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3 years ago

Take 100 POINTS!!!! look the picture

Physics

1 answer:

Savatey [412]3 years ago

6 0

Explanation:

Biological barriers are living organisms that help protect the body from pathogens. Trillions of harmless bacteria normally live on the human skin and in the urinary, reproductive, and gastrointestinal tracts. These bacteria use up food and surface space that help prevent pathogenic bacteria from colonizing the body

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particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine

AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is $1062.7\frac{m}{s^{2}}$

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

$F=ma$ (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

$F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}$

with q1 and q2 the charge of the particles, r the distance between them and k the constant $k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}$ . So:

$F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}$

$F=62.7 N$

Using that value on (1) and solving for a

$a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}$

4 0

4 years ago

Which of the following types of energy is potential energy A. gravitational energy

katovenus [111]

There are many forms of energy, but they can all be put into two categories: kinetic and potential. Kinetic energy is motion––of waves, electrons, atoms, molecules, substances, and objects. Potential energy is stored energy and the energy of position––gravitational energy

4 0

3 years ago

What is the acceleration of the car at segment C?

Masteriza [31]

-30 I think sorry if I’m wrong

7 0

3 years ago

PLEASE HELP ME 45 POINTS

sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

$\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a$ (1)

Mass B

$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

Where:

$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

$(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g$

$a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g$

If we know that $m_{A} = 5\,kg$ , $m_{B} = 2\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the acceleration of the masses is:

$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

$a = 4.203\,\frac{m}{s^{2}}$

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

$T = 28.02\,N$

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)