Convection is the type of heat transfer that occurs in which of the following?

Someone please help ! this is worth 40 points!!

makvit [3.9K]

Explanation:

non uniform motion
F=ma, 1=m(1), m=1Kg

5 0

2 years ago

A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H

Aleks04 [339]

Answer:

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)² = 40,218.75 joules

Final KE = (1/2) (1430 kg) (11.0 m/s)² = 86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec = 4,978.1 watts .

Explanation:

Dont report my answer please

3 0

2 years ago

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v remain same at both side. so we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let final pressure is P and temp $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide 2 equation by 3rd equation

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, temperature on left side is 1.48 times the temperature on right

6 0

2 years ago

Sherry and Ellen are talking about the factors that affect light scattering in the atmosphere. Sherry says one factor is the len

Delvig [45]

Sherry who says one factor is the length of the path of sunlight is correct.

<h3>Factors affecting light scattering</h3>

There are two main factors which affects light scattering, and they include the following;

the size of the particles
wavelength of the light

length of the path of sunlight is equivalent to wavelength of the light.

Thus, we can conclude that Sherry who says one factor is the length of the path of sunlight is correct.

Learn more about light scattering here: brainly.com/question/1381101

#SPJ1

3 0

1 year ago

Why is dose equvialent in rem or sv used to define the Effective Dose Equivalent and the Committed Dose Equivalent?

Komok [63]

Good question next question

8 0

3 years ago