Answer:
0.68 m
Explanation:
Since the volume of water is 310 L, and we know that 1 litre = 1 dm³. So the volume of water is V = 310 dm³. Since this volume of water is the volume of water the cubical steel tank can contain, it equals the volume of the cubical steel tank.
We know that the volume of the cubical steel tank V = L³ where L is the length of side of the cube on the inside.
So the length of side of the cube L = ∛V = ∛310 dm³ = 6.77 dm = 6.77 dm × 1m/10 dm = 0.677 m ≅ 0.68 m
So, the smallest possible inside length of the tank is 0.68 m
You are driving your car at a speed of 19.0 m/s and you hit the brakes. The car accelerates at -3.50 m/s2. (a) How long does it take the car to cover 10.0 m? (b) What is the final velocity of the car?
Recall the formulas,
[tex]x_f=x_i+v_it+\dfrac12at^2[/tex]
[tex]v_f=v_i+at[/tex]
(a)
[tex]10.0\,\mathrm m=\left(19.0\dfrac{\rm m}{\rm s}\right)+\dfrac12\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
[tex]\implies \boxed{t\approx0.555\,\mathrm s}[/tex]
(b)
[tex]v_f=19.0\dfrac{\rm m}{\rm s}+\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)(0.555\,\mathrm s)[/tex]
[tex]\implies \boxed{v_f\approx17.1\dfrac{\rm m}{\rm s}}[/tex]
Two pounds of water vapor at 30 psia fill the 4-ft3 left chamber of a partitioned system. The right chamber has twice the volume of the left and is initially evacuated. Determine the pressure of the water after the partition has been removed and enough heat has been transferred so that the temperature of the water is 40oF.
Answer:
3.38atm
Explanation:
Using data from the steam table we have that
Moles of water vapour = 907.19 / 18
= 50.4 moles
So
p1 = 30 psi = 30 x 0.68 = 2.04 atm
v1 = 4ft³= 113.2 L
Then from
PV= nRT
Then to find T we use
T1 = p1 V1 / n R
= 2.04 x 113.2 / 50.4 x 0.0821
= 55.8 K
Then to find volume two
v2 = 2v1 + v1
So
3 v1 = 339.6 K
The pressure two we use
P2 = n R T2 / V2
= 50.4 x 0.0821 x 277.6 / 339.6
So we have
= 3.38 atm =
Put Newton’s 1st Law in your own words
Answer:Newton’s law also states that larger bodies with heavier masses exert more gravitational pull, which is why those who walked on the much smaller moon experienced a sense of weightlessness, as it had a smaller gravitational pull. To help explain his theories of gravity and motion, Newton helped create a new, specialized form of mathematics.
Explanation:Here I dont know
2 differences between calorimeter and thermometer ?
Answer:
Calorimeter is used to measure heat in and represents that in units of joules per kelvin units J/˚C or kJ/K
A calorimeter is can be used to measure the amount of heat released or involved in a chemical reaction.
Whereas thermometer can only measures temperature or hotness of a substance. It cannot be used to measure the thermal rate or amount of heat energy of a reaction. Unit measurement used by thermometer is Celsius (°C).
Explanation:
Assuming it is a van der Waals gas, calculate the critical temperature, pressure and volume for CO2. (a = 3.610 atm L2 mol-2, b = 0.0429 L mol-1)
pc = ___ atm
Tc = ___ K
Vc = ___ L/mol
Answer
To get critical pressure
We use
Pc = a/(27b²)
So
= 3.610/(27 X 0.0429²)
We have
= 72.7 atm
Critical temperaturewe
We use
Tc = 8a/27Rb
= 8 x 3.610/(27 x 0.0812 x 0.0429)
= 307 K
Critical volume
We use
Vc =3b =
3 x 0.0429
= 0.129L/mol
2. If a cyclist in the Tour de France traveled southwest a distance of 12,250 meters in one hour, what would the velocity of the cyclist be?
12,25 km/h
≈ 3,4 m/s
Explanation:v = d/t
= 12250m/h
= 12,25km/h
or
v = d/t
= 12250m/h
1h = 60m×60s = 3600s
= 12250m/3600s
≈ 3,4 m/s
Finally, consider the expression (6.67 x 10^-11)(5.97 x 10^24)/(6.38 x 10^6)^2 Determine the values of a and k when the value of this expression is written in scientific rotation. Enter a and k, separated by commas.
Explanation:
We need to find the value of following expression :
[tex]\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.38 \times 10^6)^2}[/tex]
Firstly, solving the numerator of the above expression :
[tex]=\dfrac{39.8199\times 10^{-11+24}}{40.7044\times 10^{12}}\\\\=\dfrac{39.8199\times 10^{13}}{40.7044\times 10^{12}}\\\\=9.7827[/tex]
Rounding off the result = 9.78
In scientific notation : [tex]9.78\times 10^0[/tex]
The value of a = 9.78 and k = 0.
What are the benefits of testing your heart rate during exercise
Answer: Heart rate provides an objective measurement of how hard your body is working. The higher the exercise intensity, the higher your heart rate will be.
Explanation:
Answer:
You're getting exercise and training your heart!
Explanation:
(7) A 2500 lbm car moving at 25 mi/hr is accelerated at a constant rate of 15 ft/s2 up to a speed of 50 mi/hr. What is the force and total time required?
Answer:
The force is [tex]F = 1164.6\ lbf[/tex]
The time is [tex]\Delta t = 2.44 \ s[/tex]
Explanation:
From the question we are told that
The mass of the car is [tex]m = 2500 \ lbm[/tex]
The initial velocity of the car is [tex]u = 25 \ mi/hr[/tex]
The final velocity of the car is [tex]v = 50 \ mi/hr[/tex]
The acceleration is [tex]a = 15 ft/s^2 = \frac{15 * 3600^2}{ 5280} = 36818.2 \ mi/h^2[/tex]
Generally the acceleration is mathematically represented as
[tex]a = \frac{v-u}{\Delta t}[/tex]
=> [tex]36818.2 = \frac{50 - 25 }{ \Delta t}[/tex]
=> [tex]t = 0.000679 \ hr[/tex]
converting to seconds
[tex]\Delta t = 0.0000679 * 3600[/tex]
=> [tex]\Delta t = 2.44 \ s[/tex]
Generally the force is mathematically represented as
[tex]F = m * a[/tex]
=> [tex]F = 2500 * 15[/tex]
=> [tex]F = 37500 \ \frac{lbm * ft}{s^2}[/tex]
Now converting to foot-pound-second we have
[tex]F = \frac{37500}{32.2}[/tex]
=> [tex]F = 1164.6\ lbf[/tex]
Which scientist determined that electrons had predicted zones orbiting the nucleus?
Answer:
Rutherford
Explanation:
Because
Schrödinger I just took the unit review
if a cart goes around a turn at 20 km/h ,what remains constant
1.position
2.velocity
3.direction
4.speed
Answer: 4.speed
Explanation:
In this case, we know that the cart remains at a constant 20km/h.
Now, one could say that "the velocity remains constant, because it always is 20km/h"
But remember that velocity is a vector, so this has a direction, and if the cart is going around a turn, then the direction of motion is changing, which tell us that there is acceleration.
But the module of the velocity, the speed, remains constant at 20km/h.
Then the correct option is 4, speed.
Difference between calorimeter and thermometer ?
Answer:
A calorimeter is can be used to measure the amount of heat released or involved in a chemical reaction. Whereas thermometer can only measures temperature or hotness of a substance. It cannot be used to measure the thermal rate or amount of heat energy of a reaction.
Explanation:
4. Lead has a density of 11.5g/cmº. A rectangular block of lead measures 7cm x5cmx2cm.
a) Find the volume of the block of lead.
b) Find the mass of the block of lead
Answer:
(a) 70cm³
(b) 805 grams
Explanation:
(a) V = L×B×H
= 7cm×5cm×2cm
= 35cm×2cm
= 70cm³
(b) Mass = Volume × Density
= 70cm³ × 11.5g/cm³
= 805 grams
What quantity do units represent in a value?
A. size
Question:
What quantity do units represent in a value? A. Size B. Direction C. Magnitude D. Dimension
Answer:
D. Dimension
Explanation:
A unit is simply the measurement given to a dimension. In other words, units represent dimensions. For example, consider a ruler that is 20m long.
i. The value is 20m.
ii. The magnitude of the measurement is 20
iii. The unit of the measurement is meters(m) and that essentially represents the dimension (length) of the ruler.
Consider two balls in motion at the same time. Joe drops the first ball from rest at height h. Directly below Joe, on the ground, Hayley simultaneously tosses a second ball upward with speed v0.a. If the two balls collide at the moment the second ball is instantaneously at rest, what is the height of the collision?b. What is the relative speed of the balls when they collide?
Answer:
A. To find height of collision let's find
Speed of collision first
Using
Vf= √2g(h-g)
But for Harvey's ball we have
V²-2gy= 0
y=( v²/2g) so this is height of collision
B.
To find relative speed of the ball
Using V at y to find Vsr
Vs= √2g(h-v²/2g)
= √2gh- v²
A car going initially with a velocity 13.5 m/s accelerates at a rate of 1.9 m/s 2 for 6.2 s. It then accelerates at a rate of -1.2 m/s2 until it stops. a) Find the car’s maximum speed b) Find the total time from the start of the first acceleration until the car is stopped c) What’s the total distance the car travels?
(a) For the first 6.2 s, the car has velocity at time t given by
[tex]v(t)=13.5\dfrac{\rm m}{\rm s}+\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]
so that after 6.2 s, it attains a velocity of
[tex]v(6.2\,\mathrm s)=25.28\dfrac{\rm m}{\rm s}[/tex]
For any time t after 6.2 s, its velocity is given by
[tex]v(t)=25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]
which tells us the velocity only falls from this point onward. This means the maximum speed is 25.28 m/s, or about 25.3 m/s.
(b) Solve for t (after 6.2 s) that makes v(t) = 0 :
[tex]25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t=0[/tex]
[tex]\implies t\approx21.067\,\mathrm s[/tex]
It takes the car about 21.2 s to come to a rest, so the car travels a total of about 6.2 s + 21.2 s = 27.4 s.
(c) For the first 6.2 s, the car undergoes a displacement at time t of
[tex]x(t)=\left(13.5\dfrac{\rm m}{\rm s}\right)(6.2\,\mathrm s)+\dfrac12\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)(6.2\,\mathrm s)^2[/tex]
[tex]\implies x\approx120.218\,\mathrm m[/tex]
For time t beyond 6.2 s, its displacement is
[tex]x(t)=120.218\,\mathrm m+\left(25.28\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
The car comes to a rest after 21.2 s (accelerating at a rate of -1.2 m/s^2), so that its total displacement is
[tex]x(21.2\,\mathrm s)\approx386.49\,\mathrm m[/tex]
so the car travels a total distance of about 387 m.
A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting a low point eight more times in 38.0 s. After many oscillations, he finally comes to rest 25.0 m below the level of the bridge. Calculate the spring stiffness constant and the unstretched length of the bungee cord.
Explanation:
It is given that,
Mass of a bungee jumper is 65 kg
The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is
[tex]T=\dfrac{38}{8}\\\\T=4.75\ s[/tex]
After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.
For an oscillating object, the time period is given by :
[tex]T=2\pi \sqrt{\dfrac{m}{k}}[/tex]
k = spring stiffness constant
So,
[tex]k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m[/tex]
When the cord is in air,
mg=kx
x = the extension in the cord
[tex]x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m[/tex]
So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m
The spring stiffness constant is 116.7 N/m and the the unstretched length of the bungee cord is 19.54 m.
The given parameters;
mass of the bungee jumper, m = 65 kgtime of motion, t = 38 sdistance to come to rest, d = 25 mThe period of oscillation of the bungee jumper is calculated as follows;
[tex]T = \frac{t}{n} \\\\T = \frac{38}{8} \\\\T = 4.75 \ s[/tex]
The spring stiffness constant is calculated as follows;
[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\\sqrt{\frac{m}{k} } = \frac{T}{2\pi} \\\\k = m \times \frac{T^2}{4\pi^2} \\\\k = 65 \times \frac{(4.75)^2}{4\pi ^2} \\\\k = 116.7 \ N/m[/tex]
The extension of the cord is calculated as follows;
[tex]F = kx\\\\mg = kx\\\\x = \frac{mg}{k} \\\\x = \frac{65 \times 9.8}{116.7} \\\\x = 5.46 \ m[/tex]
The unstretched length of the bungee cord is calculated as;
[tex]\Delta x = l_2-l_1\\\\l_1 = l_2 - \Delta x\\\\l_1 = 25 - 5.46\\\\l_1 = 19.54 \ m[/tex]
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Which statement best defines inertia? An object's motion is unaffected by any external forces acting upon it. An object responds to a force by tending to move in the direction of that force. An object opposes any motion, naturally returning to a state of rest on its own. An object opposes any change in its velocity, either to its direction or to its speed.
Answer:
An object opposes any change in its velocity, either to its direction or to its speed.
Explanation:
Edmentum Answer
A ball with a mass of 3.7 kg is thrown downward with an initial velocity of 8 m/s from a high building. How fast will it be moving after 3 seconds?
Answer:
v=37.4 m/s
Explanation:
It is given that,
Mass of a ball, m = 3.7 kg
Initial velocity of the ball is u = 8 m/s
We need to find its velocity after 3 seconds. It is moving downwards. The equation of motion is this case is
v=u+gt
[tex]v=8+9.8\times 3\\\\v=37.4\ m/s[/tex]
So, the velocity of the ball after 3 seconds is 37.4 m/s.
What happens to the temperature as altitude increases in the exosphere? Does it increase or decrease the higher it goes?
Answer:
it gets colder the higher you go
Your friend says, “chemical changes are caused by an input in energy. In physical changes, there is no transfer of energy” is your friend correct? Why or why not?
Answer:
Ok, let's suppose the simplest of the physical changes:
We have an object that is not moving (so it is not accelerated)
and there is change, now the object moves.
Because there was a change, means that there was an acceleration, and by the second Newton's law.
Force equals mass times acceleration:
F = m*a
There must be a force.
So suppose that you pushed the object, then some energy that you had, you transferred it to the object, that now is moving and now has kinetic energy.
Now, is kinda true that in a closed system the total energy is always constant, but it depends on what is our system.
So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.
So no, your friend is not correct.
Both chemical and physical changes involve energy transfer, so your friend is not correct.
A chemical change occurs when a new substance is formed or created through a process of chemical reaction which is reversible.
The addition or removal of heat energy can affect the rate of chemical reaction. This addition or removal is known as energy transfer process.
Examples of chemical changes include;
Acid-base reaction.Rusting of iron in presence of moisture and oxygen. Cooking any food.A physical change on the other hand doesn't involve formation of new substance and it is can be reversible or irreversible.
The addition or removal of heat energy can affect the rate of physical changes.
Examples of such physical changes include;
vaporization of liquid (liquid to gas),freezing of liquid (liquid to solid), and condensation of gas (gas to liquid).Thus, both process (chemical and physical changes) involve energy transfer, so your friend is not correct.
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When charges qa, qb, and qc are placed respectively at the corners a, b, and c of a right triangle, the potential at the midpoint of the hypotenuse is 20 V. When the charge qa is removed, the potential at the midpoint becomes 15 V. When, instead, the charge qb is removed (qa and qc both in place), the potential at the midpoint becomes 12 V. What is the potential at the midpoint if only the charge qc is removed from the array of charges?
Answer:
8v
Explanation:
First we apply super position principle
Vt= v1 + v2+ v3
Remove qa
But vt= 20v
So V = v2+v3
V1= 20-15
= 5v
Remove qb
V= v1+v3
V=8v
So the potential when qa and qc are remove is the potential due to qb
Which is 8v
What is the binary number system?
Explanation:
binary number is a number expressed in the base-2 numeral system or binary numeral system, which uses only two symbols: typically "0" and "1".
Check Concepts
4.
35. Which of the following do you calculate
when
you
divide the total distance trav-
eled by the total travel time?
A) average speed
B) constant speed
C) variable speed
D) instantaneous speed
Answer:
I think its A.........
A 10 kg object is dropped from rest. a. How far will it drop in 2s? b. How long will it take a 5 kg object to drop the same distance?
Answer:
Explanation:
Using the equation of motion S = ut + 1/2at² to get the height of drop where
u is the initial velocity of the object = 0m/s
a is the acceleration due to gravity = +9.81m/s² (downwards motion of object)
t is the time it takes the object to drop = 2secs
Substituting the given parameters into the formula to get the height of drop S, we will have;
S = 0(2)+1/2(9.81)(2)²
S = 0+9.81*2
S = 19.62m
Hence the object will drop at a distance of 19.62m
b.) To determine the time it will take a 5 kg object to drop the same distance, we will use the same formula S = ut+1/2at²
Since the 5kg object also drops at the same distance, then S = 19.62m
Substituting this values into the equation we will have;
19.62 = 0(t) + 1/2(9.81)t²
19.62 = 4.905t²
t² = 19.62/4.905
t² = 4
t =2secs
This shows that it will take 5kg object 2secs to fall from the same distance. This means that no matter the mass of the object, it will take them the same time to fall at the same distance because they are all falling under the same influence of gravity.
If one object (a) is moving at 60m/s^2, and the other object (b) is moving at 65m/s^2, at what time will the faster moving object be 10m ahead of the other object?
Answer:
a is moving at 60m and the other object
Two equal charges are 2m2m apart. If the charges and the distance are divided by two, how is the force between the charges affected? g
Answer:
The force between the charges are not affected.
Explanation:
Given;
distance between two equal charges, R = 2m
The force between the charges is given by;
[tex]F = \frac{kq^2}{R^2}\\\\F_1 = \frac{kq_1^2}{R_1^2}\\\\When\ the \ charges \ and \ the \ distance \ are \ divided \ by \ two \ (q_2 = \frac{q_1}{2}, \ R_2 = \frac{R_1}{2} )\\\\ F_2 = \frac{kq_2^2}{R_2^2}\\\\F_2 = \frac{k(q_1/2)^2}{(R_1/2)^2}\\\\F_2= \frac{4k*q_1^2}{4*R_1^2}\\\\F_2 = \frac{k*q_1^2}{R_1^2}\\\\F_2 = F_1[/tex]
Therefore, the force between the charges are not affected.
A stone is thrown vertically upward with a speed of 28.0 m/s how much time is required to reach this height
Is the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth?
Answer:
Yes
Explanation:
This is because the interaction between piece of paper and earth.is gravitational while that of piece of paper and rod is electrostatic
Yes, the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth.
Gravitational force is the force by which earth attracts other objects by mass.The electrostatic force is the force of an object due to charge.Electrostatic forces are much stronger than gravitational forces. because gravity depends on mass, atoms have less masses so that the gravitational forces between them is close to zero. Whereas, the electrostatic force related to charges is bigger.Therefore, the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth.
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You are standing on top of the Empire State Building (height 330 [m]) when Superman flies past. He is headed straight down with a steady speed of 35 [m/s]. (Superman can do things like this.) At the instant he goes past, you drop a 10 [kg] lead ball over the edge. At what height above the sidewalk does the ball pass Superman?
Answer:
The ball will be overtake superman after travelled distance 248.5 m.
Explanation:
Given that,
Speed of superman = 35 m/s
Height = 330 m
Mass of ball = 10 kg
Let the height at which they are at same position be S.
For superman,
We need to calculate the time
Using equation of motion
[tex]S=ut[/tex]...(I)
For ball,
[tex]S=u't+\dfrac{1}{2}gt^2[/tex]
From equation (I) and (II)
[tex]ut=u't+\dfrac{1}{2}gt^2[/tex]
[tex]ut=0+\dfrac{1}{2}\times g\times t^2[/tex]
[tex]u=\dfrac{1}{2}\times g\times t[/tex]
[tex]t=\dfrac{2u}{g}[/tex]
Put the value into the formula
[tex]t=\dfrac{2\times35}{9.8}[/tex]
[tex]t=7.1\ sec[/tex]
We need to calculate the distance travelled
Using formula of distance
[tex]d=v\times t[/tex]
Put the value into the formula
[tex]d=35\times 7.1[/tex]
[tex]d=248.5\ m[/tex]
Hence, The ball will be overtake superman after travelled distance 248.5 m.