Ignoring air resistance, if a 20 kg ball and a 400 kg crate were both dropped from the

Albert uses as his unit of length (for walking to visit his neighbors or plowing his fields) the albert (a), the distance albert

True [87]

To solve this problem, we know that:

1 Albert = 88 meters

1 A = 88 m

The first thing we have to do is to square both sides of the equation:

(1 A)^2 = (88 m)^2

1 A^2 = 7,744 m^2

Since it is given that 1 acre = 4,050 m^2, so to reach that value, 1st let us divide both sides by 7,744:

1 A^2 / 7,744 = 7,744 m^2 / 7,744

(1 / 7,744) A^2 = 1 m^2

Then we multiply both sides by 4,050.

(4050 / 7744) A^2 = 4050 m^2

0.523 A^2 = 4050 m^2

Therefore 1 acre is equivalent to about 0.52 square alberts.

7 0

2 years ago

What type of power plant burns material to make electricity?

Alecsey [184]

<h2>Hello!</h2>

The answer is: D. Coal

Coal power plants burn coal in to get steam, the steam flows into a turbine which is coupled to an electrical generator.

Coal power plants work burning high amounts of coal into a boiler, generation a lot of steam under extreme pressures. The steam is obtained when the water is heated by the burning coal, then the steam is cooled, being transformed in liquid water again (due the condensation process) and it's sent back on a cyclical process.

Have a nice day!

6 0

2 years ago

Read 2 more answers

The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a

jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A $V_a=382\ V$

Potential at $V_c=785\ V$

when particle starts from A it reaches with velocity $v_b$ at Point while when it starts from C it reaches at point B with velocity $2v_b$

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

$q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1$

Change in Kinetic Energy of particle moving under the Potential From C to B

$q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2$

Divide 1 and 2 we get

$\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}$

on solving we get

$V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c$

$V_b=\frac{743}{3}=247.67\ V$

4 0

2 years ago

base your answer to this question on the information below and on your knowledge of physics. A toy launcher that is used to laun

Nadya [2.5K]

Answer: v = 2.24 m/s

Explanation: The Law of Conservation of Energy states that total energy is constant in any process and, it cannot be created nor destroyed, only transformed.

So, in the toy launcher, the energy of the compressed spring, called Elastic Potential Energy (PE), transforms into the movement of the plastic sphere, called Kinetic Energy (KE). Since total energy must be constant:

$KE_{i}+PE_{i}=KE_{f}+PE_{f}$

where the terms with subscript i are related to the initial of the process and the terms with subscript f relates to the final process.

The equation is calculated as:

$\frac{1}{2}kx^{2}+0=0+\frac{1}{2}mv^{2}$

$\frac{1}{2}kx^{2}=\frac{1}{2}mv^{2}$

$\frac{1}{2}50(0.1)^{2}=\frac{1}{2}(0.1)v^{2}$

$v^{2}=\frac{50(0.1)^{2}}{0.1}$

$v=\sqrt{50(0.1)}$

$v=\sqrt{5}$

v = 2.24

The maximum speed the plastic sphere will be launched is 2.24 m/s.

3 0

2 years ago

The siren on an ambulance emits a sound of frequency 2.80×103Hz. If the ambulance is traveling at 26.0 m/s (93.6 km/h or 58.2 mi

Norma-Jean [14]

To solve this problem it is necessary to apply the concepts related to the described wavelength through frequency and speed. Mathematically it can be expressed as:

$\lambda = \frac{v}{f}$

Where,

$\lambda =$ Wavelength

f = Frequency

v = Velocity

Our values are given as,

$f = 2.8*10^3Hz$

$v = 340m/s \rightarrow$ Speed of sound

Keep in mind that we do not use the travel speed of the ambulance because we are in front of it. In case it approached or moved away we should use the concepts related to the Doppler effect:

Replacing we have,

$\lambda = \frac{340}{2.8*10^3}$

$\lambda = 0.1214m$

Therefore the frequency that you hear if you are standing in from of the ambulance is 0.1214m

5 0

2 years ago