When a flowerpot falls from a window sill 36.5 m
above the sidewalk, then the velocity of the flowerpot is 26.7 m/s.
From Newton's third equation of motion,
v^2 = u^2 + 2gh
where,
h is the height of the object or body from ground
u is the initial velocity of the body or object
v is the final velocity of the body or object
g is the acceleration due to gravity
Now, as we know that
Flowerpot is at rest. So, u = 0
g = 9.81m/s^2
h = 36.5m
By substituting all the values, we get
v^2 = 2 × 9.81 × 36.5
= 716.13
v = 26.7m/s
Thus, we concluded that when a flowerpot falls from a window sill 36.5 m
above the sidewalk, then the velocity of the flowerpot is 26.7 m/s.
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Answer:
156.26N
Explanation:
The data needed are incomplete. Let the acceleration of the body be 3.5m/s²
Other given parameters
Mass = 1.35×10^1 = 13.5kg
coefficient of friction between the tires and the road = 0.850
Acceleration due to gravity = 9.8m/s²
According to Newton's second law:
Fnet = ma
Fnet = Fapp - Ff
Fapp is the applied force
Ff is the frictional force = umg
The equation becomes:
Fapp - Ff = ma
Fapp-umg = ma
Fapp - 0.85(13.5)(9.8) = 13.5(3.5)
Fapp - 109.0125 = 47.25
Fapp = 47.25+109.0125
Fapp = 156.2625N
Hence the applied force that caused the acceleration is 156.26N
Note that the acceleration of the car was assumed. Any value of acceleration can be used for the calculation.
Given Information:
Radius = ra = 2.60 cm = 0.026 m
Density = J = 15.0 nC/m
change in potential difference = ΔV = 200 V
Required Information:
Distance = d = ?
Answer:
distance = 0.088 m
Explanation:
As we know
ΔV = Vb - Va = J/4πε₀*ln(rb/ra)
Where ra and rb is the point where potential difference is Va and Vb respectively
1/4πε₀ = 9x10⁹ N.m²/C²
We want to find the distance d = rb - ra
ΔV = J/4πε₀*ln(rb/ra)
200 = 9x10⁹*15x10⁻⁹*ln(rb/ra)
200/135 = ln(rb/ra)
1.48 = ln(rb/ra)
taking e on both sides yields
e^(1.48) = rb/ra
4.39 = rb/ra
rb = 4.39*0.026
rb = 0.114 m
Therefore, the required distance is
d = rb - ra
d = 0.114 - 0.026
d = 0.088 m
Therefore, the other probe must be placed 0.088 m from the surface so that the voltmeter reads 200 V
Answer:
The magnitude of the maximum possible torque exerted on the coil is 5.73 x 10⁻³ Nm
Explanation:
Given;
number of turns of the circular coil, N = 49.5 turns
radius of the coil, r = 5.10 cm = 0.051 m
magnitude of the magnetic field, B = 0.535 T
current in the coil, I = 26.5 mA = 0.0265 A
The magnitude of the maximum possible torque exerted on the coil is calculated as;
τ = NIAB
where;
A is the area of the coil
A = πr² = π(0.051)² = 0.00817 m²
Substitute the given values and solve for the maximum torque
τ = (49.5) x (0.0265) x (0.00817) x (0.535)
τ = 0.00573 Nm
τ = 5.73 x 10⁻³ Nm
