Answer:
The weight of the water is 37.14 N.
Explanation:
Given;
density of water, ρ = 62.4 lbm/ft³
volume of water, V = 1 gallon = 0.1338 ft³
The mass of the water is calculated as;
m = ρV
[tex]m = 62.4 \ \frac{lbm}{ft^3} \ \times \ 0.1338 \ ft^3\\\\m = 8.349 \ lbm[/tex]
1 lbm = 0.454 kg
8.349 lbm = ?
= 3.79 kg
The weight of the water is calculated as;
W = mg
where;
g is acceleration due to gravity = 9.8 m/s²
W = (3.79)(9.8)
W = 37.14 N.
Therefore, the weight of the water is 37.14 N.
As an example, a 4.00-kg aluminum ball has an apparent mass of 2.10 kg when submerged in a particular liquid: calculate the density of the liquid.
Answer:
1.2825 * 10^3 kg/m³
Explanation:
Given that :
Mass of aluminum ball (m1) = 4kg
Apparent mass of ball (m2) = 2.10 kg
Density of aluminum (d1) = 2.7 * 10^3 kg/m³
Density of liquid (d2) =?
Using the relation :
d1 / d2 = m1 / (m2 - m1)
(2.7 * 10^3) / d2 = 4 / (4 - 2.10)
2700 / d2 = 4 / 1.9
4 * d2 = 2700 * 1.9
4 * d2 = 5130
d2 = 5130 / 4
d2 = 1282.5 kg/m³
Hence, density of liquid = 1.2825 * 10^3
Question 1(Multiple Choice Worth 2 points)
(06.01 LC)
How can the luster of a mineral be described? by the way this is Science
A Bland
B Dim
C Dull
D Twinkle
Unit Test
Active
11
12
13
14
15
TIME REMAININO
01:47:35
Which best describes the energy of a sound wave as it travels through a medium?
It increases
It decreases
It remains the same
It depends on the medium
Answer:
It depends on the medium is answer.
Explanation:
I hope it's helpful!
The energy of sound waves depends on the nature of medium it travels through. As the speed of the wave increases in the medium its energy increases. Hence, option d is correct.
What are sound waves ?Sound waves are type of mechanical waves passing through a medium. Sound waves are longitudinal waves hence, the oscillation of particles is in the same direction of of the wave propagation.
The energy of a wave is directly proportional to the frequency of the wave and inversely proportional to the wavelength. As the density of the medium increases, the speed of the wave decreases and thereby energy too.
Sound waves travels with higher speed through solids. As their energy increases, the waves moves with higher speed. Hence, the energy of sound waves depends on the nature of the medium.
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Acid rain is an example of which type of chemical weathering?
Given the information below, estimate the total distance travelled during these 6 seconds using a left endpoint approximation. time (sec) velocity (ft/sec) 0 30 1 54 2 56 3 34 4 8 5 2 6 22
Answer:
184 feets
Explanation:
Given the data:
time (sec) __ velocity (ft/sec)
0 __________30
1 __________ 54
2 __________56
3 __________34
4 __________ 8
5 __________ 2
6 __________22
Using left end approximation:
(0,1) ___ f(0) = 30
(1,2) ___ f(1) = 54
(2,3) ___f(2) = 56
(3,4) ___f(3) = 34
(4,5) ___f(4) = 8
(5,6) __ f(5) = 2
Hence, the Total distance traveled during the 6 second interval is:
Change ; dT = 1
1 * (30 + 54 + 56 + 34 + 8 + 2) = 184
A system consists of two particles. Particle 1 with mass 2.0 k g 2.0kg is located at ( 2.0 m , 6.0 m ) (2.0m,6.0m) and has a velocity of ( 3.1 m / s , 2.6 m / s ) (3.1m/s,2.6m/s). Particle 2 with mass 4.5 k g 4.5kg is located at ( 4.0 m , 1.0 m ) (4.0m,1.0m) and has a velocity of ( 1.1 m / s , 0.6 m / s ) (1.1m/s,0.6m/s). Determine the position and the velocity of the center of mass of the system.
Answer:
[tex]Position:(3.38m ,2.53m)[/tex]
[tex]Velocity=(1.72m/s,1.22m/s)[/tex]
Explanation:
From the question we are told that
Mass of particle 1
[tex]M_1=2.0kg[/tex]
Co-ordinate of particle 1 (2.0m,6.0m)
Velocity of Particle 1
[tex]V_1= (3.1 m / s , 2.6 m / s )[/tex]
Mass of particle 2
[tex]M_2=4.5kg[/tex]
Co-ordinate of particle 2 (4.0m,1.0m)
Velocity of Particle 1
[tex]V_2=( 1.1 m / s , 0.6 m / s )[/tex]
Generally the Position is mathematically given as
[tex]X =\frac{ ((M_1 * Cx_1) + (M_2 *Cx_2)}{(M_1 + M_2)}[/tex]
[tex]X=\frac{2*2+4.5*4}{2+4.5}[/tex]
[tex]X=3.38[/tex]
[tex]Y=\frac{(M_1* Cy_1 + M_2* Cy_2)}{(M_1 + M_2)}[/tex]
[tex]Y =\frac{(2 * 6 + 4.5 * 1)}{(2 + 4.5)}[/tex]
[tex]Y= 2.53 m[/tex]
Therefore the position is given as
[tex]Position:(3.38m ,2.53m)[/tex]
Solving for Velocity
Generally the velocity of the system is mathematically Given as
[tex]V_x =\frac{ (M_1 * vx_1 + M_2 * Vx_2)}{(M_1 + M_2)}[/tex]
[tex]V_x=\frac{ (2 * 3.1 + 4.5 *1.1)}{(2 + 4.5)}[/tex]
[tex]V_x=1.72m/s[/tex]
For Y
[tex]V_y =\frac{(M_1* vy_1 + M_2* Vy_2)}{(M_1 + M_2)}[/tex]
[tex]V_y=\frac{ (2 * 2.6) + (4.5*0.6)}{(2 + 4.5)}[/tex]
[tex]V_y=1.22m/s[/tex]
Therefore Velocity
[tex]V=(1.72m/s,1.22m/s)[/tex]
What is the easiest way to add energy to matter.
Answer:
Explanation:
Heat is probably the easiest energy you can use to change your physical state. The atoms in a liquid have more energy than the atoms in a solid.
Answer:
The easiest way is to heat matter which raises its internal energy andso the mass as well. Another possibility is to accelerate matter as in case of protons or heavy ions in the LHC, Protons ,for example , are accelerated to the energies of 6500 GeV which means increase of the rest mass by a factor of almost 7000.
A 75kg encounters 10n of friction and slides down a hall with an acceleration of 3.60 find the coefficients of kinetic between the box and floor
Answer:
Explanation:
Given,
Mass of block = 75kg
Force of friction=10N
Acceleration of box = 3.60m/s^2
Acceleration due to gravity = 9.8m/s^2
Let tetha represent the angle of inclination.
Wherefore, we have mgsinθ - force of friction = ma........ 1
Substitute the values into equation 1
75×9.8×sinθ -10 = 75×3.60
735sinθ = 10+270
735sinθ = 280
Divide both sides by 735
Sinθ = 280/735
Sinθ = 0.3809
θ = sin^-1 0.3809
θ = 22.389°
We can now solve for the coefficient of friction by considering the formula:
Fs = us×R
Where R = mgcosθ
Fs = 10
10 = Us×75×9.8×cos22.389
10 = Us× 735×0.9246
10 = Us × 679.595
Divide both sides by 679.595
10/679.595 = Us
Us = 0.01471
Hence the coefficient of friction is 0.01471
the actual ratio of weight to mass on earth is 9:8 N:1 kg what is the weight on earth of
an object whose mass is 2 kg
Answer:
The object weighs 19.6 N
Explanation:
Weight
It's the force exerted on a body by gravity.
This is often expressed in the formula
W = mg
Where
W = weight
m = mass of the object
g = gravitational acceleration
The ratio of weight to mass on Earth is 9.8 N: 1 kg. This gives us the value of [tex]g=9.8 \ m/s^2[/tex].
An object with a mass of m = 2 kg has a weight of:
[tex]W = 2\ Kg *9.8 \ m/s^2[/tex]
W = 19.6 N
The object weighs 19.6 N
Describe how electric potential energy, kinetic energy, and work change when two charges of opposite sign are placed near each other.
Answer:
As opposites attract, q is pulled to Q. And that force from Q is working on q, force over distance. Which means the potential energy q started with is being converted into kinetic energy. q is accelerating and picking up speed.
Explanation:
A helicopter is flying 8 kilometers above the ground. It has a gravitational potential energy of 376 Joules. What would the mass be?
Answer:
0.0048kg
Explanation:
Given parameters:
Height of flight = 8km = 8000m
Gravitational potential energy = 376J
Unknown:
Mass = ?
Solution:
To solve this problem;
Gravitational potential energy = mgh
m is the mass of the body
g is the acceleration due to gravity
h is the height of the helicopter
Insert the parameters and solve;
376 = m x 9.8 x 8000
376 = 78400m
m = 0.0048kg
How much heat is needed to rise the temperature by 10 oc of mass 5kg if substance of specific heat capacity 300j/kg of What is thermal of a substance
Answer: 15000 J
Explanation:
Given ;
Temperature change, dθ = 10
Mass of substance, m = 5 kg
Specific heat capacity, C = 300 j/kg°C
Quantity of heat, Q required is obtained using the relation ;
Q = mCdθ
Q = 5kg * 300 j/kg°C * 10°C
Q = 5 * 300 J * 10
Q = 15000 J
__________ energy always moves from a warmer object to a cooler object.
A) electromagnetic
B) nuclear
C) thermal
8.There are many ways to measure range of motion: goniometers, inclinometers, tape measures and even using apps on phones. In your opinion, do you think one tool is more accurate than another
Answer:
the relative error is similar in all of them, which is why they all have the same precision.
Explanation:
For this question we must be careful since there are several important concepts to take into account:
* the absolute error that is directly related to the precision or error in the mean given by the instrument
* the relative error to take into account the precision of the instrument and the magnitude of the measurement
The precisions of the different measuring instruments are
metric conta ± 0.1 cm
micrometric screw ±0.001 cm
vernier ± 0.005 cm or 0.001 cm
goniometers ± 1º
I tilt meters ± 1º
Mobile applications varies a lot, but on the order of ±0.01cm
We can see that the tape measure, the micrometric screw and the vernier are the only ones that give us a direct measurement, in the others some calculation must be made to obtain the distance reading, for which the error must also be propagated for the calculation.
We must also take into account that the vernier and micrometric screw are for short measurements only a few centimeters, the meters allow a medium to about 20 m, for measurements of more distance the other instruments are needed
If we only take into account the absolute error, the device with a smaller error is the most accurate, but this is not very correct.
The precision must be related to the magnitude of the measurement carried out, that is why the error or relative uncertainty was defined.
Let's take an example with the tape measure.
If we measure a distance of 1 cm the relative error is ±0.1, for a distance of 10 cm the relative error is ±0.01 which is very good
if we use a vernier to measure 2 cm the error is ±0.0025
if we use an inclinometer or a goniometer to measure a distance of 100 m in error it is of the order of ±0.09 m
With a cell phone it depends on the form of measurement, but all programs involve the measurement of time of a pulse of light, and assume a constant speed of light regardless of the refractive index of the medium that changes this speed. In principle this could be a very precise method, but you must know the calculation procedure and the approximations used.
As we can see, to give a correct answer we must use the relative error in this case the instruments that use the optical measurement method should be the most accurate, but the software for the calculation can involve large approximations.
Of the other instruments, the relative error is similar in all of them, which is why they all have the same precision.
A 10248 kg car is pulled by a low tow truck that has an acceleration of 2.0m/s what is the net force on the car
Answer: 20496N
Explanation:
The formula to calculate the net force will be given as:
Net force = Acceleration × Mass
Since 10248 kg car is pulled by a low tow truck that has an acceleration of 2.0m/s, the net force would be:
= 10248 × 2
= 20496N
1. To get to school, a girl walks 1 km North in 15 minutes. She
then walks 200 m South-west in 160 seconds.
What is the girl's average velocity for her walk to school?
The girl's average velocity is 1.13 m/s
The formula used to determine average velocity is:
Velocity = distance/ timeV = D/tThis means to determine the girl's average velocity is necessary to find out the total distance and the total time.
Distance:
1 km north and 200 m SouthwestAs you can see there are two different distances and they are in different units (meters vs kilometers), let's convert the kilometers to meters so both distances are in the same unit.
1 x 1000 = 1000 (1 km = 1000 meters)Add the two distances:
1000 meters + 200 meters = 1200 meters
Time
15 minutes and 160 seconds
Let's convert the minutes into seconds:
15 x 60 = 900
Add the time:
900 + 160 = 1060
Finally, calculate the average velocity:
1200 meters / 1060 seconds = 1.13 meters per secondLearn more about velocity in: https://brainly.com/question/862972
When you look at a star, where in space are you looking?
A. Into the galaxies
B. In the Milky Way Galaxy beyond Pluto
C. Towards the sun
D. At the failed star known as Jupiter
Answer:
A
Explanation:
into the galaxies of the
Answer:A
Explanation:I’m taking the test rn
Neon has 3 naturally occurring isotopes, Neon-20, Neon-21, and Neon-22. Neon's average atomic mass on the periodic table is 20.179 amu. Based on this information which isotope is most abundant? Explain why abundance of an isotope matters when calculating the average atomic mass of an isotope.
An object is released from a rocket moving upwards at a speed of 8.9 m / s.
What is the downward acceleration of this object?
Answer:
9.81 m/s²
Explanation:
It is given that the upward speed of the object is 8.9 m/s.
An object is released.
Now, we know that if the object is released then it will move under the control of gravity which means that the initial velocity of the rocket will not affect the acceleration of the object.
Hence, the downward acceleration of the object will be equal to g i.e. 9.81 m/s².
Wind is ________ rushing in to fill an area of low pressure.
Wind is the stream of air rushing in to fill an area of low pressure. Winds are generated by the differential distribution of temperature resulting in a pressure difference.
What are winds?The natural movement of air or other gases in relation to a planet's surface is referred to as wind. There are many different sizes of winds, ranging from thunderstorm flows that last only a few minutes to local breezes.
Large-scale atmospheric circulation is primarily brought on by the planet's rotation and the difference in temperature between the equator and the poles (Coriolis effect).
Thermal low circulations over topography and high plateaus can cause monsoon circulations in the tropics and subtropics. Local winds can be defined in coastal regions by the sea breeze/land breeze cycle; in regions with varying terrain, mountain and valley.
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A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5 kg and 0.5 kg, respectively. The piston diameter is 10 cm. The local atmospheric pressure is 1 bar. T
Answer:
The force acting on the shaft is 1324.75 N
Explanation:
Given that,
Cross sectional area of shaft [tex]A_{sh}=0.8\ cm^{2}[/tex]
Gas pressure [tex]P_{gas}=3\ bar=3\times10^5\ Pa[/tex]
Total mass M=24.5+0.5=25 kg
Diameter of piston, d=10 cm
We need to calculate the cross section area of piston
Using formula of area
[tex]A_{p}=\pi\times\dfrac{d^2}{4}[/tex]
Put the value into the formula
[tex]A_{p}=\pi\times\dfrac{0.1^2}{4}[/tex]
[tex]A_{p}=0.00785\ m^{2}[/tex]
We need to calculate the weight of the piston and shaft
Using formula of weight
[tex]W=mg[/tex]
Put the value into the formula
[tex]W=25\times9.81[/tex]
[tex]W=245.25\ N[/tex]
We need to calculate the force due to gas pressure
Using formula of force
[tex]F_{gas}=P_{gas}\times A_{p}[/tex]
Put the value into the formula
[tex]F_{gas}=3\times10^{5}\times0.00785[/tex]
[tex]F_{gas}=2355\ N[/tex]
We need to calculate the force due to atmospheric pressure,
Using formula of force
[tex]F_{atm}=P_{atm}\times A_{p}[/tex]
Put the value into the formula
[tex]F_{atm}=1\times10^5\times0.00785[/tex]
[tex]F_{atm}=785\ N[/tex]
We need to calculate the force acting on the shaft
from free body diagram
[tex]F_{gas}=F+F_{atm}+W[/tex]
Put the value into the formula
[tex]2355=F+785+245.25[/tex]
[tex]F=2355-785-245.25[/tex]
[tex]F=1324.75\ N[/tex]
Hence, The force acting on the shaft is 1324.75 N
The complete question is :
The figure (attached) shows a gas contained in a vertical piston-cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5kg and 0.5kg respectively. The piston diameter is 10cm. The local atmospheric pressure is 1 bar. The piston moves smoothly in the cylinder and g=9.81 m/s2.
Solution :
Given
The cross sectional area of the shaft, [tex]$A_{s} = 0.8 \ cm^2$[/tex]
Gas pressure, [tex]$P_{gas} = 3 \ bar = 3 \times 10^5 \ Pa$[/tex]
The total mass, m = 24.5 + 0.5
= 25 kg
Diameter of the piston, d = 10 cm
The cross sectional area of the piston, [tex]$A_{p} = \frac{\pi}{4} \times (0.1)^2$[/tex]
Weight of the piston and shaft, W = mg
= 25 x 9.81
= 245.25 N
Force due to the gas pressure,
[tex]$F_{gas} = P_{gas} \times A_{p}$[/tex]
[tex]$F_{gas} = 3 \times 10^5 \times 0.00785 $[/tex]
= 2355 N
Force due to atmospheric pressure,
[tex]$F_{atm} = P_{atm} \times A_{p}$[/tex]
[tex]$= 1 \times 10^5 \times 0.00785$[/tex]
= 785 N
Now [tex]$F_{gas} = F + F_{atm} + W$[/tex]
⇒ 2355 = F + 785 + 245.25
⇒ F = 1324.75 N
A 0.290 kg block on a vertical spring with a spring constant of 5.00 ✕ 103 N/m is pushed downward, compressing the spring 0.110 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?
___m
Answer:
The height at point of release is 10.20 m
Explanation:
Given:
Spring constant : K= 5 x 10 to the 3rd power n/m
compression x = 0.10 m
Mass of block m= 0.250 kg
Here spring potential energy converted into potential energy,
mgh = 1/2 kx to the 2 power
For finding at what height it rise,
0.250 x 9.8 x h = 1/2 x 5 x 10 to the 3 power x (0.10)to the 2 power) - ( g= 9.8 m/8 to the 2 power
h= 10.20
Therefore, the height at point of release is 10.20 m
how are all the spheres similar
Consider two object has mass m1=5.0kg m2=0.5kg and the specific heat c1=c2=4200j/kg oc respectively assume that first object has temperature 20oc and second object has temperature 90oc After they place these two object touch each other determine the mixture temperature of the substance
Answer: [tex]T=26.36^{\circ}C[/tex]
Explanation:
Given
mass of the first object [tex]m_1=5\ kg[/tex]
mass of the second object [tex]m_2=0.5\ kg[/tex]
specific heat of the two is [tex]c_1=c_2=4200\ J/kg/^{\circ}C[/tex]
The temperature of the first [tex]T_1=20^{\circ}[/tex]
The temperature of the second [tex]T_2=90^{\circ}[/tex]
Heat flows from high temperature to low temperature
Suppose T is the common temperature
[tex]m_1c_1(T-T_1)=m_2c_2(T_2-T)\\5\times 4200\times (T-20)=0.5\times 4200\times (90-T)\\5T-100=45-0.5T\\5.5T=145\\T=26.36^{\circ}C[/tex]
Is the force exerted by the scale on the superball greater than, less than, or equal to the force exerted by the scale on the clay
Answer: A glob of very soft clay is dropped from above onto a digital scale. The clay sticks to the scale on impact. A graph of the clay's velocity vs. time, clay (t), is given, with the upward direction defined as positive.
The experiment is then repeated, but instead of using the clay glob, a superball with identical mass is dropped from the same height onto the scale.
Both the clay and the superball hit the scale 2.9s after they are dropped. Assume that the duration of the collision is the same in both cases and the force exerted by the scale on the clay and the force exerted by the scale on the superball are constant.
Explanation:
Car A is traveling at 18.0 m/s and car B at 25.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with an acceleration of 1.80 m/s2. How long does it take car A to overtake car B
Answer:
car A reaches and immediately overtakes the car B at 22.56 s.
Explanation:
After car A accelerate at 1.8 m/s2, it travels a distance x(A) and car B will have travels a distance x(B), let's recall that the initial distance between them is 300 m, so we have:
[tex]x_{A}=300+x_{B}[/tex]
Now, we can rewrite this equation in terms of speed and time
[tex]V_{iA}t+\frac{1}{2}at^{2}=300+V_{iB}t[/tex]
Where:
V(iA) is the initial speed of car A
V(iB) is the initial speed of car B
t is the time when car A reaches the car B
a is the acceleration
[tex]18t+\frac{1}{2}1.8t^{2}=300+25t[/tex]
[tex]0.9t^{2}-7t-300=0[/tex]
Solving this quadratic equation for t, and taking just the positive value, we will have:
t=22.56 s
Therefore, car A reaches and immediately overtakes the car B at 22.56 s.
I hope it helps you!
a body is undergoing non uniform circular motion work done by tangential force on body is
Answer:
non-zero
Explanation:
The tangential force acts as a circular moment on the body and increases its velocity, and also increases its kinetic energy.
In unequal circular movement work done on the object is non-zero. If the object's velocity increases in the tangential direction, the force is acting in the same direction.
When did the object have the highest average
speed? *
5 points
Time (s)
30
Distance (m)
90
240
60
120
580
180
1080
O Between 0 and 30 seconds
Between 30 and 60 seconds
Between 60 and 120 seconds
Between 120 and 180 seconds
Answer:
Between 120 and 180 seconds
A car is traveling to the right with a speed of 2.0 m/s on an icy road when the brakes are applied. The car begins sliding with constant acceleration for 3.0 m until it comes to a stop. How long does it take the car to slide to a stop? Answer using a coordinate system where rightward is positive. Round the answer to two significant digits.
Answer:
t = 3.0s
Explanation:
U = 2.0m/s , V = 0 (stop) , S = 3m , t =?
From V^2 - U^2 = 2aS
=) a = -4/6 = -0.667m/s^2
Now again by V-U = at
We have t = -U/a = 2/0.667 = 3s
Required time is t = 3.0s
the orbit of an object is caused by what two factors *
1. The initial force of motion and an inward-acting force
2. 2 unbalanced forces
3, A push or pull on an object
The initial force of motion and inward acting force