Answer: Energy can neither be created nor destroyed, rather it is converted from one form to another
Explanation:
The principle of conversation of energy explains how energy is conserved in nature by being converted from one form to another such that no energy is created nor destroyed.
Practical examples include:
- electrical pressing iron that converts electrical energy to heat energy
- solar panels that converts solar energy to electrical energy
- Car batteries that converts chemical energy to light energy etc
Explanation :
A circuit is the representation of the path of the flow of current. The circuit can be either closed or open.
When the switch is off the circuit is closed circuit and when the switch is not connected the circuit is open.
The items that are sufficient to make a circuit are as follows :
- Voltage source like a battery.
- Resistors or electrical equipment like heater, motor etc.
Other components can be ammeter, voltmeter, ac source, variable resistors etc.
The acceleration of the crate after it begins to move is 0.5 m/s²
We'll begin by calculating the the frictional force
Mass (m) = 50 Kg
Coefficient of kinetic friction (μ) = 0.15
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (N) = mg = 50 × 10 = 500 N
<h3>Frictional force (Fբ) =?</h3>
Fբ = μN
Fբ = 0.15 × 500
<h3>Fբ = 75 N</h3>
- Next, we shall determine the net force acting on the crate
Frictional force (Fբ) = 75 N
Force (F) = 100 N
<h3>Net force (Fₙ) =?</h3>
Fₙ = F – Fբ
Fₙ = 100 – 75
<h3>Fₙ = 25 N</h3>
- Finally, we shall determine the acceleration of the crate
Mass (m) = 50 Kg
Net force (Fₙ) = 25 N
<h3>Acceleration (a) =?</h3>
a = Fₙ / m
a = 25 / 50
<h3>a = 0.5 m/s²</h3>
Therefore, the acceleration of the crate is 0.5 m/s²
Learn more on friction: brainly.com/question/364384
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
For the answer to the question above, first find out the gradient.
<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>
<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>
<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>
<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>
<span>y is the velocity of the car, x is the time.
</span>I hope this helps.
