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3 years ago

What is the combined resistance for two resistors with resistances of 10 Ohms and 23.7 Ohms if they are connected in series

Physics

1 answer:

Agata [3.3K]3 years ago

3 0

Answer: 33.7Ω

Explanation:

Since there are two resistors connected in series, the total resistance (Rtotal) of the circuit is the sum of each resistance.

i.e Rtotal = R1 + R2

R1 = 10Ω

R2 = 23.7Ω

Hence, Rtotal = 10Ω + 23.7Ω

Rtotal = 33.7Ω

Thus, the combined resistance for two resistors is 33.7Ω

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starting from rest , a formula one car accelerates uniformly at 25m\s2 for 30secs. what distance does it cover in the last one s

Anestetic [448]

The distance covered in the last second of motion is 737.5 m

Explanation:

The motion of the car is a uniformly accelerated motion, so we can use the suvat equations.

First of all, we have to find the velocity of the car when the last second of motion starts, that is the velocity of the car after t = 29 s. We can use the equation:

v = u + at

where

u = 0 is the initial velocity

$a=25 m/s^2$ is the acceleration

Substituting t = 29 s,

$v=0+(25)(29)=725 m/s$

Now we can find the distance covered in the last second of motion by using

$s=ut+\frac{1}{2}at^2$

where

u = 725 m/s is the velocity at the beginning of the last second

t = 1 s is the time interval considered

$a=25 m/s^2$ is the acceleration

Substituting,

$s=(725)(1)+\frac{1}{2}(25)(1)^2=737.5 m$

Note that the acceleration of $25 m/s^2$ is not realistic for a car, but I still have used the data of the problem.

Learn more about acceleration:

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6 0

3 years ago

Whats the differnece between a series circuit and a parallel circuit<br><br>

Romashka-Z-Leto [24]

Answer:

A series circuit has a direct flow of current and because of that, the current is constant throughout the circuit, while the voltage is what changes.. In a parallel circuit, the current travels through multiple paths, so the current is divided among those paths. The voltage, on the other hand, is constant.

Explanation:

Definition of series and parallel circuits.

brainliest please <3

7 0

3 years ago

A 97.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 48.0 m/s. If both are initially at rest and if t

OverLord2011 [107]

Answer:

s₁ = 0.022 m

Explanation:

From the law of conservation of momentum:

$m_1u_1 + m_2u_2 = m_1v_1+m_2v_2$

where,

m₁ = mass of hockey player = 97 kg

m₂ = mass of puck = 0.15 kg

u₁ = u₂ = initial velocities of puck and player = 0 m/s

v₁ = velocity of player after collision = ?

v₂ = velocity of puck after hitting = 48 m/s

Therefore,

$(97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s$

negative sign here shows the opposite direction.

Now, we calculate the time taken by puck to move 14.5 m:

$s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s$

Now, the distance covered by the player in this time will be:

$s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)$

4 0

3 years ago

Two point charges of -7uC and 4uC are a distance of 20

aivan3 [116]

Answer:

Approximately 0.979 J.

Explanation:

Assume that the two charges are in vacuum. Apply the coulomb's law to find their initial and final electrical potential energy $\mathrm{EPE}$ .

$\displaystyle \mathrm{EPE} = \frac{k \cdot q_1 \cdot q_2}{r}$ ,

where

The coulomb's constant $k = 8.99\times 10^{9}\; \rm N\cdot m^{2} \cdot C^{-2}$ ,
$q_1$ and $q_2$ are the sizes of the two charges, and
$r$ is the separation of (the center of) the two charges.

Note that there's no negative sign before the fraction.

Make sure that all values are in SI units:

$q_1 = -7\rm \;\mu C = -7\times 10^{-6}\; C$ ;
$q_2 = 4\rm \;\mu C = 4\times 10^{-6}\; C$ ;

Initial separation: $\rm 20\; cm = 0.20\; cm$ ;
Final separation: $\rm 90\; cm = 0.90\; cm$ .

Apply Coulomb's law:

Initial potential energy:

$\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.20}\\&= \rm -1.2586\; J\end{aligned}$ .

Final potential energy:

$\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.90}\\&= \rm -0.279689\; J\end{aligned}$ .

The final potential energy is less negative than the initial one. In other words, the two particles gain energy in this process. The energy difference (final minus initial) will be equal to the work required to move them at a constant speed.

$\begin{aligned}\text{Work required} &= \text{Final EPE} - \text{Initial EPE}\\&= \rm -0.279689\; J - (-1.2586\; J)\\&\approx 0.979\; J\end{aligned}$ .

8 0

3 years ago

Read 2 more answers

When should you use a piece of pipe as a leverage extension on the handle on a wrench?

Vesnalui [34]

Answer:

Explanation:

We must never use a piece of pipe as a leverage extension on the handle on a wrench.

Hence option d is correct.

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3 years ago