The absence of any mechanical linkage between the throttle pedal and the throttle body requires the use of a ______

The circuit below represents four resistors connected to a 12-volt source. What is the total current in the circuit? 4.0Ω 6.0Ω 1

Tpy6a [65]

Answer:

For series connected resistors I = 0.5A

For parallel connected resistors I = 8.5A

Explanation:

Since the diagram is not available, our solution will be divided into two;

According to ohm's law which states that "the current passing through a,metallic conductor at constant temperature is directly proportional to the potential difference across its ends. Mathematically, E = IRt where;

E is the electromotive force

I is the total current in the circuit

Rt is total equivalent resistance.

Where E = 12volts

Rt can be gotten depending on the arrangements of the resistors which can either be in series or parallel.

If the resistors are in series, their equivalent resistance gives;

Rt = 4.0Ω+6Ω+8Ω+6Ω

Rt = 24Ω

The current I will be;

I = E/Rt = 12/24

I = 0.5A

If the connection is in series, the total current in the circuit will be 0.5A.

For resistance in parallel;

1/Rt = 1/4Ω+1/6Ω+1/8Ω+1/6Ω

1/Rt = 6+4+3+4/24

1/Rt = 17/24

Rt = 24/17Ω

I = E/Rt

I = 12/(24/17)

I = 12×17/24

I = 8.5A

If the connection is in parallel, the total current in the circuit will be 8.5A

8 0

2 years ago

Read 2 more answers

At what time t is the turtle second time a distance of 10.0 cm from its starting point?

skad [1K]

Answer:

10 cm.

Explanation:

5 0

2 years ago

A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i

victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

$F= \frac{kqQ}{r^2}$

Where,

$k = 8.99 * 10^9 N m^2 / C^2.$

q = charges of the objects

r= distance/radius

Our values are previously given, so

$q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59$

Replacing,

$F=\frac{kqQ}{r^2}$

$F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}$

$F= 4.8423N$

The force acting on the block are given by,

$F-mgsin\theta = ma$

$a = \frac{F-mgsin\theta}{m}$

$a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}$ $a = 10.31m/s^2$

Therefore the box is accelerated upward.

3 0

2 years ago

In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.05 kg. The masses of the pulley and string are negligible by comparis

Rus_ich [418]

M1 descending
−m1g + T = m1a

m2 ascending
m2g − T = m2a

this gives :
(m2 − m1)g = (m1 + m2)a

a = (m2 − m1)g/m1 + m2
= (5.60 − 2)/(2 + 5.60) x 9.81
= = 4.65m/s^2

5 0

2 years ago

You and your friend throw balloons filled with water from the roof of a several story apartment house. You simply drop a balloon

Aleks [24]

Answer:

Height = 53.361 m

Explanation:

There are two balloons being thrown down, one with initial speed (u1) = 0 and the other with initial speed (u2) = 43.12

From the given information we make the following summary

$u_{1}$ = 0m/s