Answer:
Wc = m*g = 50kg * 9.8N/kg = 490 N = Wt.
of crate.
Fc = 490N 60o = Force of crate.
Fp = 490*sin60 = 424.4 N. = Force
parallel to incline.
Fv = 490*cos60 = 245 N. = Force perpendicular to the incline.
Ff = u*Fv = u*245 = Force of friction.
Fap-Fp-Ff = m*a
Fap-424.4-u*245 = m*0 = 0
Fap = 424.4 + 245u. = Force applied.
L = 200m/sin60 = 231 m. = Length of
incline.
Work = Fap * L=(424.4+245u) * 231.
Explanation:
What is a lever?....
Answer:
A lever is a simple machine in the subject "S.T.E.M"
Explanation:
Levers use the following forces in order to be used: push or pull.
examples are crow bars,shovels, brooms and ect.
Entomology is the application of__________ to be used in a criminal investigation??
A/ soil science
B/insect science
C/plant science
D/glass science
eleven wa=eighs 47 kg. her height is 1.63 m. what is her bmi
Answer:
17.7kg/[tex]m^{2}[/tex]
Explanation:
BMI =
= [tex]\frac{weight (kg)}{height (m) . height (m)}[/tex]
= 47kg/(1.63m×1.63m)
= 17.689kg/[tex]m^{2}[/tex]
≈ 17.7kg/[tex]m^{2}[/tex]
The ___ of an object changes when you take it to a different planet.
Answer:
weight
Explanation:
weight depends on gravity
The magnetic field of a straight, current-carrying wire is:
What is the vertical component of a ball thrown at a 27 degree angle at 16 m/s?
y = y 0 + v 0 y t − 1 2 g t 2 . If we take the initial position y 0 to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity: v 0 y = v 0 sin θ 0 = ( 30.0 m / s ) sin 45 ° = 21.2 m / s .
un cuerpo suspendido en el aire es capaz de:
a) liberar energía al caer
b) puede ejecutar un trabajo
c) posee energía antes de caer
d) todas son correctas
Answer:
Explanation:
d
Which factor is not a type of spectra?
A.
emission
B.
absorption
C.
continuous
D.
discontinuous
A 22 kg body is moving through space in the positive direction of an x axis with a speed of 190 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 13 kg, moves away from the point of explosion with a speed of 130 m/s in the positive y direction. A second part, with a mass of 2.2 kg, moves in the negative x direction with a speed of 460 m/s. What are the (a) x-component and (b) y-component of the velocity of the third part
Answer: Our notation is as follows : the mass of the original body is M=20.0kg ; its initial velocity is
ν
0
=(200m/s)
i
^
; the mass of one fragment is m
1
=10.0kg ; its velocity is
ν
1
=(100m/s)
j
^
; the mass of the second fragment is m
2
=4.0kg ; its velocity is
ν
2
=(−500m/s)
i
^
; and , the mass of the third fragment is m
3
=6.00kg . Conservation of linear momentum requires
M
ν
0
=m
1
ν
1
m
2
ν
2
+m
3
ν
3
.
The energy released in the explosion is equal to ΔK , the change in kinetic energy .
(a) Using the above momentum -conservation equation leads to
ν
3
=
m
3
M
ν
0
−m
1
ν
1
−m
2
ν
2
=
6.00kg
(20.0kg)(200m/s)
i
^
−(10.0kg)(100m/s)
j
^
−(4.0kg)(−500m/s)
i
^
=(1.00×10
3
m/s)
i
^
−(0.167×10
3
m/s)
j
^
.
The magnitude of
ν
3
is
ν
3
=
(1000m/s)
2
+(−167m/s)
2
=1.01×10
3
m/s
It points at θ=tan
−1
(−167/1000)=−9.48
∘
(that is at 9.5
∘
measured clockwise from the +x axis) .
(b) The energy released is ΔK :
ΔK=K
f
−K
i
=(
2
1
m
1
ν
1
2
+
2
1
m
2
ν
2
2
+
2
1
m
3
ν
3
2
)−
2
1
Mν
0
2
=3.23×10
6
J
Explanation:
How do you find the magnitude of the air inside a balloon?
Answer:
This demonstration is often done following a discussion of the ideal gas equation of state, PV=nRT.
We begin by weighing a balloon, then blowing it up and weighing it again. In the photo shown on right, the mass indication increased from 3.4 to 3.5 grams. At this point, it is important to note that the scale measures force, even though it reports a conclusion about mass based on the force measurement.
One assumption made in reaching the conclusion is that the buoyant force on the object being weighed is negligible. In the case of the balloon, this is incorrect. The buoyant force on this balloon is equal to the weight of the air displaced.
Since the volume of air inside the balloon is essentially the same as the volume of air displaced, we should expect that the buoyant force would support the weight of the air inside the balloon: The reported mass should not go up at all, because the force required of the scale should not change.
The increase in reported mass of .1 gram is attributed to the higher density of the air inside the balloon: The tension in the balloon compresses the air inside, as attested by the pressure required to blow the balloon up. Evidently, for this experiment, the pressure inside is greater than atmospheric by about 2%.
In the picture at right, the balloon is being pressed into a pan of liquid nitrogen. (The pan is the styrofoam lid of a small lunch box.) The balloon floats lightly on the liquid nitrogen unless pressed down. Pressing down places more surface area in contact with the cold nitrogen and speeds the demonstration. It is interesting to note the buoyant force by this liquified constituent of air.
The balloon shrinks dramatically, as indicated below. When left in contact with the liquid nitrogen long enough (perhaps 5 minutes) the oxygen inside the balloon liquifies, and then the nitrogen liquefies also. Close observation of the photo at the upper left corner of the pan shows some liquid nitrogen bubbles may forming above the dark spot in the center of the pan. One can also make out a faint line at the upper left corner of the pan which is the liquid nitrogen surface. The balloon still floats, riding rather high on that surface. Evidently, some of the balloon contents remain in the gas phase, making the mass of the balloon less than the mass of the displaced liquid nitrogen.
Next, we take the shrunken balloon and place it back on the scale, as above. In this instance, the reported mass is 8.7 grams, an increase of 5.2 grams.
A look at the figure on the right shows a faint line near the bottom of the cold balloon. Above that line, the balloon contains gas; below the liquid. That line represents the top surface of the liquid air inside the balloon. With this evidence, the easy thing to say would be, "Of course, liquids are heavier than gases," but that would be incorrect. We assert that the amount of air inside the balloon has not changed and that the mass of that air is not dependent on temperature.
If these assertions are true, then the force of gravity on the balloon has not changed. The scale reading is determined by the force which it must exert on the balloon in order to keep it stationary. Evidently, the required force is larger when the balloon is shrunken. The reason is that the buoyant force (upward) has decreased to practically zero, leaving the scale alone to balance the downward force by gravity.
From the data, we can say that the change in the buoyant force is equal to the weight associated with the apparent change in mass. The weight of 5.2 grams is about .052 newtons. The buoyant force is less now because the balloon displaces less air. If we could measure the change in volume of the balloon as DV, then the buoyant force would be (r g DV) upwards, where r is the density of air that was displaced by the balloon, and g is the gravitational field strength, 9.8 Newton/kg.
Note that the .052 newton force is not the weight of the air inside the balloon. Rather, it is the weight of the air that was displaced by the balloon. If we ignore the compression of air inside the balloon, the two numbers are the same. However, the two samples are completely different.
We can estimate the volume of the balloon by assuming that the hand in the photograph is about .1meters across. For purposes of estimation, we say that the volume shrank to almost zero when the balloon was cold so that the change in volume was nearly equal to the original volume. Plugging in numbers gives fair agreement with the book value of 1kg/cubic meter for the density of air.
The value for the density of air is secondary to two main features of this demonstration:
Large changes in temperature produce the large changes in volume that are indicated by the ideal gas equation.
The mass of air in a volume equal to the volume of a balloon can be determined provided that the buoyant force is understood.
SOMEONE PLEASE HELP ME IM GOING INSANE 4 Select the correct answer. A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled 2.0 x 107 meters and is only 10 meters away from its starting point. What is the numerical value of the satellite's average velocity after that one hour? O A. -3.77 x 10-2 meters/second OB. -2.77 x 10-3 meters/second OC. -2.62 x 10-2 meters/second D. -5.55 x 103 meters/second
Hi there!
Since we are dealing with such large distances, we can approximate.
Recall that the equation for speed is:
displacement / time = velocity
OR:
d/t = v
Begin by converting one hour to seconds:
1 hr = 3600 sec
Now, we can solve for velocity:
(-2.0 × 10⁷) / 3600 = -5555.6 m/s ⇒ D: -5.55 × 10³ m/s
The steepest street in the world is Baldwin Street in Dunedin, NZ. It is inclined at an angle of 380 , with the horizontal. A child slides down the street with a constant velocity on a sled with high friction runners. What is the coefficient of friction between the sled runners and the street?
Newton's second law allows to find the result for the friction coefficient of the street is:
The friction coeficinwete is: μ = 0.78
Newton's second law establishes a relationship between the net force, the mass and the acceleration of the body.
∑ F = m a
Where bold indicates vectors, m is to mass and acceleration.
In the attached we see a free body diagram, it is a diagram of the forces without the details of the body, the x-axis is parallel to plane also shown with the positive in the direction of movement, going down the plane and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight.
Sin θ = [tex]\frac{W_x}{W}[/tex]
cos θ = [tex]\frac{W_y}{W}[/tex] / W
Wₓ = W sin θ
[tex]W_y[/tex] = W cos θ
We write Newton's second law for each axis.
y-axis
N- [tex]W_y[/tex] = 0
N = mg cos θ
x-axis
Wₓ - fr = ma
Since they indicate that the body goes down at a constant speed, the acceleration is zero.
W sin θ = fr
The friction force is the macroscopic representation of the interactions between the two surfaces and the formula.
fr = μ N
we substitute.
fr = μ mg cos θ
mg sin θ = μ cos θ
μ = tan θ
Let's calculate.
μ = tan 38.0
μ = 0.78
In conclusion using Newton's second law we can find the results for the friction coefficient of the street is:
The frivtion coefficient is: μ = 0.78
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what is the fate of the energy in ultraviolet light that is incident upon glass?
Answer:
we can say when UV light is incident. Essentially, when it strikes glass or incident upon gloss, it will be absorbed. The energy has to go somewhere. Um, and in this case, this radiation is then converted to thermal energy and the glass raises and temperature.
Explanation:
Got the information from the internet
a. A pure sample was analysed and was found to contain 6.25 g of lithium, 5.40 g of carbon, & 21.60
g of oxygen. Calculate the empirical formula. Show your work out. If you do not show any work out,
marks will not be given. [3]
b. The molar mass of the compound is 116 g/mol, what is its molecular formula?
PLS URGENTLY ANS
The empirical formula of the compound is Li₂CO₃
First, we will determine the respective number of moles of each element present in the compound by dividing by their atomic masses
Lithium Carbon Oxygen
6.25÷6.94 5.40÷12.01 21.60÷16.00
0.9006 0.4496 1.35
Now, divide through by the smallest number of moles, that is, 0.4496
Lithium Carbon Oxygen
0.9006÷0.4496 0.4496÷0.4496 1.35÷0.4496
2 1 3
Hence, the empirical formula of the compound is Li₂CO₃
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Explain how potential and kinetic energy are at play when we talk about Newton’s second law of motion?
Answer:
represents the energy the object possesses by virtue of its motion. ... This type of energy is generally known as kinetic energy. Thus, Equation (16) states that any work done on point object by an external force goes to increase the object's kinetic energy.
Explanation:
The relation between some bodies are involved by Object's position and motion, i.e., through potential and kinetic energy in which it actually explained by Newton's second law of motion.
What is Newton's second law of motion?
Definition:
Newton's second law states that force is equal to the rate of change of momentum, i.e., the mass and velocity.
It can be written as F = mv where m is mass and v is velocity. Potential energy can be described by a body's position while kinetic energy is possessed by a body's motion. Both forms of energy are influenced by forces and are equal to the total momentum. Momentum can be described by explaining the mass and velocity of the object.Learn more about Newton's second law of motion,
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A young woman walks up 55 steps to the top of a water slide. She slides
down, feet first, all the way to the bottom gaining speed as she goes. If each
step of the slide is.10m tall, how fast will she be going at the bottom?
Answer:
potential energy PE = M g h
KE at bottom = 1/2 M V^2
Regardless of the slope of the slide the change in energy is the same
1/2 V^2 = g h
V = (2 g h)^1/2 = (2 * 9.8 m/s^2 * 10 m)^1/2 = 14 m / s
Perhaps the question says that h = 55 * .1 = 5.5 m
Then V = (2 * 9.8 * 5.5) = 10.4 m/s
The temperature of a body is measured using the resistance of a wire. To calibrate the device the following measurements were taken (ohms is the unit of electrical resistance).
Resistance in melting ice 240 ohms.
Resistance in boiling water 250 ohms.
a. What is the change in resistance for a change in temperature of 1 Degree Celcius?
b. The wire is placed into some hot water and the resistance is mesured to be 246 ohms. What is the temperature of the water?
Answer:
A)250-240
=10+960
=970
exoplanets are difficult to detect because they are:
Explanation:
Exoplanets are very hard to see directly with telescopes. They are hidden by the bright glare of the stars they orbit. So, astronomers use other ways to detect and study these distant planets.
Exoplanets are difficult to detect because they are a star 'wobbles' as it orbits the center of mass, changing the wavelength of light it emits.
Which will be different on the moon than it is on Earth?
weight
mass
Answer:
Below
Explanation:
On the moon, an objects weight will be different than it is on earth. This is because we cannot change the mass of an object, the mass of an object is the measure of matter an object has. However the weight is depended on the gravitational pull of whatever planet you are on. In this case, weight will be lighter on the moon than it is on earth because the moon's gravitational pull is 1.62 m/s^2 while earths is around 9.8 m/s^2.
Hope that helps!
part 2 of 2
With what speed does the stone strike the
water?
Answer in units of m/s.
Hi there!
We can begin by finding the final VERTICAL velocity of the ball.
Use the following kinematic equation: vf² = vi² + 2ad
The ball does not have any initial velocity in the vertical direction so:
vf² = 2ad
Plug in given values:
vf² = 2(9.81)(44) = 29.38 m/s
Find the TOTAL speed using the pythagorean theorem (involving both the horizontal velocity which remains the same and vertical velocity)
v = √(21²) + (29.38²) = 36.115 m/s
If two stars are in a binary system with a combined mass of 5.5 solar masses and an orbital period of 12 years, what is the average distance between the two stars
The average distance between the two stars is 792 light years
Let the mass of the first star be [tex]m_1[/tex]
Let the mass of the second star be [tex]m_2[/tex]
The combined mass of the two stars, [tex]m_1+m_2=5.5[/tex] solar masses
The orbital period of the stars, P = 12 years
Average distance between the two stars, D = ?
The average distance between the two stars can be calculated using Kepler's equation
[tex]D=(m_1+m_2)P^2[/tex]
Substitute [tex]m_1+m_2=5.5[/tex] and P = 12 into the formula [tex]D=(m_1+m_2)P^2[/tex]
[tex]D=5.5(12^2)[/tex]
D = 5.5(144)
D = 792 light years
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If the mass of the Earth is doubled, how will the moon’s orbit be affected?
What Would Happen to the Orbit of the Moon if it Were Twice as Massive? ... The orbit would not change.
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Which of the followng is Newton's Second Law?
a. An object in motion will stay in motion unless acted upon by an unbalanced outside force
b .Force = mass times acceleration
c.When one object puts a force on the another object, the second object puts a force back on the first, equal in magnitude and opposite in direction
Answer:
A
Explanation:
A stone is dropped off a tall building. It takes 4.5 s to hit the ground Determine the height of the building.
Explanation:
time =4.5sec
gravity =9.8m/s square
so to determime the height we can use the formula
1/2 of gravity times time square
which results in 99.25
Please find attached photograph for your answer.
Hope it helps.
Do comment if you have any query.
A 2.00kg block is attached to a horizontal ideal spring with a spring constant k=100Nm. The block-spring system is set on a horizontal surface with negligible friction. A graph of the potential energy U as a function of time t for this system is shown. The maximum displacement xMAX of the block from its equilibrium position and the maximum speed vmax of the block during the motion represented by the graph are most nearly
We have that for the Question, it can be said that the maximum velocity is
[tex]V_m = 1.414m/s[/tex]
From the question we are told
A 2.00kg block is attached to a horizontal ideal spring with a spring constant k=100Nm.
The block-spring system is set on a horizontal surface with negligible friction.
Generally the equation for the Potential energy is mathematically given as
[tex]P.E=\frac{1}{2}Rx_m^2\\\\2=\frax{1}{2}*100x_m^2\\\\x_m^2 = 0.04m\\\\x_m = 0.2m[/tex]
The PE is converted to KE, Therefore
[tex]KE = 2\\\\\frac{1}{2}MV_m^2 = 2\\\\V_m^2 = \frac{4}{M}\\\\V_m^2 = \frac{4}{2}\\\\V_m = \sqrt{2}\\\\V_m = 1.414m/s[/tex]
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Complete Question
A 2.00 kg block is attached to a horizontal ideal spring with a spring constant k = 100 N The block-spring system is set on a horizontal surface with negligible friction. A graph of the potential energy U as a function of time t for this system is shown. The maximum displacement IMAX of the block from its equilibrium position and the maximum speed Umat of the block during the motion represented by the graph are most nearly A IMAX = 2.0 m and UMAX 1.4" B UMAX = 1.4 and UMAX=0.20" с MAX 0.20 m and UMAX = 1.4" D IMAX 0.40 m and UMAX 1.4 E IMAX 0.04 m and UMAX 2.0"
2. In a race, if a runner starts and stops at the same position, what is their
displacement? *
Answer:
It is the same
Explanation:
I Jsut know
how much does it cost to charge an electric car at a public charging station
Answer:
Rs 16.5 lakh is a charge of an electric car at a public charging station
why is clay soil chosen for earthenware? Explain it.
There are many reasons actually...
◇ Clayey soil is easy to mould than the other soils.
The other soils (like sandy soil & loamy soil) are grainy or dry in nature & cannot be moulded. In the ancient times, when technology & the know-hows were still not introduced, people used clayey soul to make earthern pots to store stuff. This practice still continues even today.
◇ It keeps water cool.
Ever wondered why water stored in the earthenware remains cool even on a sunny day? It's because the heat required for evaporation is taken from water inside the pot & thus it cools the water stored inside.
◇ Clayey soil is also moist, durable, more plastic like & easy to retain making it ideal for potters to make earthenware.
__________
Clay soil is used to make earthenwares because, clay can be easily molded to desired sizes and shapes. Clay absorbs water moiety and make the pot very cool inside.
What is clay?Clay is a type of soil made by the sedimentation process near water resources. They are rich in minerals. Clay is rich in silicates and intermolecular force in the soil is comparatively less and can be easily molded any shape and size.
When clay is fired to a high enough temperature, the glass-forming materials in the ceramic form a liquid glass. This glass bridges the gaps between the clay particles. As the pottery cools at the end of the firing process, the glass between the clay particles solidifies again.
Clay pots make the water cooler inside. Hence, clay is used to make earthen wares.
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In a titration, 50.00 cm3 of 0.300 mol/dm3
sodium hydroxide solution is exactly neutralized by 25.0
cm3 of a dilute solution of hydrochloric acid.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Calculate the concentration of the hydrochloric acid. Show your work out
PLS URGENT ANS
Answer:
Step 1: Calculate the amount of sodium hydroxide in moles
Volume of sodium hydroxide solution = 50.0 ÷ 1,000 = 0.05 dm3
Rearrange:
Concentration in mol/dm3 = amount of solute in molvolume in dm3
Amount of solutein mol = concentration in mol/dm3 × volume in dm3
Amount of sodium hydroxide = 0.300 × 0.05
= 0.01 5mol
Step 2: Find the amount of hydrochloric acid in moles
The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
So the mole ratio NaOH:HCl is 1:1
Therefore 0.015 mol of NaOH reacts with 0.015 mol of HCl
Step 3: Calculate the concentration of hydrochloric acid in mol/dm3
Volume of hydrochloric acid = 25.00 ÷ 1000 = 0.025 dm3
Concentration in mol/dm3 = amount of solute in molvolume in dm3
Concentration in mol/dm3 = 0.015/0.025
= 0.6 mol/dm3
Step 4: Calculate the concentration of hydrochloric acid in g/dm3
Relative formula mass of HCl = 1 + 35.5 = 36.5
Mass = relative formula mass × amount
Mass of HCl = 36.5 × 0.6
= 21.9 g
So concentration = 21.9 g/dm3
The internal energy is the total kinetic energy and __________ energy of all the particles that make up a system. What word completes the sentence?
Answer:
Potential
Explanation:
The internal energy is the total amount of kinetic energy and potential energy of all the particles in the system. When energy is given to raise the temperature, particles speed up and gain kinetic energy.
Hope this helps :)