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3 years ago

A force of 30 N is applied to an object with a mass of 15 kg. What is the resulting acceleration?

Physics

2 answers:

svet-max [94.6K]3 years ago

5 0

Answer:

We know that Force = mass × acceleration

By substituting the values we get,

30 N = 15 kg × a (where a is acceleration)

Or we can write it as

15 kg × a = 30 N

Transposing 15 to RHS,

a = 30 ÷ 15 m/s²

Therefore, acceleration = 2 m/s²

pls give brainliest for the answer

Gnoma [55]3 years ago

4 0

Just remember when dealing with Newton II (where F=m*a), that F is the net (or resultant) force on the object. 30/15=2m/s

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4 years ago

Below is a circuit schematic of sources and resistors (Figure 3). VS = 10V , R1 = 100Ω, R2 = 50Ω, R3 = 25Ω, IS = 2A. Calculate t

lorasvet [3.4K]

Answer:

$V_3\approx 4.28\,\,V$

$I_1=0.0572\,\,amps$

$I_3\approx 0.171\,\,amps$

Explanation:

Notice that this is a circuit with resistors R1 and R2 in parallel, connected to resistor R3 in series. It is what is called a parallel-series combination.

So we first find the equivalent resistance for the two resistors in parallel:

$\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega$

By knowing this, we can estimate the total current through the circuit,:

$Vs=I\,*\,(\frac{100}{3} +25)\\10=I\,*\,\frac{175}{3} \\I=\frac{30}{175} \,amps$

So approximately 0.17 amps

and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:

$V_3=\frac{30}{175} *\,25=\frac{30}{7} \approx 4.28\,\,V$

So now we know that the potential drop across the parellel resistors must be:

10 V - 4.28 V = 5.72 V

and with this info, we can calculate the current through R1 using Ohm's Law:

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Please give me a brainlest

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Two gerbils run in place with a linear speed of 0.50 m/s on an exercise wheel that is shaped like a hoop. Find the angular momen

Brilliant_brown [7]

Answer:

$L = 0.018 kg m^2/s$

Explanation:

Angular speed and linear speed is related to each other as

$v = R\omega$

here we know that

$v = 0.50 m/s$

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now we have

$\omega = \frac{v}{R}$

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$\omega = 6.25 rad/s$

Now we know that moment of inertia of the system is given as

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$L = I\omega$

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3 years ago