Answer:
a)15077 W
b)20.2185 horse power
Explanation:
P=F*V
F=ma
a=Vf-VS/t
Vf=28m/s
t=5.2
a=28/5.2
a=5.384 m/s²
F=100kg*5.384m/s²
F=538.4 N
P=F*V
P=538.4N*28m/s
P=15077 W=20.2185 horse power
1W=0.00134 Horse power
Answer:
t = Δa / v
Explanation:
To know which option is not true, we shall fine a relationship between acceleration (a), velocity (v), time (t) and radius (r). This is illustrated below:
Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as shown below:
Acceleration = change in velocity / time
a = Δv / t ..... (1)
But
Δv = v₂ – v₁
Substitute the value of Δv into equation (1)
a = Δv / t
a = v₂ – v₁ / t ....... (2)
From equation (1), make Δv the subject of the equation.
a = Δv / t
Cross multiply
Δv = at .... (3)
From equation (1), make t the subject of the equation.
a = Δv / t
Cross multiply
at = Δv
Divide both side by a
t = Δv /a ...... (4)
From circular motion, centripetal's force is given by:
F = mv²/r
F = ma꜀
Therefore,
ma꜀ = mv²/r
Cancel out m
a꜀ = v²/r
SUMMARY:
a = Δv / t
a = v₂ – v₁ / t
Δv = at
t = Δv /a
a꜀ = v²/r
Considering the options given in question above, t = Δa / v is not a true statement.
Answer:
v = 36.667 m/s
Explanation:
Knowing the rotational inertia as
Lₙ = 550 kg * m²
r = 1.0 m
m = 30.0 kg
To determine the minimum speed v must have when she grabs the bottom
Lₙ = I * ω
I = ¹/₂ * m * r²
I = ¹/₂ * 30.0 kg * 1.0² m
I = 15 kg * m²
Lₙ = I * ω ⇒ ω = Lₙ / I
ω = [ 550 kg * m² /s ] / ( 15 kg * m² )
ω = 36.667 rad /s
v = ω * r
v = 36.667 m/s
Its the amount of matter in something.
for example, a large truck as compared to a mouse, the truck would have much greater of a mass.
Orient the semi-circle arc such that it is symmetric with respect to the y-axis. Now, by symmetry, the electric field in the x-direction cancels to zero. So the only thing of interest is the electric field in the y-direction.
dEy=kp/r^2*sin(a) where k is coulombs constant p is the charge density r is the radius of the arc and a is the angular position of each point on the arc (ranging from 0 to pi. Integrating this renders 2kq/(pi*r^3). Where k is 9*10^9, q is 9.8 uC r is .093 m
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