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2 years ago

7

An 8 Ω resistor is connected to a 6 V battery and another resistor in series. If the current in the circuit is 0.5 A, what is th

e resistance of the other resistor?

1 answer:

Semmy [17]2 years ago

4 0

8+R=V/I
8+R=6/0.5=12 ohm
then R=4 ohm

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The name of C (S) + o2 (g) CO2 (g)

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Answer:

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2 years ago

Definition: The force applied when using a simple machine.

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the force applied when using a simple machine is called the effort force

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The main purpose of an air bag is to stop a passenger during a car accident in a greater amount of time than if the air bag were

Simora [160]

Answer:

a) 45571 N

b) 22786 N

c) 4557 N

Explanation:

Since the goal of the airbag is helping the person to stop after the collision in a greater time, this means that the change in momentum must finish when this is just zero.
In other words, the change in momentum, must be equal to the initial one, but with opposite sign.

$\Delta p = - p_{o} = -m*v = -55 kg*29m/s = -1595 kgm/s (1)$

Now, just applying the original form of Newton's 2nd Law, we know that this change in momentum must be equal to the impulse needed to stop the person:

$\Delta p = F* \Delta t (2)$

So, as we know the magnitude of Δp from (1) and we have different Δt as givens, we can get the different values of F (in magnitude) required to stop the person for each one of them, as follows:

$F_{1} = \frac{\Delta p}{\Delta t_{1}} = \frac{1595kgm/s}{0.035s} = 45571 N (3)$

$F_{2} = \frac{\Delta p}{\Delta t_{2}} = \frac{1595kgm/s}{0.07s} = 22786 N (4)$

$F_{3} = \frac{\Delta p}{\Delta t_{3}} = \frac{1595kgm/s}{0.35s} = 4557 N (5)$

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3 years ago

A force that is doing work on a ball when that ball is falling through the air.

Ronch [10]

Answer:

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2 years ago

A neutron consists of one "up" quark of charge +2e/3 and two "down" quarks each having charge -e/3. If we assume that the down q

Anon25 [30]

Answer:

The magnitude of the electrostatic force is 120.85 N

Explanation:

We can use Coulomb's law to find the electrostatic force between the down quarks.

In scalar form, Coulomb's law states that for charges $q_1$ and $q_2$ separated by a distance d, the magnitude of the electrostatic force F between them is:

$F = k \frac{|q_1q_2|}{d^2}$

where $k$ is Coulomb's constant.

Taking the values:

$d = 4.6 \ 10^{-15} m$

$q_1 = q_2 = - \frac{e}{3} = - \frac{1.6 \ 10^{-19} \ C}{3}$

and knowing the value of the Coulomb's constant:

$k = 8.99 \ 10 ^{9} \frac{N m^2}{C^2}$

Taking all this in consideration:

$F = 8.99 \ 10 ^{9} \frac{N m^2}{C^2} \frac{ (- \frac{1.6 \ 10^{-19} \ C}{3} ) ^2}{(4.6 \ 10^{-15} m)^2}$

$F = 120.85 \ N$

8 0

3 years ago