2. What is the kinetic energy of an 80 kg football player running at 8m/s? *

what is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 3 m/sec

s344n2d4d5 [400]

Data:
m (mass) = 1 Kg
s (speed) = 3 m/s
Kinetic energy = ? (Joule)

Formula (Kinetic energy)
$E_{k} = \frac{m*s^2}{2}$

Solving:
$E_{k} = \frac{m*s^2}{2}$
$E_{k} = \frac{1*3^2}{2}$
$E_{k} = \frac{1*9}{2}$
$E_{k} = \frac{9}{2}$
$\boxed{\boxed{E_{k} = 4.5\:J}}\end{array}}\qquad\quad\checkmark$

3 0

3 years ago

Someone got this paper?

Snezhnost [94]

First question (upper left):
1/Req = 1/12 + 1/24 = 1/8
Req = 8 ohms
Voltage is equal through different resistors, and V1 = V2 = 24 V.
Current varies through parallel resistors: I1 = V1/R1 = 24/12 = 2 A. I2 = 24/24 = 1 A.

Second question (middle left):
V1 = V2 = 6 V (parallel circuits)
I1 = 2 A, I2 = 1 A, IT = 2+1 = 3 A.
R1 = V1/I1 = 6/2 = 3 ohms, R2 = 6/1 = 6 ohms, 1/Req = 1/2 + 1/1, Req = 2/3 ohms

Third question (bottom left):
V1 = V2 = 12 V
IT = 3 A, meaning Req = V/It = 12 V/3 A = 4 ohms
1/Req = 1/R1 + 1/R2, 1/4 = 1/12 + 1/R2, R2 = 6 ohms
I1 = V/R1 = 1 A, I2 = V/R2 = 2 A

Fourth question (top right):
1/Req = 1/20 + 1/20, Req = 10 ohms
IT = 4 A, so VT = IT(Req) = 4*10 = 40 V
Parallel circuits, so V1 = V2 = VT = 40 V
Since the resistors are identical, the current is split evenly between both: I1 = I2 = IT/2 = 2 A.

Fifth question (middle right):
1/Req = 1/5 + 1/20 + 1/4, Req = 2 ohms
IT = VT/Req = 40 V/2 ohms = 20 A
V1 = V2 = V3 = 40 V
The current of 20 A will be divided proportionally according to the resistances of 5, 20, and 4, the factors will be 5/(5+20+4), 20/(5+20+4), and 4/(5+20+4), which are 5/29, 20/29, and 4/29.
I1 = 20(5/29) = 100/29 A
I2 = 20(20/29) = 400/29 A
I3 = 20(4/29) = 80/29 A

Sixth question (bottom right):
V2 = 30V is given, but since these are parallel circuits, V1 = VT = 30 V.
Then I1 = V1/R1 = 30 V/10 ohms = 3 A.
I2 = 30 V/15 ohms = 2 A.
IT = 3 + 2 = 5 A
1/Req = 1/10 + 1/15, Req = 6 ohms

6 0

3 years ago

A long copper rod of diameter 2.0 cm is initially at a uniform temperature of 100°C. It is now exposed to an air stream at 20°C

Lisa [10]

Answer:

t = 4.0 min

Explanation:

given data:

diameter of rod = 2 cm

T_1 = 100 degree celcius

Air stream temperature = 20 degree celcius

heat transfer coefficient = 200 W/m2. K

WE KNOW THAT

copper thermal conductivity = k = 401 W/m °C

copper specific heat Cp = 385 J/kg.°C

density of copper = 8933 kg/m3

charateristic length is given as Lc

$Lc = \frac{V}{A_s}$

$Lc = \frac{\frac{\pi D^2}{4} L}{\pi DL}$

$Lc = \frac{D}{4}$

$Lc = \frac{0.02}{6} = 0.005 m$

Biot number is given as $Bi = \frac{hLc}{k}$

$Bi = \frac{200*0.005}{401}$

Bi = 0.0025

As Bi is greater than 0.1 therefore lumped system analysis is applicable

so we have

$\frac{T(t) - T_∞}{Ti - T_∞} = e^{-bt}$ ............1

where b is given as

$b = \frac{ hA}{\rho Cp V}$

$b = \frac{ h}{\rho Cp Lc}$

$b = \frac{200}{8933*385*0.005}$

b = 0.01163 s^{-1}

putting value in equation 1

$\frac{25-20}{100-20} = e^{-0.01163t}$

solving for t we get

t = 4.0 min

6 0

3 years ago

A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr

const2013 [10]

Answer:

$388 cm^3$

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

$pV=const.$