Which quantity is a scalar quantity?

Wheat ground into flour is an example of a ___ change?

EleoNora [17]

Oml... Its physical... Unless if your turning that wheat into bread by using fire it would be chemical.

Yeeeeeeeetus

6 0

3 years ago

A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 13° above the horizontal. (a) if

Blababa [14]

Answer: Therefore, x component: Tcos(24Â°) - f = 0 y component: N + Tsin(24Â°) - mg = 0 The two equations I get from this are: f = Tcos(24Â°) N = mg - Tsin(24Â°) In order for the crate to move, the friction force has to be greater than the normal force multiplied by the static coefficient, so... Tcos(24Â°) = 0.47 * (mg - Tsin(24Â°)) From all that I can get the equation I need for the tension, which, after some algebraic manipulation, yields: T = (mg * static coefficient) / (cos(24Â°) + sin(24Â°) * static coefficient) Then plugging in the values... T = 283.52. Reference https://www.physicsforums.com/threads/difficulty-with-force-problems-involving-friction.111768/

7 0

3 years ago

Legacy issues $570,000 of 8.5%, four-year bonds dated January 1, 2019, that pay interest semiannually on June 30 and December 31

choli [55]

Answer:

1) Determine the total bond interest expense to be recognized.

Total bond interest expense over life of bonds:

Amount repaid:

8 payments of $24,225: $193,800

Par value at maturity: $570,000

Total repaid: $763800 (193,800 + 570,000)

Less amount borrowed: $508050

Total bond interest expense: $255750 (763800 - 508,050)

2)Prepare a straight-line amortization table for the bonds' first two years.

Semiannual Interest Period End; Unamortized Discount; Carrying Value

01/01/2019 61,950 508,050

06/30/2019 54,206 515,794

12/31/2019 46,462 523,538

06/30/2020 38,718 531,282

12/31/2020 30,974 539,026

3) Record the interest payment and amortization on June 30:

June 30 Bond interest expense, dr 31969

Discount on bonds payable, Cr (61950/8) 7743.75

Cash, Cr ( 570000*8.5%/2) 24225

4) Record the interest payment and amortization on December 31:

Dec 31 Bond interest expense, Dr 31969

Discount on bonds payable, Cr 7744

Cash, Cr 24225

6 0

3 years ago

A high voltage transmission line with a resistance of 0.51 Ω/km carries a current of 1099 A. The line is at a potential of 1300

Illusion [34]

<h2>Answer:</h2>

$\boxed{P=96.09MW}$

<h2>Explanation:</h2>

First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:

$R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega$

So we can calculate the power loss as:

$P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}$

Finally, the power loss due to resistance in the line is 96.09MW

5 0

3 years ago

A person with mass mp = 75 kg stands on a spinning platform disk with a radius of R = 1.59 m and mass md = 186 kg. The disk is i

Ad libitum [116K]

Answer:

1. $I_0=424.7208\,kg.m^2$

2. $I_f=256.1808\,kg.m^2$

3. $\omega_f=3.3158\,rad.s^{-1}$

4. $\Delta KE=558.8432\,J$

5. $a_c=5.8271\,m.s^{-2}$

6. $\omega_0=2\,rad.s^{-1}$

Explanation:

Given:

mass of the person, $m_p=75\,kg$

radius of the disk, $r=1.59\,m$

mass of the disk, $m_d=186\,kg$

initial angular speed of the disk, $\omega_0=2\,rad.s^{-1}$

distance of the person from the center of the disk, $d=0.53\,m$

1.

Initial moment of inertia of the system when the man stands at the rim of disk:

Moment of inertia of the disc:

$I_D=\frac{1}{2} .m_d.R^2$

$I_D=\frac{1}{2} \times 186\times 1.59^2$

$I_D=235.1133\,kg.m^2$

Now for the person, we treat the mass to be a point revolving around R:

$I_P=m_p.R^2$

$I_P=75\times 1.59^2$

$I_P=189.6075\,kg.m^2$

∴We have the moment of inertia of the system in this case as:

$I_0=I_D+I_P$

$I_0=235.1133+189.6075$

$I_0=424.7208\,kg.m^2$

2.

Moment of inertia when the person stands at 0.53 m from the center of the disk:

Moment of inertia of the disk will be constant:

$I_D=235.1133\,kg.m^2$

For the person, we treat the mass to be a point revolving around radius 0.53 m:

$I_P=75\times 0.53^2$

$I_P=21.0675\,kg.m^2$

∴We have the moment of inertia of the system

$I_f=I_D+I_P$

$I_f=235.1133+21.0675$

$I_f=256.1808\,kg.m^2$

3.

The final angular velocity of the disk:

Using the conservation of angular momentum:

$I_0.\omega_0=I_f.\omega_f$

$424.7208\times 2=256.1808\times \omega_f$

$\omega_f=3.3158\,rad.s^{-1}$

4.

Change in Kinetic Energy:

∵ $KE=\frac{1}{2} I.\omega^2$

∴ $\Delta KE= \frac{1}{2} (I_f.\omega_f^2-I_0.\omega_0^2)$

$\Delta KE=\frac{1}{2} (256.1808\times 3.3158^2-424.7208\times 2^2)$

$\Delta KE=558.8432\,J$

5.

Centripetal acceleration of the person when she is at R/3:

Centripetal acceleration is given as:

$a_c=r'.\omega^2$

we have ω=3.3158 radian per second at R=0.53 m

$a_c=\frac{R}{3} .\omega^2$

$a_c=0.53\times 3.3158^2$

$a_c=5.8271\,m.s^{-2}$

6.

If the person now walks back to the rim of the disk:

Then by the law of conservation of angular momentum the initial angular speed of $\omega_0=2\,rad.s^{-1}$ will be restored.

8 0

3 years ago