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Mamont248 [21]
3 years ago
15

Is v2 = v1t+a dimensionally correct? Explain please!

Physics
1 answer:
Lady bird [3.3K]3 years ago
4 0
You want v2 = v1 + at
v is measured in m/s, a in m/s2, and t in s.
the dimensions multiply like algebraic quantities. 
so because v2 is measured in m/s, then (v1 + at) has to come out in m/s
 the units for (v1 + at) are (m/s) + (m/s2)(s)
time "s" cancels out one acceleration "s", so it comes ut to (m/s) + (m/s), which = (m/s). 
if you had (v1t + a), then you would have (m/s)(s) + (m/s2) which = (m) + (m/s2), which doesn't work.

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From mechanics, you may recall that when the acceleration of an object is proportional to its coordinate, d2xdt2=−kmx=−ω2x , suc
marysya [2.9K]

Answer:

    q = q₀ sin (wt)

Explanation:

In your statement it is not clear the type of circuit you are referring to, there are two possibilities.

1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor

                    ΔV = ΔV_{C}

we assume that the source has a voltage of the form

                    ΔV = ΔV₀o sin wt

The capacitance of a capacitor is

                   C = q / ΔV

                  q = C ΔV sin wt

the current in the circuit is

                    i = dq / dt

                    i = c ΔV₀ w cos wt

if we use

                  cos wt = sin (wt + π / 2)

we make this change by being a resonant oscillation

we substitute

                  i = w C ΔV₀ sin (wt + π/2)

With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current

2) Another possible circuit is an LC circuit.

In this case the voltage alternates between the inductor and the capacitor

                     V_{L} + V_{C} = 0

                      L di / dt + q / C = 0

the current is

                      i = dq / dt

                       

they ask us for a solution so that

                    L d²q / dt² + 1 / C q = 0

                     d²q / dt² + 1 / LC q = 0

this is a quadratic differential equation with solution of the form

                    q = A sin (wt + Ф)

to find the constant we derive the proposed solution and enter it into the equation

                di / dt = Aw cos (wt + Ф)

                d²i / dt²= - A w² sin (wt + Ф)

                 - A w² + 1 /LC  A = 0

                  w = √ (1 / LC)

To find the phase factor, for this we use the initial conditions for t = 0

in the case of condensate for t = or the charge is zero

                 0 = A sin Ф

                  Ф = 0

             

                  q = q₀ sin (wt)

6 0
3 years ago
Write any two uses of simple machines.
Nana76 [90]
Simple machines could be used to reduce effort or extend the ability of people to perform tasks beyond their normal capabilities.
Examples include pulley, lever, and incline plane
4 0
2 years ago
An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 16-W LED bulb can replace a 100-W inc
Yanka [14]

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

P=VI

Express this in terms of 'I'. This gives,

I=\frac{P}{V}

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, P_1=16\ W, V=110\ V

So, current drawn is given as:

I_1=\frac{16\ W}{110\ V}=0.145\ A

For incandescent bulb, P_2=100\ W, V=110\ V

So, current drawn is given as:

I_2=\frac{100\ W}{110\ V}=0.909\ A

For CFL bulb, P_3=24\ W, V=110\ V

So, current drawn is given as:

I_3=\frac{24\ W}{110\ V}=0.218\ A

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

5 0
3 years ago
If a suspended object A is attracted to a charged object B, can we conclude that A is charged?
spayn [35]
Not necessarily, object A could also be neutral, and becoming a dipole due to object B's charge. A charged object can induce a dipole in a neutral object, and that object would then become attracted without being charged.
8 0
3 years ago
A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it
makkiz [27]

Answer:D

Explanation:

Given

mass of object m=5 kg

Distance traveled h=10 m

velocity acquired v=12 m/s

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy=mgh=5\times 9.8\times 10=490 J

Final Energy=\frac{1}{2}mv^2+W_{f}

=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}

where W_{f} is friction work if any

490=360+W_{f}

W_{f}=130 J

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

4 0
3 years ago
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