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babymother [125]

3 years ago

7

A student connects four AA batteries (1.5 V each) in series to light up a light bulb. The circuit has a resistance of 35 2. How

much current
flows through the circuit? (AKS 10a)
A.47 A
B. 5.83 A
C.25 A
D.17 A

1 answer:

allsm [11]3 years ago

8 0

Answer:

D

Explanation:

6/35=0.17

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When two things are the same temperature:

Rus_ich [418]

Answer:

its obviously D if all the other says "incorrect answer" XD

Explanation:

4 0

3 years ago

The only force acting on a 1.9 kg canister that is moving in an xy plane has a magnitude of 3.9 N. The canister initially has a

Hunter-Best [27]

Answer:

The work done on the canister is 15.34 J.

Explanation:

Given;

mass of canister, m = 1.9 kg

magnitude of force acting on x-y plane, F = 3.9 N

initial velocity of canister in positive x direction, $v_i$ = 3.9 m/s

final velocity of the canister in positive y direction, $v_j = 5.6 \ m/s$

The change in the kinetic energy of the canister is equal to net work done on the canister by 3.9 N.

ΔK.E = $W_{net}$

ΔK.E $= K.E_f -K.E_i$

The initial kinetic energy of the canister;

$K.E_i = \frac{1}{2} mv_i^2\\\\K.E_i = \frac{1}{2} m(\sqrt{v_i^2 +v_j^2 + v_z^2}\ )^2\\\\K.E_i = \frac{1}{2} *1.9(\sqrt{3.9^2 +0^2 + 0^2}\ )^2 = 14.45 \ J$

The final kinetic energy of the canister;

$K.E_f =\frac{1}{2} mv_j^2 \\\\K.E_f = \frac{1}{2} m(\sqrt{v_i^2 +v_j^2 + v_z^2}\ )^2\\\\K.E_f = \frac{1}{2} *1.9(\sqrt{0^2 +5.6^2 + 0^2}\ )^2 = 29.79 \ J$

ΔK.E = 29.79 J - 14.45 J

ΔK.E = $W_{net}$ = 15.34 J

Therefore, the work done on the canister is 15.34 J.

5 0

3 years ago

A defibrillator containing a 18.4 μF capacitor is used to shock the heart of a patient by holding it to the patient's chest. Jus

gregori [183]

Answer:

The energy E is 1603.008 J.

Explanation:

Given that,

Capacitor = 18.4 μF

Voltage = 13.2 kV

We need to calculate the energy

Using formula of energy

$E=W=\int{Q}\ dV$ .....(I)

We know that,

$Q=CV$

Put the value of Q in equation (I)

$E=\int{CV}dV$

On integration

$E=\dfrac{CV^2}{2}$

Put the value into the formula

$E=\dfrac{18.4\times10^{-6}\times(13.2\times10^{3})^2}{2}$

$E=1603.008\ J$

Hence, The energy E is 1603.008 J.

3 0

3 years ago

A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav

gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

$v^2=v_o^2+2gy$

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}$

With this value you can find the spring constant k from the equation (1):

$mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}$

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

$W=F_e\\\\mg=kx$

you solve the last expression for x:

$x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m$

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}$

Next, you calculate x from the equation (1):

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m$

The net fire is stretched 2.72 m

5 0

3 years ago

Find the current flowing through the following circuit

kap26 [50]

Answer:

0.5

Explanation:

calculating the total resistance for the both the parallel and the series resistors then later using the ohm's law formulae V=IR to calculate current.

4 0

3 years ago