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3 years ago

9

A loop circuit has a resistance of R1 and a current of 2.2 A. The current is reduced to 1.6 A when an additional 2.8 Ω resistor

is added in series with R1. What is the value of R1? Assume the internal resistance of the source of emf is zero. Answer in units of Ω

1 answer:

kvasek [131]3 years ago

7 0

Answer:

Value of $R_1$ is 7.466 ohm

Explanation:

Let emf of the battery is e

In first case when only $R_1$ is there current through battery is 2.2 A

So $2.2=\frac{e}{R_1}$

$e=2.2R_1$ ......eqn 1

In second case current is reduced to 1.6 A when additional resistance of 2.8 ohm is added in series

So $1.6=\frac{e}{R_1+2.8}$

$e=1.6R_1+4.48$ ....eqn 2

Fro eqn 1 and eqn 2

$2.2R_1=1.6R_1+4.48$

$0.6R_1=4.48$

$R_1=7.466ohm$

So value of $R_1$ is 7.466 ohm

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When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____

valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

$D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''$

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

$D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26$

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

3 0

4 years ago

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

kati45 [8]

Answer:

The coefficient of kinetic friction $\mu_k = 0.724$

Explanation:

From the question we are told that

The length of the lane is $l = 36.0 \ m$

The speed of the truck is $v = 22.6\ m/s$

Generally from the work-energy theorem we have that

$\Delta KE = N * \mu_k * l$

Here N is the normal force acting on the truck which is mathematically represented as

$\Delta KE$ is the change in kinetic energy which is mathematically represented as

$\Delta KE = \frac{1}{2} * m * v^2$

=> $\Delta KE = 0.5 * m * 22.6^2$

=> $\Delta KE = 255.38m$

$255.38m = m * 9.8 * \mu_k * 36.0$

=> $255.38 = 352.8 * \mu_k$

=> $\mu_k = 0.724$

6 0

3 years ago

Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

3 0

3 years ago

A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the

jasenka [17]

Answer:

1.8 cm

Explanation:

$m$ = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

$q$ = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

$\Delta V$ =Potential difference through which the ion is accelerated = 215 V

$v$ = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

$(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}$

$r$ = Radius of the path followed by ion

$B$ = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

$qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm$

5 0

3 years ago

A ball falls from the top of a building. As it falls, its speed increases. Which type of energy is the ball gaining as it falls?

Elina [12.6K]

<h2>Hello!</h2>

The answer is: B. Kinetic energy

<h2>Why?</h2>

Since the ball is falling, speed increases because the gravity acceleration is acting. When speed increases, the kinetic energy increases too, so the ball is gaining kinetic energy.

The gravity acceleration is equal to $9.81\frac{m}{s^{2}}$ , it means that when falling, the ball will increase it's speed 9.81m every second.

We can calculate the kinetic energy by using the following formula:

$KE=\frac{1}{2}*m*v^{2}$

Where:

$m=mass\\v=velocity$

Have a nice day!

<h2 />

5 0

3 years ago