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Vlad1618 [11]
3 years ago
10

You're carrying a 4.0-mm-long, 22 kg pole to a construction site when you decide to stop for a rest. You place one end of the po

le on a fence post and hold the other end of the pole 35 cm from its tip.
How much force must you exert to keep the pole motionless in a horizontal position?
Physics
1 answer:
Tju [1.3M]3 years ago
6 0

Answer:

    F = 118.14 N

Explanation:

For this exercise we will use the rotational equilibrium condition

      ∑ τ = I α = 0

Since the pole  is in balance.

 

Let's fix a reference system located the point where the post rests on the fence, select the counterclockwise rotation as positive

         -F (L -0.35) + W L / 2 = 0

         F = W L / 2 (L-0.35)

         L = 4.0 m

Let's calculate

         F = 22 9.8 2 / 3.65

         F = 118.14 N

You might be interested in
The magnetic field in a solenoid is . A circular wire of radius 8 cm is concentric with a solenoid of radius 2 cm and length d =
defon

Answer:

6.03 mV

Explanation:

length of solenoid, L = 2 m, N = 12000, di/dt = 40 A/s,

Magnetic field due to solenoid

B = μ0 n i = μ0 N i / L

dB/dt = μ0 N / L x di / dt

dB /dt = (4 x 3.14 x 10^-7 x 12000 x 40) / 2 = 0.3 T/s

Induced emf, e = rate of change of magnetic flux

e = dΦ / dt = A x dB / dt

e = 3.14 x 0.08 x 0.08 x 0.3 = 6.03 x 10^-3 V = 6.03 mV

7 0
3 years ago
A crane is used to lower weights into a lake for an underwater construction project. Determine the tension in the cable of the c
atroni [7]

Answer:

Tension in the cable = 591292.8 N or 591.3 kN

Explanation:

Tension in the cable = Weight of object in air - upthrust/bouyancy due to the liquid

Weight of object in air = mass * acceleration due to gravity, g

Since mass is not given, it is obtained from the formula, mass = density * volume

Volume of the sphere = 4/3πr³ where r, radius = 2 m; π = 22/7

volume = 4/3 * 22/7 * (2m)³

volume of sphere = 33.52 m³

Density of concrete sphere = specific gravity * density of water

where, specific gravity = 2.8, density of water = 1000 kg/m³

density of concrete =  2.8 * 1000 kg/m³

density of concrete = 2800 kg/m³

acceleration due to gravity, g = 9.8 m/s²

Thus, weight of concrete = 2800 kg/m³ * 33.52 m³ * 9.8 m/s² = 919788.8 N

Upthrust = density of water * volume of sphere * g

Upthrust = 1000 kg/m³ * 33.52 m³ * 9.8 m/s² = 328496 N

The tension in the cable is the calculated as below;

Tension in the cable = Weight of object in air - upthrust/bouyancy due to the liquid

Tension in cable = 919788.8 N - 328.496 N = 591292.8 N

Therefore, tension in the cable is 591292.8 N or 591.3 kN

5 0
3 years ago
PLEASE HELP ASAP!! CORRECT ANSWER ONLY PLEASE!!
adelina 88 [10]
A line on a graph has a downward slope
8 0
3 years ago
a ball is thrown straight up into the air with a speed of 13 m/s. if the ball has a mass of 0.25 kg, how high does the ball go?
evablogger [386]
<h2>Hello!</h2>

The answer is: 8.62m

<h2>Why?</h2>

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

E=PE+KE

Where:

PE=m.g.h

KE=\frac{1*m*v^{2}}{2}

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

<em>h </em>is the reached height of the object

<em>v</em> is the velocity of the object

Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:

KE_{1}+PE_{1}=KE_{2}+PE_{2}

Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.

\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}

So,

h_{2}=\frac{1*m*v_{1} ^{2}}{2*m*g}\\h_{2}=\frac{1*v_{1} ^{2}}{2g}=\frac{(\frac{13m}{s})^{2} }{2*\frac{9.8m}{s^{2}}}\\h_{2}=8.62m}

Hence,

The height at the second moment (maximum height) is 8.62m

Have a nice day!

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4 years ago
The numbers of times you do lift without resting during a muscle training workout is called
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