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Reil [10]
3 years ago
15

The

Physics
1 answer:
yKpoI14uk [10]3 years ago
5 0

13.6g

136g

Explanation:

Given parameters:

Density of mercury = 13.6g/cm³

Unknown:

Mass of :

1cm³ of mercury

10cm³ of mercury

Solution:

Density is the mass per unit volume of a substance. It is the amount of matter in a given volume.

   Density = \frac{mass}{volume}

Since the unknown in the problem is mass;

    Mass = density x volume

Mass of  1cm³ of mercury = 13.6 x 1 = 13.6g

Mass of  10cm³ of mercury = 10 x 13.6 = 136g

learn more:

Density problems brainly.com/question/3433940

#learnwithBrainly

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An amusement park ride called the Rotor debuted in 1955 in Germany. Passengers stand in the cylindrical drum of the Rotor as it
steposvetlana [31]

Answer:

μs = 0.36

Explanation:

  • While the drum is rotating, the riders, in order to keep in a circular movement, are accelerated towards the center of the drum.
  • This acceleration is produced by the centripetal force.
  • Now, this force is not a different type of force, is the net force acting on the riders in this direction.
  • Since the riders have their backs against the wall, and the normal force between the riders and the wall is perpendicular to the wall and aiming out of it, it is easily seen that this normal force is the same centripetal force.
  • In the vertical direction, we have two forces acting on the riders: the force of gravity (which we call weight) downward, and the friction force, that will oppose to the relative movement between the riders and the wall, going upward.
  • When this force be equal to the weight, it will have the maximum possible value, which can be written as follows:

       F_{frmax} = \mu_{s}* F_{n}  = m * g  (1)

  • where μs= coefficient of static friction (our unknown)
  • As  we have already said Fn = Fc.
  • The value of the centripetal force, is related with the angular velocity ω and the radius of the drum r, as follows:

      F_{n} = m* \omega^{2} * r  (2)

  • Replacing (2) in (1), simplifying and rearranging terms, we can solve for μs, as follows:

       \mu_{s} = \frac{g}{\omega^{2} r}  (3)

  • Prior to replace ω for its value, is convenient to convert it from rev/min to rad/sec, as follows:

       \omega = 26.0 \frac{rev}{min} * \frac{1min}{60 sec} *\frac{2*\pi rad }{1 rev} = 2.72 rad/sec (4)

  • Replacing g, ω and r in (3):
  • \mu_{s} = \frac{g}{\omega^{2} r} = \frac{9.8m/s2}{(2.72rad/sec)^{2} *3.7 m} = 0.36 (5)

3 0
3 years ago
Electromagnetic induction
vredina [299]

Answer:

Electromagnetic Induction or Induction is a process in which a conductor is put in a particular position and magnetic field keeps varying or magnetic field is stationary and a conductor is moving. This produces a Voltage or EMF (Electromotive Force) across the electrical conductor.

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What is one factor that helps keep the planets and other solar system objects in orbit around the sun?
pshichka [43]
The answer is D<span>.the gravitational force between the sun and each object in the solar system</span>
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A car is brought to rest in a distance of 484m using a constant acceleration of -8.0m/s^2. What was the velocity of the car when
Agata [3.3K]

Answer:

88 m/s

Explanation:

To solve the problem, we can use the following SUVAT equation:

v^2-u^2=2ad

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance covered

For the car in this problem, we have

d = 484 m is the stopping distance

v = 0 is the final velocity

a=-8.0 m/s^2 is the acceleration

Solving for u, we find the initial velocity:

u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s

6 0
3 years ago
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
belka [17]

Answer:

F_a=5.67\times 10^{-5}\ N

<u />F_b=3.49\times 10^{-5}\ N

F_c=9.16\times 10^{-5}\ N

Explanation:

Given:

  • mass of particle A, m_a=363\ kg
  • mass of particle B, m_b=517\ kg
  • mass of particle C, m_c=154\ kg
  • All the three particles lie on a straight line.
  • Distance between particle A and B, x_{ab}=0.5\ m
  • Distance between particle B and C, x_{bc}=0.25\ m

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}

F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})

F_a=5.67\times 10^{-5}\ N

<u>Force on particle C due to particles B & A:</u>

<u />F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}<u />

F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )

F_c=9.16\times 10^{-5}\ N

<u>Force on particle B due to particles C & A:</u>

<u />F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}<u />

<u />F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2}  )<u />

<u />F_b=3.49\times 10^{-5}\ N<u />

3 0
3 years ago
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