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3 years ago

How do you calculate the total resistance of a series circuit

Physics

1 answer:

CaHeK987 [17]3 years ago

5 0

Add all the resistances across the circuit together the calculate the total resistance

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If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for

steposvetlana [31]

Answer:

$\ d_{out} = 100 \ m.$

Explanation:

Given data:

$F_{in} = 50 \ \rm N$

$F_{out} = 10 \ \rm N$

$d_{in} = 20 \ m$

Let the distance traveled by the object in the second case be $d_{out}.$

In the given problem, work done by the forces are same in both the cases.

Thus,

$W_{in} = W_{out}$

$F_{in}.d_{in} = F_{out}.d_{out}$

$\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}$

$\ d_{out} = \frac{50 \times 20}{10}$

$\ d_{out} = 100 \ m.$

5 0

3 years ago

How much speed is needed for an object to travel 250 km in 1.5 hrs? *<br> Your answer

NemiM [27]

Answer: 166.67km/hr

Explanation:

Given the following :

Distance traveled = 250km

Time taken = 1.5 hours

Recall :

Speed = Distance traveled / time taken

Speed = 250 km / 1.5 hours

Speed = 166. 67 km/hr

Speed in m/s:

166.67km/hr = (166.67 × 1000)m / 3600 s

= 166670m / 3600s

= 46.3m/s

7 0

4 years ago

How much work would have to be done to bring a 1150kg automobile traveling at 86km/h to a stop?

krok68 [10]

Explanation:

We have,

Mass of an automobile is 1150 kg

The automobile traveling at 86 km/h and then it comes to stop.

86 km/h = 23.88 m/s

It is required to find work done by the automobile.

Concept used : Work energy theorem

Th change in kinetic energy of an object is equal to the work done by it. The work done is then given by :

$W=\dfrac{1}{2}m(v^2-u^2)$

Here, v = 0

$W=-\dfrac{1}{2}mu^2\\\\W=-\dfrac{1}{2}\times 1150\times (23.88)^2\\\\W=-327896.28\ J$

$W=3.27\times 10^5\ J$

Therefore, the work done by the automobile is $-3.27\times 10^5\ J$ .

7 0

3 years ago

Because scientists cannot actually precisely measure a process like soil erosion over the entire planet, they must _____.

Pani-rosa [81]

Your answer is C. I hope this helps you.

4 0

4 years ago

Car A (1750 kg) is travelling due south and car B (1450 kg) is travelling due east. They reach the same intersection at the same

Contact [7]

Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as

$\underset{V_{A}}{\rightarrow}$ = velocity of car A before collision = 0 i - $V_{A}$ j

$\underset{V_{B}}{\rightarrow}$ = velocity of car B before collision = $V_{B}$ i + 0 j

$\underset{V_{AB}}{\rightarrow}$ = velocity of combination after collision = (35.8 Cos31.6) i - (35.8 Sin31.6) j = 30.5 i - 18.8 j

$M_{A}$ = mass of car A = 1750 kg

$M_{B}$ = mass of car B = 1450 kg

Using conservation of momentum

$M_{A}$ $\underset{V_{A}}{\rightarrow}$ + $M_{B}$ $\underset{V_{B}}{\rightarrow}$ = ( $M_{A}$ + $M_{B}$ ) ( $\underset{V_{AB}}{\rightarrow}$ )

(1750) (0 i - $V_{A}$ j) + (1450) ( $V_{B}$ i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)

(1450) $V_{B}$ i - (1750) $V_{A}$ j = 97600 i - 60160 j

Comparing the coefficient of "i" and "j" both side

(1450) $V_{B}$ = 97600 and - (1750) $V_{A}$ = - 60160

$V_{B}$ = 67.3 km/h and $V_{A}$ = 34.4 km/

6 0

4 years ago