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1 year ago

12

A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at -1.2 m/s2 , which force is closest to the net force

acting on the vehicle?
Answer choices:
A. -9600N
B. -1300N
C. -900N
D. -94N

2 answers:

larisa [96]1 year ago

8 0

Answer:

D

Explanation:

boyakko [2]1 year ago

7 0

Answer:

Explanation:

D

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What force would be needed to accelerate a 0.040-kg golf ball at 20.0 m/s?

Naily [24]

Answer:

any amount of force will do it as time is not mentioned here

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2 years ago

1. A particular lever is 90.0% efficient. If 50.0 J of work are done on the lever, then how much work does the lever do on its l

laila [671]

Answer:

Explanation:

Using the efficiency formula;

Efficiency = Work done by the machine (output)/work done on the machine (input) ×100%

Efficiency =w/50 ×100

90 = 100w/50

Cross multiply

90×50 = 100W

4500 = 100W

W = 4500/100

W = 45Joules

Hence the lever does 45Joules of work on its load

2) Mechanical Advantage= Load/Effort

Given

MA = 4

Load = 500N

4 = 500/Effort

Effort = 500/4

Effort =125N

Hence the effort required to lift the load is 125N

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1 year ago

Given three capacitors, c1 = 2.0 μf, c2 = 1.5 μf, and c3 = 3.0 μf, what arrangement of parallel and series connections with a 12

Lesechka [4]

Answer:

Connect C₁ to C₃ in parallel; then connect C₂ to C₁ and C₂ in series. The voltage drop across C₁ the 2.0-μF capacitor will be approximately 2.76 volts.

$-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-$ .

Explanation:

Consider four possible cases.

<h3>Case A: 12.0 V.</h3>

$-\begin{array}{c}-{\bf 2.0\;\mu\text{F}-}\\-1.5\;\mu\text{F}- \\-3.0\;\mu\text{F}-\end{array}-$

In case all three capacitors are connected in parallel, the $2.0\;\mu\text{F}$ capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.

<h3>Case B: 5.54 V.</h3>

$-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-$

In case the $2.0\;\mu\text{F}$ capacitor is connected in parallel with the $1.5\;\mu\text{F}$ capacitor, and the two capacitors in parallel is connected to the $3.0\;\mu\text{F}$ capacitor in series.

The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.

The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,

$\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_3}+ \dfrac{1}{C_1+C_2}} = \frac{1}{\dfrac{1}{3.0}+\dfrac{1}{2.0+1.5}} = 1.62\;\mu\text{F}$ .

What will be the voltage across the 2.0 μF capacitor?

The charge stored in two capacitors in series is the same as the charge in each capacitor.

$Q = C(\text{Effective}) \cdot V = 1.62\;\mu\text{F}\times 12\;\text{V} = 19.4\;\mu\text{C}$ .

Voltage is the same across two capacitors in parallel.As a result,

$\displaystyle V_1 = V_2 = \frac{Q}{C_1+C_2} = \frac{19.4\;\mu\text{C}}{3.5\;\mu\text{F}} = 5.54\;\text{V}$ .

<h3>Case C: 2.76 V.</h3>

$-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-$ .

Similarly,

the effective capacitance of the two capacitors in parallel is 5.0 μF;
the effective capacitance of the three capacitors, combined: $\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_2}+ \dfrac{1}{C_1+C_3}} = \frac{1}{\dfrac{1}{1.5}+\dfrac{1}{2.0+3.0}} = 1.15\;\mu\text{F}$ .

Charge stored:

$Q = C(\text{Effective}) \cdot V = 1.15\;\mu\text{F}\times 12\;\text{V} = 13.8\;\mu\text{C}$ .

Voltage:

$\displaystyle V_1 = V_3 = \frac{Q}{C_1+C_3} = \frac{13.8\;\mu\text{C}}{5.0\;\mu\text{F}} = 2.76\;\text{V}$ .

<h3 /><h3>Case D: 4.00 V</h3>

$-2.0\;\mu\text{F}-1.5\;\mu\text{F}-3.0\;\mu\text{F}-$ .

Connect all three capacitors in series.

$\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_1} + \dfrac{1}{C_2}+\dfrac{1}{C_3}} =\frac{1}{\dfrac{1}{2.0} + \dfrac{1}{1.5}+\dfrac{1}{3.0}} =0.667\;\mu\text{F}$ .

For each of the three capacitors:

$Q = C(\text{Effective})\cdot V = 0.667\;\mu\text{F} \times 12\;\text{V} = 8.00\;\mu\text{C}$ .

For the $2.0\;\mu\text{F}$ capacitor:

$\displaystyle V_1=\frac{Q}{C_1} = \frac{8.00\;\mu\text{C}}{2.0\;\mu\text{F}} = 4.0\;\text{V}$ .

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2 years ago

Classify the following as alkali metals, alkaline earth metals, transition elements, or inner transitional elements: calcium, go

Digiron [165]

Alkali metals : sodium , potassium
alkaline earth : magnesium , calcium
the rest are transition elements... i don't know about "inner transition"

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2 years ago

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A TV satellite broadcasts at a frequency of 5000 MHz, (1 MHz = 1 million Hertz). What is the wavelength of this radiation?

Nitella [24]

Answer:

$\lambda=0.06\ m$

Explanation:

Given:

frequency of the broadcast, $f=5000\ MHz=5\times 10^9\ Hz$

we have the speed of the radiation equal to the speed of light, $c=3\times 10^8\ m.s^{-1}$

The broadcast waves are the electromagnetic waves but it can travel only upto a hundred kilometers without any loss of information carried by it.

<u>The relation between the frequency and the wavelength:</u>

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

$\lambda=\frac{3\times 10^8}{5\times 10^9}$

$\lambda=0.06\ m$

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2 years ago

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