Answer:
The mass of the apples in the box are 1145.95 lb.
Explanation:
Given that,
Width = 3.5 ft
length = 3.5 ft
Depth = 2.75 feet
Density = 49.3 lb/ft³
We need to calculate the volume
Using formula of volume
![V=l\times h\times b](https://tex.z-dn.net/?f=V%3Dl%5Ctimes%20h%5Ctimes%20b)
Put the value
![V=3.5\times3.5\times2.75](https://tex.z-dn.net/?f=V%3D3.5%5Ctimes3.5%5Ctimes2.75)
![V=33.6875\ ft^3](https://tex.z-dn.net/?f=V%3D33.6875%5C%20ft%5E3)
We need to calculate the mass
Using formula of density
![\rho=\dfraxc{m}{V}](https://tex.z-dn.net/?f=%5Crho%3D%5Cdfraxc%7Bm%7D%7BV%7D)
![m=\rho\timesV](https://tex.z-dn.net/?f=m%3D%5Crho%5CtimesV)
Put the value into the formula
![m=49.3\times33.6875](https://tex.z-dn.net/?f=m%3D49.3%5Ctimes33.6875)
![m=1660.79\ lb](https://tex.z-dn.net/?f=m%3D1660.79%5C%20lb)
But porosity is 31 % so apples is 69% mass of total container.
So, Total mass in pounds is
![M=\dfrac{69}{100}\times1660.79](https://tex.z-dn.net/?f=M%3D%5Cdfrac%7B69%7D%7B100%7D%5Ctimes1660.79)
![M=1145.95\ lb](https://tex.z-dn.net/?f=M%3D1145.95%5C%20lb)
Hence, The mass of the apples in the box are 1145.95 lb.
An electric field is an electric property associated with each point in space when charge is present in any form.
Explanation:
w = E
efficiency = (Woutput /Winput) x 100%= Eo/Ei x100%
= (2021 j / 7.4822 kj) x 100%
= 27.01% == B
The speed of the car is 60 m/s
Explanation:
The speed of an object is a scalar quantity telling "how fast" the object is moving, regardless of its direction. It can be calculated as follows:
![speed=\frac{d}{t}](https://tex.z-dn.net/?f=speed%3D%5Cfrac%7Bd%7D%7Bt%7D)
where
d is the distance covered
t is the time taken
For the car in this problem, we have
is the distance covered
is the time taken
Subsituting into the equation, we find the speed:
![speed=\frac{1.08\cdot 10^5}{1800}=60 m/s](https://tex.z-dn.net/?f=speed%3D%5Cfrac%7B1.08%5Ccdot%2010%5E5%7D%7B1800%7D%3D60%20m%2Fs)
Learn more about speed:
brainly.com/question/8893949
#LearnwithBrainly
Given:
density of air at inlet, ![\rho_{a} = 1.20 kg/m_{3}](https://tex.z-dn.net/?f=%5Crho_%7Ba%7D%20%3D%201.20%20kg%2Fm_%7B3%7D)
density of air at inlet, ![\rho_{b} = 1.05 kg/m_{3}](https://tex.z-dn.net/?f=%5Crho_%7Bb%7D%20%3D%201.05%20kg%2Fm_%7B3%7D)
Solution:
Now,
![\dot{m} = \dot{m_{a}} = \dot{m_{b}}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%20%3D%20%5Cdot%7Bm_%7Ba%7D%7D%20%3D%20%5Cdot%7Bm_%7Bb%7D%7D)
(1)
where
A = Area of cross section
= velocity of air at inlet
= velocity of air at outlet
Now, using eqn (1), we get:
![\frac{v_{b}}{v_{a}} = \frac{\rho_{a}}{\rho_{b}}](https://tex.z-dn.net/?f=%5Cfrac%7Bv_%7Bb%7D%7D%7Bv_%7Ba%7D%7D%20%3D%20%5Cfrac%7B%5Crho_%7Ba%7D%7D%7B%5Crho_%7Bb%7D%7D)
= 1.14
% increase in velocity =
=114%
which is 14% more
Therefore % increase in velocity is 14%