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4 years ago

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an explanation of conduction(science explanation) pleases helppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp

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1 answer:

beks73 [17]4 years ago

3 0

Answer:

the process by which heat or electricity is directly transmitted through a substance when there is a difference of temperature or of electrical potential between adjoining regions, without movement of the material.

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A 2kg mass is initially at rest on a frictionless surface. A 45N force acts at an angle of 300

raketka [301]

Answer:

a) The work done by the force is 136.400 joules.

b) The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

Explanation:

The correct statement is shown below:

<em>A 2-kg mass is initially at reat on a frictionless surface. A 45 N force acts at an angle of 30º to the horizontal for a distance of 3.5 meters:</em>

<em>a)</em><em> Find the work done by the force.</em>

<em>b)</em><em> Find the speed of the mass at the end of the 3.5 meters.</em>

a) Given that a external constant force is acting on the mass on a frictionless surface, the work done by the force ( $W_{F}$ ), measured in joules, is:

$W_{F} = F\cdot \Delta s \cdot \cos \theta$ (1)

Where:

$F$ - External constant force exerted on the mass, measured in newtons.

$\Delta s$ - Horizontal travelled distance, measured in meters.

$\theta$ - Direction of the external force regarding the horizontal, measured in sexagesimal degrees.

If we know that $F = 45\,N$ , $\Delta s = 3.5\,m$ and $\theta = 30^{\circ}$ , then the work done by the force is:

$W_{F} = (45\,N)\cdot (3.5\,m)\cdot \cos 30^{\circ}$

$W_{F} = 136.400\,J$

The work done by the force is 136.400 joules.

b) The final speed of the mass at the end ot the 3.5 meter is calculated by means of the Work-Energy Theorem, which means that the work done on the mass is transformed into a translational kinetic energy. That is:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$ (2)

Where:

$m$ - Mass, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the mass, measured in meters per second.

If we know that $W_{F} = 136.400\,J$ , $m = 2\,kg$ and $v_{1} = 0\,\frac{m}{s}$ , then the final speed of the mass is:

$\frac{2\cdot W_{F}}{m} = v_{2}^{2}-v_{1}^{2}$

$v_{2}^{2} =v_{1}^{2}+\frac{2\cdot W_{F}}{m}$

$v_{2} =\sqrt{v_{1}^{2}+\frac{2\cdot W_{F}}{m} }$

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\frac{2\cdot (136.400\,J)}{2\,kg} }$

$v_{2} \approx 11.679\,\frac{m}{s}$

The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

3 0

3 years ago

6. A 4 kg object hangs below a 6 kg object by a string of negligible mass. If the 6 kg object is pulled upward by a force of 440

MrRissso [65]

Answer:

T =176 N

Explanation:

from diagram

F -(m_1+m_2_g) = (m_1+m_2_g)a

440 - (6+4)g = (6+4)a

a =\frac{440-10*9.8}{10}

a =34.2 m/s^2

frrom free body diagram of mass m2 = 4kg

T -m_2g =m_2a

T = m_2(g +a)

T = 4(9.81+34.2)

T =176 N

7 0

4 years ago

A parallel-plate capacitor with circular plates of radius 0.10 m is being discharged. A circular loop of radius 0.20 m is concen

Sedaia [141]

Answer:

7.1934 x 10^12 V/m.s

Explanation:

In order to do this exercise, you need to use the correct formula. Besides that, we need to identify our data.

First we have the radius of the plates which are circular, and it's 0.1 m. The current of the loop (I) is 2.0 A, and the radius of the loop is 0.2 m.

Now with this data, we use the next formula:

I = dE/dt Eo A

Where:

dE/dt = rate of electric field

Eo = constant of permittivity of free space

A = Area of circle

Solving for dE/dT:

dE/dt = I / Eo*A

Now, the area of the circle is A = πr²

A = 3.1416 * (0.1)² = 0.031416 m²

Now solving the electric field:

dE/dt = 2 / (8.85x10^-12 * 0.031416)

dE/dt = 7.1934 x 10^12 V/m.s

6 0

3 years ago

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an elect

klio [65]

Complete Question

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?

Answer:

The magnetic field strength is $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

$\Delta V q = \frac{1}{2}mv^2$

Making v the subject

$v = \sqrt{[\frac{2 \Delta V * q }{m}] }$

Where m is the mass of electron

v is the velocity of electron

q charge on electron

$\Delta V$ is the potential difference

Substituting values

$v = \sqrt{\frac{2 * 5.9 *10^3 * 1.60218*10^{-19} }{9.10939 *10^{-31]} }$ f

$= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[ $F = B q v$ ] must be equal to the centripetal force[ $\frac{mv^2}{r}$ ] and this is mathematically represented as

$Bqv = \frac{mv^2}{r}$

making B the subject

$B = \frac{mv}{rq}$

r is the radius with a value = 5.4cm = $= \frac{5.4}{100} = 5.4*10^{-2} m$

Substituting values

$B = \frac{9.1039 *10^{-31} * 4.556 *10^7}{5.4*10^-2 * 1.60218*10^{-19}}$

$= 0.0048 T$

7 0

4 years ago

When you lift a box, what is the effort force?

aleksandrvk [35]

Answer:

effort force: The force used to move an object over a distance. resistance force: The force which an effort force must overcome in order to do work on an object via a simple machine. ideal mechanical advantage: The factor by which a mechanism multiplies the force put into it.

Explanation:

5 0

3 years ago