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photoshop1234 [79]
3 years ago
14

A 8.57-m ladder with a mass of 21.4 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upw

ard with a force of 258 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.63 rad/s2 about an axis passing through the bottom end of the ladder. The ladder's center of gravity lies halfway between the top and bottom ends. (a) What is the net torque acting on the ladder
Physics
1 answer:
Andrei [34K]3 years ago
4 0

Answer:

1311.5\ \text{Nm}

Explanation:

l = Length of ladder = 8.57 m

m = Mass of ladder = 21.4 kg

F = Force on ladder = 258 N

\alpha = Angular acceleration = 1.63\ \text{rad/s}^2

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Net torque is given by

\tau=lf-\dfrac{l}{2}mg\\\Rightarrow \tau=8.57\times 258-\dfrac{8.57}{2}\times 21.4\times 9.81\\\Rightarrow \tau=1311.5\ \text{Nm}

The net torque acting on the ladder is 1311.5\ \text{Nm}.

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D.) White Dwarf

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Here, M = mass of the Sun.

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3 years ago
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14. Three identical light bulbs are connected in series, then are disconnected and arranged in parallel. For each of the scenari
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In order to make things easier to describe and explain, let's call
the resistance of each bulb 'R', and the battery voltage 'V'.

a).  In series, the total resistance is 3R.
In parallel, the total resistance is R/3.
Changing from series to parallel, the total resistance of the circuit
decreases to 1/9 of its original value.

b).  In series, the total current is  V / (3R) .
In parallel, the total current is  3V / R .
Changing from series to parallel, the total current in the circuit
increases to 9 times its original value.

c).  In series, the power dissipated by the circuit is 

                                   (V) · V/3R  =  V² / 3R .

In parallel, the power dissipated by the circuit is

                                   (V) · 3V/R  =  3V² / R .

Changing from series to parallel, the power dissipated by
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8 0
3 years ago
The law of conservation of matter states that matter can be neither created nor destroyed. Your friend shows you the following c
MrRa [10]

Yes, our friend is right, because there is no contradiction to the law of conservation of mass in the above equation. It just the mass of the product is equal to the mass of reactants.. and that is shown in the equation you have presented earlier

7 0
3 years ago
A 2.4 kg box has an initial velocity of 3.6 m/s upward along a plane inclined at 27◦ to the horizontal. The coefficient of kinet
Vika [28.1K]

Answer:

d= 1.18 m

Explanation:

In abscense of  friction, total mechanical energy must be constant, i.e.,

ΔK + ΔU = 0

As we are told that there exists a kinetic friction between the box and the plane, we need to take into account the work done by the friction force in the equation, as follows:

ΔK + ΔU = Wnc (1)

If we take as our zero gravitational potential energy reference, the height at which the box is sent upward, we can write the following equations for the different terms in (1):

ΔK = Kf- K₀ = 0 - 1/2*m*v₀² = -1/2*2.4kg* (3.6)²(m/s)² = -15.6 J

ΔU = Uf - U₀ = m*g*h = *m*g*d*sin θ = 2.4 kg*9.81 m/s²*d*0.454 = 10.7*d J

Wnc = Ff. d* cos (180º) = μk*N*d*cos(180º) (2)

The friction force always opposes to the displacement, so the angle between force and displacement is 180º.

The normal force, as is always perpendicular to the surface, takes the value needed to equilibrate the component of the weight perpendicular to the incline, as follows:

N = m*g*cosθ =  2.4 kg*9.81 m/s²*cos 27º = 21 N

Replacing in (2):

Wnc = 0.12*21*cos (180º) = -2.52*d J

Replacing in (1):

-15.6 J + 10.7*d J = -2.52*d J

Solving for d:

d = 1.18 m

 

7 0
4 years ago
A fleeing student covered a distance of 210 meters in 35 seconds. what is the student's speed in meters/second? what is the stud
snow_lady [41]

Explanation:

Given that,

A student covered a distance of 210 meters in 35 seconds.

We need to find the student's speed in meters/second and also in meters/minute.

Speed, v = distance (d)/time (t)

So,

v=\dfrac{210\ m}{35\ s}\\\\=6\ m/s

We know that, 1 minute = 60 seconds

6\dfrac{m}{s}=6\times \dfrac{m}{(\dfrac{1}{60})\ \text{minutes}}\\\\=360\ \text{meters/minutes}

Hence, the student's speed is 6 m/s or 360 meters/minute.

6 0
3 years ago
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