Ohm's Law states V = IR
So,
I = V/R
The answer is B. 10/5=2 amps
Ep = mgh
= (35)(9.8)(3.5)
= 1200.5 J
Ek = (1/2)(mv^2)
= (0.5)(35)(5^2)
= 437.5 J
Answer:
(a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
Explanation:
Given that,
Charge = 10.1 μC
Capacitor C₁ = 1.10 μF
Capacitor C₂ = 1.92 μF
Capacitor C₃ = 1.10 μF
Potential V₁ = 51.5 V
Let V₁ and V₂ be the potentials on the two plates of the capacitor.
(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor
Using formula of potential difference
Put the value into the formula
The potential on the second plate
(b). We need to calculate the equivalent capacitance of the two capacitors
Using formula of equivalent capacitance
Put the value into the formula
Hence, (a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
Answer:
Explanation:
Think about the equation for velocity. Its the change in position over the change in time. And now think about the formula for slope, its the change in y over the change in x. Now, if you have a position vs time graph, the slope is position over time, the direct formula for velocity. Therefore, the slope of a position vs time graph gives the velocity in that interval. You could take the derivative at that point and get the instantaneous velocity.
In the first argument, the only way to get a negative slope is if youre moving to the left of your initial position as you pass through time. So the claim isnt necessarily true. When you move away from the origin, that simply means youre moving through time.
In the second argument, a horizontal slope means the value is 0. That means there is a value of 0 for your velocity, aka standing still.
In the third argument, a positive slope means youre moving to the right of your initial position as you go through time.
The fourth one is incorrect. Speed is the magnitude of velocity, and the slope can be determined an any point besides vertical slopes, which would require time to literally stop.
I dont understand the very last part of what you posted, so try to make a decision based on what I just explained to you
lets assume
L = length of the object = 5 cm
h = height of the object = 3 cm
w = width of the object = 2 cm
V = Volume of the object
Since the dimensions of object suggest it to be a cuboid
Volume of the object is given as
V = L w h
inserting the values
V = 5 x 3 x 2
V = 30 cm³
m = mass of the object given = 10 g
ρ = density of the object = ?
density of the object is given as
density of object = mass of object / Volume of object
ρ = m/V
inserting the values
ρ = 10/30
ρ = 0.33 g/cm³