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SVETLANKA909090 [29]
3 years ago
12

Physics double pivot question​

Physics
1 answer:
andriy [413]3 years ago
3 0

Explanation:

Assuming the wall is frictionless, there are four forces acting on the ladder.

Weight pulling down at the center of the ladder (mg).

Reaction force pushing to the left at the wall (Rw).

Reaction force pushing up at the foot of the ladder (Rf).

Friction force pushing to the right at the foot of the ladder (Ff).

(a) Calculate the reaction force at the wall.

Take the sum of the moments about the foot of the ladder.

∑τ = Iα

Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0

Rw (3.0 sin 60°) = mg (1.5 cos 60°)

Rw = mg / (2 tan 60°)

Rw = (10 kg) (9.8 m/s²) / (2√3)

Rw = 28 N

(b) State the friction at the foot of the ladder.

Take the sum of the forces in the x direction.

∑F = ma

Ff − Rw = 0

Ff = Rw

Ff = 28 N

(c) State the reaction at the foot of the ladder.

Take the sum of the forces in the y direction.

∑F = ma

Rf − mg = 0

Rf = mg

Rf = 98 N

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a 230 kg roller coaster reaches the top of the steepest hill with a speed of 6.2 km/h. It then descends the hill, which is at an
igor_vitrenko [27]

Answer: 81.619 kJ

Explanation:

Given

Mass of roller coaster is m=230\ kg

It reaches the steepest hill with speed of u=6.2\ km/h\ or \ 1.72\ m/s

Hill to bottom is 51 m long with inclination of 45^{\circ}

Height of the hill is h=51\sin 45^{\circ}=36.06\ m

Conserving energy to get kinetic energy at bottom

Energy at top=Energy at bottom

\Rightarrow K_t+U_t=K_b+U_b\\\Rightarrow \dfrac{1}{2}mu^2+mgh=K_b+0\\\\\Rightarrow K_b=0.5\times 230\times 1.72^2+230\times 9.8\times 36.06\\\Rightarrow K_b=340.216+81,279.24\\\Rightarrow K_b=81,619.456\ J\\\Rightarrow K_b=81.619\ kJ

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Describe the motion represented by a horizontal line on a distance-time graph.
denpristay [2]

Answer:

hi, this is the answer

Explanation:

A horizontal line on a distance-time graph shows no change in distance, therefore there is no motion.

The object is stationary. ...

Constant speed is motion that occurs with the same ratio of distance to time throughout the entire length of the motion.

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The United States spends over $20 billion a year on space exploration through NASA. Do you think that this has been worth the co
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Sunday, July 20, marked 45 years since the United States put the first two astronauts safely on the moon. The cost for the Mercury, Gemini and Apollo programs was more than $25 billion at the time more like $110 billion in today’s world. The ensuing U.S. space efforts have cost an additional $196 billion for the shuttle and $50 billion for the space station. NASA’s total inflation-adjusted costs have been more than $900 billion since its creation in 1958 through 2014 (more than $16 billion per year). Looking back, have we gotten our money’s worth from the investment?

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A charge of 12 c passes through an electroplating apparatus in 2. 0 min. what is the average current?
Studentka2010 [4]

A charge of 12 c passes through an electroplating apparatus in 2.0 min, then the average current is 0.1 ampere.

<h3>What is an electric charge?</h3>

Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. You can have a positive or negative electric charge.

Electric current is defined as the charge per unit of time.

The mathematical relation between current and the electric charge

I =Q/T

where I is the current flowing

Q is the total electric charge

T is the time period for which the current is flowing

As given in the problem A charge of 12 c passes through an electroplating apparatus in 2.0 min

Let us first convert the time period of minutes into seconds

1 min = 60 seconds

2 min = 2*60 seconds

         =120 seconds

By using the above relation between electric current and electric charge

and by substituting the respective values of the charge and the time period

I =Q/T

I = 12c/120 seconds

I = 0.1 Ampere

Thus, the average current flowing through the apparatus would be 0.1 Ampere.

Learn more about an electric charge from here

brainly.com/question/8163163

#SPJ4

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