Hey! So referring to the data the thing we can clearly see is that in a vacuum, everything, regardless of its mass, falls at the same speed.
Acceleration is often confused with speed, or velocity, but the difference is, acceleration by definition is the rate of which an object falls with respect to its mass and time.
Every single thing in the world falls at the same acceleration, this is because of gravity. The difference is the speed of which it falls. In space, there is not any gravity, and so, the objects are able to fall at the same speed regardless of their mass.
<h3>Hello there!</h3>
Here, you are looking for the amount of heat put in for water, at a mass of 187 grams, to change by 80 degrees.
The equation commonly accepted to find the answer to questions like these is the specific heat formula.
The equation is Q = mc∆T, where Q is the amount of energy put in to raise the temperature by a certain amount, m is the mass, c is the specific heat capacity, and ΔT is the amount of temperature change.
The information given:
m = 187 grams
c = specific heat capacity of water, or in this case 1 calorie, or 4.184 joules (which is what we will be using)
ΔT = 80 degrees
Now just plug everything in to solve.
Q = 187 * 4.184 * 80
Q = 62592.64
So you have your answer: 62592.64 joules.
Hope this helped!
Answer:
each resistor draws 1/3 of an amp or 0.33333 amps
Explanation:
V = I * R
V = 10 volts
R = 30 ohms
10 = I * 30 Divide by 30
10/30 = I
I = 0.33333
Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>