'
Given that,
Initially
V = V and I = 120mA
Then, V' = 3V and I' =?
So,
V / I = V' / I'
V / 120 = 3V / I'
Cross multiply
V × I' = 120 × 3V
Divide both sides by V
I' = 120 × 3V / V
I' = 360mA
So, the current in the wire when the voltage was tripled is 360mA, the current was also tripled