Answer:
Distillation separates a liquid from a solution. For example, water can be separated from salty water by simple distillation. This method works because the water evaporates from the solution, but is then cooled and condensed into a separate container. The salt does not evaporate and so it stays behind.
A spring stretches 5 cm when a 300-N mass is suspended from it. Calculate the spring constant in N / m .
Answer:
Spring constant in N / m = 6,000
Explanation:
Given:
Length of spring stretches = 5 cm = 0.05 m
Force = 300 N
Find:
Spring constant in N / m
Computation:
Spring constant in N / m = Force/Distance
Spring constant in N / m = 300 / 0.05
Spring constant in N / m = 6,000
How to calculate this operation? m=10kg and L=2500J/kg? What is the Energy?
I hope you will give me the correct answer because the last answer was too rude. His known by SOD!
Answer:
25000J
Explanation:
Formula : Q = m×c×Δt
Q=Heat energy
m= mass
c=specific heat capacity
ΔT = change in temperature.
Q=M x C for this question.
Specific heat capacity = 2500 J/Kg
Mass = 10kg
2500 x 10 = 25000 J
When you are driving on the freeway and following the car in front of you, how close is too close? Let's do an estimation.
1. Pick a car model (preferably the one you drive, but can also be any car of your dream), and find its stopping distance at highway speeds (you can usually find this type of data online).
2. Assuming that the car in front of you suddenly does a hard brake. For simplicity, assume that its braking performance is about the same as yours. Then also assume a reasonable amount of reaction time on your part (the time delay between seeing the brake lights lit up and applying your own brake). In order for you not to run into the car your are following, what's the closest distance you need to keep between the two cars?
3. Redo the same calculation if the vehicle in front of you is a typical big-rig truck. Find its braking data online.
4. There is a rule of thumb which says that you must stay one car length behind the car in front of you for every 10 mi/h of driving speed. From your calculation, does this rule make sense?
Answer:
1) v= 90km/h d = 70 m, 2) x₁ = v t_r, x₁ = 6.25 m, 3) x₁=6.25 no change
4) x = 22 m
Explanation:
1) for the first part, you are asked to find the minimum safety distance with the vehicle in front
The internet is searched for the stopping distance for two typical speeds on the highway
v (km/ h) v (m/s) d (m)
90 25 70
100 27.78 84
the safe distance is this distance plus the distance traveled during the person's reaction time, which can be calculated with infirm movement
v = x / t_r
x₁ = v t_r
the average reaction time is t_r = 0.25s for a visual stimulus and t_r 0.17 for an auditory stimulus
therefore the safe distance is
x_total = x₁ + d
2) The distance is the sum of the distance traveled in the reaction
x₁ = v t_r
for v = 90 km / h
x₁ = 25 0.25
x₁ = 6.25 m
for v = 100 km / h
x₁ = 27.78 0.25
x₁ = 6.95 m
the total distance is
x_total = x₁ + d
for v = 90 km / h
x_total = 25 0.25 + 70
x_total = 76.25 m
this is the distance until the cars stop and do not collide
3) the stopping distance of a truck is
v = 90 km / h d = 100 m
in this case we see that the braking distance is much higher,
the safe distance is given by the distance traveled during the reaction, as the truck brakes slower than the car this distance does not change
4) let's analyze the empirical rule: maintain the length of a car for each increase in speed of v = 10 m / h = 4.47 m / s
for the car case at v = 90km / h = 25 m / s
according to this rule we must this to
x = 25 / 4.47 = 5.6 cars
each modern car is about 4 m long so the distance is
x = 22 m
we see that this distance is much greater than the reaction distance so it does not make much sense
Which of the following is an example of an object with kinetic energy?
a. a plane lifting off of the runway
b. a bobsled perched at the top of a run
c. a snowball tumbling down a hill
d. both a and c
Answer:
A and C
Explanation:
Both have mass and are in motion
PLEASE HELP WITH THIS ONE QUESTION
A bird lands on a bird feeder which is connected to a spring. The mass of the bird is exactly the same as the mass of the bird feeder. How does the added mass affect the period of oscillation of the bird feeder?
The period is less, but not halved.
The period is more, but not doubled.
The period is halved.
The period is doubled.
Answer:
The period is more, but not doubled.
Explanation:
Recall that the period of a mass on a spring is T=2πmk.
Is TV light is converging rays, divergent rays or parallel rays?
What factors determine electric potential?
A. Mass and distance
B. Charge and density
C. Charge and distance
O D. Mass and charge
Answer: C. Charge and distance
The idea is to get as much EMF produced from the sprinter running through it. If you were the Olympic coach on a year when there happens to be a global energy crisis, and medals were assigned based on how much EMF (or current) were produced by the sprinters, what 3 pieces of advice (one is quite obvious; the other two involve the fixed orientation of the baton and the maintained position of the baton within the circular solenoid cross-section) would you give your sprinters in order to win?
Answer:
the greater the speed, the greater the electromotive force
* The metal pole must be parallel to the field
* you must keep the ball of the field
Explanation:
To determine the advice to the runners, let's use the Farad equation to and
fem = -N [tex]\frac{ d \phi}{dt}[/tex] = -N [tex]\frac{ B A Cos \theta }{dt}[/tex]
how the runners are moving
fi = B l x
fem = -N B l v
therefore the advice we can give are:
* the greater the speed, the greater the electromotive force
* The metal pole must be parallel to the field
* you must keep the ball of the field
HELP PLEASE
60 POINTS
HAVE A GREAT REST OF YOUR DAY PEOPLE :>
Answer:
The sun and the stars
Explanation:
I hope this helps!
Answer:
Northstar & North Pole
Explanation:
yyggggggggggggggg
Which of the following requires constant agonist-antagonist muscle contraction
Answer: Dynamic balance
Explanation: Dynamic balance movements are movements in which constant agonist-antagonist muscle contractions occur in order to maintain a certain position or posture. ISSA pg 121
An object is experiencing an acceleration of 0.4 m/s^2 while traveling in a circle of 35 m. What is it’s velocity?
Answer:
v = 3.74 m/s
Explanation:
Given that,
The acceleration of the object in circular path, a = 0.4 m/s²
The radius of the circle, r = 35 m
We need to find the velocity of the object. The acceleration of an object on the circular path is given by :
[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{0.4\times35}\\\\v=3.74\ m/s[/tex]
So, the velocity of the object is equal to 3.74 m/s.
In a thunderstorm at 32.0°C, Reginald sees a bolt of lightning and hears the thunderclap 2.00s later. How far from Reginald did the lightning strike? The speed of sound through air at 32.0°C is 350.2 m/s. Show your work.
PLEASE HELP, THANKS!
Answer:
d = 700.4 m
Explanation:
Given that,
The speed of sound through air at 32.0°C is 350.2 m/s.
Reginald sees a bolt of lightning and hears the thunderclap 2.00s later.
We need to find how far from Reginald did the lightning strike. Let the distance be d. So,
Speed = distance/time
d = vt
So,
d = 350.2× 2
d = 700.4 m
So, the required distance is 700.4 m.
A 125 kg mail bag hangs by a vertical rope 3.3 m long. A postal worker then displaces the bag to a position 2.2 m sideways from its original position, always keeping the rope taut.
1) What horizontal force is necessary to hold the bag in the new position?
2) As the bag is moved to this position, how much work is done by the rope?
3) As the bag is moved to this position, how much work is done by the worker?
Answer:
1) the required horizontal force F is 1095.6 N
2) W = 0 J { work done by rope will be 0 since tension perpendicular }
3) work is done by the worker is 1029.4 J
Explanation:
Given that;
mass of bag m = 125 kg
length of rope [tex]l[/tex] = 3.3 m
displacement of bag d = 2.2 m
1) What horizontal force is necessary to hold the bag in the new position?
from the figure below; ( triangle )
SOH CAH TOA
sin = opp / hyp
sin[tex]\theta[/tex] = d / [tex]l[/tex]
sin[tex]\theta[/tex] = 2.2/ 3.3
sin[tex]\theta[/tex] = 0.6666
[tex]\theta[/tex] = sin⁻¹ ( 0.6666 )
[tex]\theta[/tex] = 41.81°
Now, tension in the string is resolved into components as illustrated in the image below;
Tsin[tex]\theta[/tex] = F
Tcos[tex]\theta[/tex] = mg
so
Tsin[tex]\theta[/tex] / Tcos[tex]\theta[/tex] = F / mg
sin[tex]\theta[/tex] / cos[tex]\theta[/tex] = F / mg
we know that; tangent = sine/cosine
so
tan[tex]\theta[/tex] = F / mg
F = mg tan[tex]\theta[/tex]
we substitute
Horizontal force F = (125kg)( 9.8 m/s²) tan( 41.81° )
F = 1225 × 0.8944
F = 1095.6 N
Therefore, the required horizontal force F is 1095.6 N
2) As the bag is moved to this position, how much work is done by the rope?
Tension in the rope and displacement of mass are perpendicular,
so, work done will be;
W = Tdcos90°
W = Td × 0
W = 0 J { work done by rope will be 0 since tension perpendicular }
3) As the bag is moved to this position, how much work is done by the worker
from the diagram in the image below;
SOH CAH TOA
cos = adj / hyp
cos[tex]\theta[/tex] = ([tex]l[/tex] - h) / [tex]l[/tex]
we substitute
cos[tex]\theta[/tex] = ([tex]l[/tex] - h) / [tex]l[/tex] = 1 - h/[tex]l[/tex]
cos[tex]\theta[/tex] = 1 - h/[tex]l[/tex]
h/[tex]l[/tex] = 1 - cos[tex]\theta[/tex]
h = [tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
now, work done by the worker against gravity will be;
W = mgh = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
W = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
we substitute
W = (125 kg)((9.8 m/s²)(3.3 m)( 1 - cos41.81° )
W = 4042.5 × ( 1 - 0.745359 )
W = 4042.5 × 0.254641
W = 1029.4 J
Therefore, work is done by the worker is 1029.4 J
Explain why the same side of the moon is always facing Earth.
Answer:
The moon keeps the same face pointing towards the Earth because its rate of spin is tidally locked so that it is synchronized with its rate of revolution (the time needed to complete one orbit). In other words, the moon rotates exactly once every time it circles the Earth.
A roller-coaster car has a mass of 1240 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 22 m, its speed is not changing. (a) At the top of the hill, what is the normal force (using the negative sign for the downward direction) FN on the car from the track if the car's speed is v = 8.7 m/s? (b) What is FN if v = 20 m/s? Use g=9.81 m/s2.
Answer:
a) N = 7.90 10³ N, b) N = -1.04 10⁴ N
Explanation:
a) For this exercise we can use Newton's second law
N -W = m (-a)
The relationship is centripetal, the negative sign of the acceleration is because it points towards the center of the circle
a = v² / r
we substitute
N = mg -m v² /r
N = m (g - v² /r)
let's calculate
v = 8.7 m / s
N = 1240 (9.81 - 8.7²/22)
N = 7.90 10³ N
b) v = 20 m / s
N = 1240 (9.81 - 20²/22)
N = -1.04 10⁴ N
a guitar string transmits waves at 509 m/s. to create a 491 Hz note, to what length should the guitar player shorten the string?
Answer:
Wavelength = 1.04 meters
Explanation:
Given the following data;
Speed = 509m/s
Frequency = 491Hz
To find the wavelength;
Wavelength = speed/frequency
Wavelength = 509/491
Wavelength = 1.04 meters
Therefore, the guitar player should shorten the length of the string to 1.04 meters.
Answer: 0.52
Explanation:
A girl with a mass of 22 kg is playing on a swing. There are three main forces acting on her at any time: gravity, force due to centripetal acceleration, and the tension in the swing's chain (ignore the effects of air resistance). At the instant shown in the image below, she is at the bottom of the swing and is traveling at a constant speed of 2 m/s. What is the tension in the swing's chain at this time? (Recall that g = 9.8 m/s^2)
A. 97.5N
B. 180.9N
C. 237.6N
D. 117.4N
Answer:
A is the answer I think pls check it
What is the earliest time at which the oscillator shown below is stationary?
Answer:
0.0s
Explanation:
I got it right in acellus
Answer: 0.0
Explanation:
Planets don't collide into
the sun because they
A. Are moving
B. Have too much mass
C. Have their own gravity
D. Are more attracted to each other
Two girls are estimating each other's power. One runs up some step
ng each other's power. One runs up some steps, and the other times her. Here are their
results:
height of one step = 20 cm
number of steps = 36
mass of runner = 45 kg
time taken = 4.2 s
a .Calculate the runner's weight. (Acceleration due to gravity g=10m
b .Calculate the increase in the girl's gravitational potential energy as she runs up the steps.
c. Calculate her power. Give your answer in kilowatts (kW).
Answer:
A. 450 N
B. 3240 J
C. 0.77 KW
Explanation:
From the question given above, the following data were obtained:
Height of one step = 20 cm
Number of steps = 36
Mass of runner = 45 kg
Time taken = 4.2 s
Next, we shall convert 20 cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
20 cm = 20 cm × 1 m /100 cm
20 cm = 0.2 m
Next, we shall determine the total height. This can be obtained as follow:
Height of one step = 0.2 m
Number of steps = 36
Total height =?
Total height = 36 × 0.2
Total height = 7.2 m
A. Determination of the runner's weight.
Mass of runner (m) = 45 kg
Acceleration due to gravity (g) = 10 m/s²
Weight (W) =?
W = m × g
W = 45 × 10
W = 450 N
B. Determination of the increase in the potential energy.
At the ground level, the potential energy (PE₁) is 0 J.
Next, we shall determine the potential energy at 7.2 m. This can be obtained as follow:
Mass of runner (m) = 45 kg
Acceleration due to gravity (g) = 10 m/s²
Total height (h) = 7.2 m
Potential energy at height 7.2 m (PE₂) = ?
PE₂ = mgh
PE₂ = 45 × 10 × 7.2
PE₂ = 3240 J
Final, we shall determine the increase in potential energy. This can be obtained as follow:
Potential energy at ground (PE₁) = 0 J
Potential energy at height 7.2 m (PE₂) = 3240 J
Increase in potential energy =?
Increase in potential energy = PE₂ – PE₁
Increase in potential energy = 3240 – 0
Increase in potential energy = 3240 J
C. Determination of the power.
Energy (E) = 3240 J
Time (t) = 4.2 s
Power (P) =?
P = E/t
P = 3240 / 4.2
P = 771.43 W
Finally, we shall convert 771.43 W to kilowatt (KW). This can be obtained as follow:
1000 W = 1 KW
Therefore,
771.43 W = 771.43 W × 1 KW / 1000 W
771.43 W = 0.77 KW
Therefore, her power is 0.77 KW
coefficient of viscosity of a glycerine is 8.4 poison explain
Answer:
coefficient of viscosity of 8.4 poison denotes that the tangential frictional force acting per unit area when divided by the velocity gradient as a result of streamline flow conditions gives 8.4.
Explanation:
Viscosity is defined as the extent to which a fluid can resist flow when a force is applied to it.
Now, coefficient of viscosity is the term in which viscosity is calculated. It is basically the tangential frictional force acting per unit area which is divided by the velocity gradient as a result of streamline flow conditions.
Thus, coefficient of viscosity of 8.4 poison denotes that the tangential frictional force acting per unit area when divided by the velocity gradient as a result of streamline flow conditions gives 8.4.
A cyclist exerts a 15.0 N force while riding 251 m in 30.0 s. What power does the cyclist develop?
Answer:
P=126W
Explanation:
Sorry if im wrong!
Answer:
125.5 watts
Explanation:
P=work/time
work=F*d
P=(F*d)/t
P=(15*251)/30
P=125.5 watts
At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 55 kg, 90 kg, and 42.5 kg respectively. Car A is moving to the right with a velocity vA = 2 m/s and car C has a velocity vC = 1.5 m/s to the left, but car B is initially at rest. The coefficient of restitution between each car is 0.8. Determine the final velocity of each car, after all impacts, assuming car A hits car B before car C does. Assume positive sign denoting forward motion and negative sign denoting backward motion.
Answer:
Vb = 0.334 m/s
Va = -1.265 m/s
Vc = 1.424 m/s
Explanation:
Favorite Answer
Initial momentum = 255(2) – 242.5(1.5) = 146.25
Final momentum = 255Va + 290Vb + 242.5 Vc = 146.25
Vb - Va = 0.8(2) = 1.6
Vc - Vb = 0.8(1.5) = 1.2
Va = Vb -1.6
Vc = Vb + 1.2
255(Vb -1.6) + 290Vb + 242.5(Vb + 1.2) = 146.25
255 Vb – 408 + 290 Vb + 242.5 Vb + 291 = 146.25
787.5 Vb = 263.25
Vb = 0.334 m/s
Va = Vb -1.6 = 0.334 – 1.6 = -1.265 m/s
Vc = Vb + 1.2 = 0.224 + 1.2 = 1.424 m/s
Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slide forward, crushing the cab, if the truck stops suddenly in an accident or even in braking. Assume, for example, that a 10 000-kg load sits on the flatbed of a 20 000-kg truck moving at 12.0 m/s. Assume that the load is not tied down to the truck, but has a coefficient of friction of 0.500 with the flatbed of the truck.
A) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck.
B) Is any piece of data unnecessary for the solution?
a) mass of the load.
b) mass of the truck.
c) velocity.
d) coefficient of static friction.
e) all are necessary.
Answer:
A)
the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B)
data that were not necessary to the solution are;
a) mass of truck and b) mass of load
Explanation:
Given that;
mass of load [tex]m_{LS}[/tex] = 10000 kg
mass of flat bed [tex]m_{FB}[/tex] = 20000 kg
initial speed of truck [tex]v_{0}[/tex] = 12 m/s
coefficient of friction between the load sits and flat bed μs = 0.5
A) the minimum stopping distance for which the load will not slide forward relative to the truck.
Now, using the expression
Fs,max = μs [tex]F_{N}[/tex] -------------let this be equation 1
where [tex]F_{N}[/tex] = normal force = mg
so
Fs,max = μs mg
ma[tex]_{max}[/tex] = μs mg
divide through by mass
a[tex]_{max}[/tex] = μs g ---------- let this be equation 2
in equation 2, we substitute in our values
a[tex]_{max}[/tex] = 0.5 × 9.8 m/s²
a[tex]_{max}[/tex] = 4.9 m/s²
now, from the third equation of motion
v² = u² + 2as
[tex]v_{f}[/tex]² = [tex]v_{0}[/tex]² + 2aΔx
where [tex]v_{f}[/tex] is final velocity ( 0 m/s )
a is acceleration( - 4.9 m/s² )
so we substitute
(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx
0 = 144 m²/s² - 9.8 m/s²Δx
9.8 m/s²Δx = 144 m²/s²
Δx = 144 m²/s² / 9.8 m/s²
Δx = 14 m
Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B) data that were not necessary to the solution are;
a) mass of truck and b) mass of load
examples of buildup of static energy?
Answer:
Materials that can lose or gain electrons in this way are called triboelectric, according to Northwestern University. One common example of this would be shuffling your feet across carpet, particularly in low humidity which makes the air less conductive and increases the effect.
Explanation:
Hope that helps :)
a race car goes around a circular track of radius 150 m at speed of 10.0 m/s. How long does it take to complete one lap?
Answer:
94.25 seconds
Explanation:
Solve for period (T) using: v=(2*pi*r)/T
rearrange: vT=2*pi*r
rearrange: T=(2*pi*r)/v
Plug in values.
T=(2*pi*150)/10
T=94.25 seconds
If a race car goes around a circular track of a radius of 150 m at speed of 10.0 m/s ,then the time taken to complete the one lap would be 94.25 seconds.
What is speed?The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.
As given in the problem a race car goes around a circular track of radius 150 m at speed of 10.0 m/s.
vT = 2 × π × r
T = (2 × π × r)/ v
T = (2 × π× 150)/10
T = 94.25 seconds
Thus, the time taken to complete the one lap would be 94.25 seconds.
To learn more about speed here, refer to the link given below ;
https://brainly.com/question/7359669
#SPJ2
What is the average kinetic energy of particles in a gas at a temperature of 245 Kelvins?
Answer:
2)
Explanation:
An official major league baseball has a mass of 0.14 kg. A pitcher throws a 40 m/s fastball which is hit by the batter straight back up the middle at a speed of 46 m/s.
a) What is the change in momentum of the ball during the collision with the bat?
b) If this collision occurs during a time of 0.012 seconds, what is the average force exerted by the bat on the ball?
Answer:
(a) The change in momentum is 12.04 kg-m/s
(b) The force exerted by the bat is 1003.33 N
Explanation:
Given that,
The mass of a ball, m = 0.14 kg
Initial speed of the ball, u = 40 m/s
Final speed of the ball, v = -46 m/s
(a) The change in momentum of the ball during the collision with the bat is given by :
[tex]\Delta p=m(v-u)\\\\=0.14(-46-40)\\\\=-12.04\ kg-m/s[/tex]
(b) Time for collision, t = 0.012 s
Now the force can be calculated as follows :
[tex]F=\dfrac{\Delta p}{t}\\\\F=\dfrac{12.04}{0.012}\\\\=1003.33\ N[/tex]
Hence, this is the required solution.
Answer:
a. = 12.04 kg*m/s
b. = 1,003.3N
Explanation:
The answer above is correct.
If your water heat has an efficiency of 95 percent, how much energy would it take to heat 45kg of water from 23 C to 60 C. (The specific heat of water is 4.18 J/g/C.) Please show your work
Answer: 6611.715 joules
Explanation:
Q = MxCxdeltaT = 6959.7 which is 100%
95% = 6611.715
An instructor wishes to determine the wavelength of the light in a laser beam. To do so, she directs the beam toward a partition with two tiny slits separated by 0.180 mm. An interference pattern appears on a screen that lies 5.30 m from the slit pair. The instructor's measurements show that two adjacent bright interference fringes lie 1.60 cm apart on the screen. What is the laser's wavelength (in nm) ?
Answer:
λ = 5.434 x 10⁻⁷ m = 543.4 nm
Explanation:
To solve this problem we can use the formula provided by Young's Double Slit experiment:
[tex]\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta xd}{L}[/tex]
where,
λ = wavelength of light = ?
Δx = distance between adjacent bright fringes = 1.6 cm = 0.016 m
d = slit separation = 0.18 mm = 0.00018 m
L = Distance between slits and screen = 5.3 m
Therefore,
[tex]\lambda = \frac{(0.016\ m)(0.00018\ m)}{5.3\ m}[/tex]
λ = 5.434 x 10⁻⁷ m = 543.4 nm