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3 years ago

Give an example of an energy conversion that produces an unwanted form of energy.

1 answer:

5 0

An example of an energy conversion that produces an unwanted form of energy is mixing acids with water. For example adding concentrated sulfuric acid with water that generates too much heat and must be controlled in systems that use this step of addition.

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A 100 kg bag of sand has a weight on 100 N. When dropped its acceleration is what?

Anettt [7]

Answer:

The acceleration of the bag of sand is $1\ m/s^2$ .

Explanation:

We have,

Mass of a bag of sand is 100 kg

Weight of the bag of sand is 100 N

It is required to find the acceleration of the bag when it is dropped. The weight of an object is given by :

$F =ma$

When it is dropped, a = g

$g=\dfrac{F}{m}\\\\g=\dfrac{100}{100}\\\\g=1\ m/s^2$

So, the acceleration of the bag of sand is $1\ m/s^2$ .

4 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

3 years ago

What is the direction of the magnetic field if an electron moving in the positive x direction experiences a magnetic force in th

Alex787 [66]

Given :

An electron moving in the positive x direction experiences a magnetic force in the positive z direction.

To Find :

The direction of the magnetic field.

Solution :

We know, force is given by :

$\vec{F}=q(\vec{v}\times \vec{B)}$

Here, q = -e.

$\vec{F}=(-e)(\vec{v}\times \vec{B)}\\\\\hat{k}=(-e)(\hat{i}\times \vec{B})$

Now, for above condition to satisfy :

$\hat{i}\times \vec{B}=-\hat{k}$

So, $\vec{B}=-\hat{j}$

Therefore, direction of magnetic field is negative y direction.

Hence, this is the required solution.

7 0

3 years ago

A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a

aleksley [76]

Answer:

T₂ =602 °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

$\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\$

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602 °C

5 0

3 years ago

What is the easiest way to increase the magnetic force acting on the rotor in an induction motor?

Schach [20]

Answer:

Explanation:

Magnets are of two major forms namely the permanent magnet and the temporary magnets. Temporary magnets magnetizes and demagnetize easily while permanent magnets does not magnetizes and demagnetize easily.

This permanents magnets are applicable in loudspeakers, generators, induction motor etc.

To increase the

The following will tend to increase the magnetic force acting on the rotor in an induction motor.

1. Increasing the strength of the bar magnet. Increase in strength of the magnet will lead to increase in the magnetic force acting on the rotor.

2. Increase in the magnetic line of force also known as the magnetic flux around the magnet will also increase the magnetic force acting on the rotor.

6 0

2 years ago