Answer:
[tex]\Delta H=687.4 J[/tex]
Explanation:
Hello!
In this case, for this melting process, we can identify two sub-processes in order to take the stainless steel from solid to liquid:
1. Heat up from 298.15 K to 1673 K.
2. Undergo the phase transition.
Both process have an associated enthalpy as shown below:
[tex]\Delta H_1=1g*0.5\frac{J}{g*K} (1673K-298.15K)=687.4J[/tex]
[tex]\Delta H_2=0.001kg*\frac{0.260J}{kg} =0.00026J[/tex]
Therefore, the required heat is:
[tex]\Delta H=\Delta H_1+\Delta H_2\\\\\Delta H=687.4J+0.00026J\\\\\Delta H=687.4J[/tex]
Notice the problem is not providing neither the mass or volume, that is why we assumed the mass is 1 g; however, it can be changed to the mass you are given.
Best regards!
A 3 kg exercise ball is held 2m above the ground. What is the gravitational potential energy?
Answer:
58.56 J
Explanation:
Since the formula for gravitational potential energy is:
GE = m x g x h - where m = mass (kg), g = acceleration due to gravity (9.81 m/s²), h = height (m)
GE= 3 x 9.81 x 2
GE = 58.86 J
Hope this helps
Including them, the mass of the balloon was 1890 kg and had a volume of 11,430 m3 . The balloon floats at a constant height of 6.25m above the ground. What is the density of the hot air in the balloon
Given :
The mass of the balloon was 1890 kg and had a volume of 11,430 m3 .
The balloon floats at a constant height of 6.25m above the ground.
To Find :
The density of the hot air in the balloon.
Solution :
We know,
Volume × ( Density of surrounding air - Density of hot air ) = mass
Putting given values in above equation, we get :
[tex]11430\times ( 1.29 - \rho_{hot \ air } ) = 1890\\\\\rho_{hot \ air } = 1.29 - \dfrac{1890}{11430}\\\\\rho_{hot \ air } = 1.125\ kg\ m^3[/tex]
Therefore, the density of hot air in the balloon is 1.125 kg m³.
What happens between particles of different charges?
Answer:
Oppositely charged objects will exert an attractive influence upon each other. In contrast to the attractive force between two objects with opposite charges, two objects that are of like charge will repel each other.
Explanation:
9. A 15kg mass is lifted upward at a constant speed to a height of 22 m. Calculate the work done by the lifting
force.
Work done by lifting is 3,300 Newton.
Given that;
Mass of thing = 15 kg
Height lifted = 22 m
Find:
Work done by lifting
Computation:
Work done = mgh
Work done by lifting = (15)(10)(22)
Work done by lifting = 3,300 Newton
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A rugby players runs 10.0 m straight down the playing field in 2.00 s. The player is then hit and pushed 5.00 m straight backward in 1.50 s. The player breaks the tackle and runs straight forward another 17.2 m in 4.10 s. Calculate the velocity for the players entire motion.
Answer:
2.92 m/s
Explanation:
For the entire motion, the average velocity (V) can be calculated thus;
V = [tex]\frac{displacement}{time}[/tex]
Total displacement of the player = 10.0 + (-5) + 17.2
= 22.2 m
Total time = 2.00 + 1.50 + 4.10
= 7.60 s
V = 22.2/7.60
= 2.92 m/s
Hence, the velocity of the players for the entire motion is 2.92 m/s.
Define each of the three heat transfer methods:
Conduction
Convection
Radiation
differentiate between capital g and small g
Answer:
Capital (G) is a universal gravitation law. and small (g) is acceleration of gravity of the each (9.8m/s^2)
Explanation:
According to the web
A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When the proton is a distance R from the nucleus its velocity has decreased to 1/2vo. How far from the nucleus will the proton be when its velocity has dropped to 1/4vo
Answer:
The value is [tex]R_f = \frac{4}{5} R[/tex]
Explanation:
From the question we are told that
The initial velocity of the proton is [tex]v_o[/tex]
At a distance R from the nucleus the velocity is [tex]v_1 = \frac{1}{2} v_o[/tex]
The velocity considered is [tex]v_2 = \frac{1}{4} v_o[/tex]
Generally considering from initial position to a position of distance R from the nucleus
Generally from the law of energy conservation we have that
[tex]\Delta K = \Delta P[/tex]
Here [tex]\Delta K[/tex] is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
[tex]\Delta K = K__{R}} - K_i[/tex]
=> [tex]\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2[/tex]
=> [tex]\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2[/tex]
=> [tex]\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2[/tex]
And [tex]\Delta P[/tex] is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
[tex]\Delta P = P_f - P_i[/tex]
Here [tex]P_i[/tex] is zero because the electric potential energy at the initial stage is zero so
[tex]\Delta P = k * \frac{q_1 * q_2 }{R} - 0[/tex]
So
[tex]\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0[/tex]
=> [tex]\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}[/tex]
=> [tex]- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )[/tex]
Generally considering from initial position to a position of distance [tex]R_f[/tex] from the nucleus
Here [tex]R_f[/tex] represented the distance of the proton from the nucleus where the velocity is [tex]\frac{1}{4} v_o[/tex]
Generally from the law of energy conservation we have that
[tex]\Delta K_f = \Delta P_f[/tex]
Here [tex]\Delta K[/tex] is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
[tex]\Delta K_f = K_f - K_i[/tex]
=> [tex]\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2[/tex]
=> [tex]\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2[/tex]
=> [tex]\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2[/tex]
And [tex]\Delta P[/tex] is the change in electric potential energy from initial position to a position of distance [tex]R_f[/tex] from the nucleus , this is mathematically represented as
[tex]\Delta P_f = P_f - P_i[/tex]
Here [tex]P_i[/tex] is zero because the electric potential energy at the initial stage is zero so
[tex]\Delta P_f = k * \frac{q_1 * q_2 }{R_f } - 0[/tex]
So
[tex]\frac{1}{2} * m * \frac{1}{8} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R_f }[/tex]
=> [tex]\frac{1}{2} * m *v_o^2 [-\frac{15}{16} ] = k * \frac{q_1 * q_2 }{R_f }[/tex]
=> [tex]- \frac{15}{32} * m *v_o^2 = k * \frac{q_1 * q_2 }{R_f } ---(2)[/tex]
Divide equation 2 by equation 1
[tex]\frac{- \frac{15}{32} * m *v_o^2 }{- \frac{3}{8} * m *v_0^2 } } = \frac{k * \frac{q_1 * q_2 }{R_f } }{k * \frac{q_1 * q_2 }{R } }}[/tex]
=> [tex]-\frac{15}{32 } * -\frac{8}{3} = \frac{R}{R_f}[/tex]
=> [tex]\frac{5}{4} = \frac{R}{R_f}[/tex]
=> [tex]R_f = \frac{4}{5} R[/tex]
Two forces with magnitudes of 6 pounds and 18 pounds are applied to an object. The magnitude of the resultant is 13 pounds. Find the measurement of the angle between the resultant vector and the vector of the 18 pound force to the nearest whole degree.
Answer: the angle between the resultant vector and the vector of the 18 pound is 28°
Explanation:
given that data in the question; as its interpreted in the diagram below;
from the cosine rule, we know that;
a² = b² + c² - 2bc
so
(13)² = (6)² + (18)² - (2 × 6 × 18 ) cos∅
169 = 36 + 324 - 216cos∅
169 = 360 - 216cos∅
216cos∅ = 360 - 169
216cos∅ = 191
cos∅ = 0.8842
∅ = cos⁻¹ ( 0.8842 )
∅ = 27.8° ≈ 28° {nearest whole number}
Therefore the angle between the resultant vector and the vector of the 18 pound is 28°
what belongs in the center section
Answer:
The second one I think
Explanation:
B
To celebrate a victory, a pitcher throws her glove straight upward with an initial speed of 5.3 m/s. How long does it take for the glove to reach its maximum height
Hello!!
For the maximum height the final velocity is zero, because can't up more.
Then, use the formula:
V = Vi + gt
Replacing, we have:
0 m/s = 5,3 m/s + (-9,8 m/s² * t)
0 m/s - 5,3 m/s = -9,8 m/s² * t
(-5,3 m/s) / -9,8 m/s² = t
t = 0,54 s
The time it will take to reach the maximum height is 0,54 seconds.
If a man has
a mass of 115 pounds what what is his mass in gram
Answer:
52,163.1
Explanation:
Thermal expansion of a gas causes a change in the ____ of the gas.
A. Radioactiveness
B. Density
C. Mass
D. Altitude
Answer:
B. Density
Explanation:
Thermal expansion of a gas will cause a definite change in the density of the gas.
It is product of an increase in the average kinetic energy of the system in which molecules are found.
Gaseous thermal expansion will increase the volume in which gases occupy. Mass of the gas will still remain the same. Since density is the mass per unit volume, as the gases expand, their density reduces more and more.Therefore, thermal expansion of gas will cause a change in the density of the gas.
At what angle should the roadway on a curve with a 50-m radius be banked to allow cars to negotiate the curve at 12 m/s even if the roadway is icy (and the frictional force is zero)
Answer:
The value is [tex]\theta = 16.38^o[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 50 \ m[/tex]
The velocity is [tex]v = 12 \ m/s[/tex]
Generally the angle is mathematically represented as
[tex]\theta = tan^{-1} [\frac{v^2}{ gr} ][/tex]
=> [tex]\theta = tan^{-1} [\frac{12^2}{ 9.8 * 50 } ][/tex]
=> [tex]\theta = tan^{-1} [ 0.2939 ][/tex]
=> [tex]\theta = 16.38^o[/tex]
During an isothermal process one mole of a monoatomic gas did 3000 J of work on its surrounding. The final volume and pressure of the gas are 25 L and 1 atm, respectively. What was the initial volume of the gas
Answer: The initial volume of the gas is 7.72 L
Explanation:
For an isothermal process the temperature is constant.
[tex]PV=nRT[/tex]
as P = pressure = 1 atm ,
V = Volume = 25 L
n = moles
R= gas constant
T = temperature
[tex]PV=1atm\times 25L[/tex]
[tex]nRT=25Latm=25\times 101.3J=2532.5J[/tex] (1Latm=101.3 J)
For isothermal reaction :
[tex]w=-2.303nRT\log\frac{V_2}{V_1}[/tex]
where , w = work done by system = -ve
n = moles = 1
[tex]V_2[/tex] = final volume = 25 L
[tex]V_1[/tex] = initial volume = ?
[tex]-3000J=-2.303\times 2532.5\log \frac{25}{V_1}[/tex]
[tex]V_1=7.72L[/tex]
Thus initial volume of the gas is 7.72 L
A ball is thrown with a speed of 100 ft/s in a direction of 30 degrees above the horizontal. determine the horizontal distance in m, the time of flight, and the height to which it rises..Note: you have to label your answer and final answer must be two decimal place
Answer:
Horizontal distance covered is 270.63ft
Time of flight is 3.125secs
Height to which it rises is 39.06feet
Explanation:
The horizontal distance covered is the range;
Range = U²sin2theta/g
Range = 100²sin2(30)/32
Range = 10000sin60/32
Range = 10000(0.8660)/32
Range = 8660/32
Range = 270.625ft
Hence the horizontal distance covered is 270.63ft
The time of flight is expressed using the formula;
T = 2Usin theta/g
g is the acceleration due to gravity
T = 2(100)sin30/32
T = 200(0.5)/32
T = 100/32
T = 3.125secs
Hence the time of flight is 3.125secs
Maximum height H = u²sin²theta/2g
H = 100²(sin30)²/2(32)
H = 10000(0.5²)/64
H = 10000(0.25)/64
H = 2500/64
H = 39.0625
Hence the height to which it rises is 39.06feet
The earth makes one complete revolution around the sun in 365.24 days. Assuming that the orbit of the earth is circular and has a radius of 93,000,000 mi, determine the speedand magnitude of the acceleration of the earth, while treating earth as a point particle.[Hint: you can find the angular speed]Please be consistent with the units.
Answer:
The magnitudes of the speed and acceleration of the Earth are approximately 66,661.217 miles per hour and 47.782 miles per square hour, respectively.
Explanation:
Given that the Earth has a circular orbit and make a revolution at constant speed around the Sun. Then, the kinematic formulas for the speed and acceleration of the planet are, respectively:
Speed
[tex]v = \frac{2\pi\cdot R}{T}[/tex] (1)
Acceleration
[tex]a = \frac{4\pi^{2}\cdot R}{T^{2}}[/tex] (2)
Where:
[tex]v[/tex] - Speed of the planet, measured in miles per hour.
[tex]a[/tex] - Acceleration of the planet, measured in miles per square hour.
[tex]R[/tex] - Radius of the orbit, measured in miles.
[tex]T[/tex] - Period of rotation, measured in hours.
If we know that [tex]R = 93,000,000\,mi[/tex] and [tex]T = 8,765.76\,h[/tex], then the magnitudes of the speed and acceleration of the planet is:
[tex]v = \frac{2\pi\cdot (93,000,000\,mi)}{8,765.76\,h}[/tex]
[tex]v \approx 66,661.217\,\frac{mi}{h}[/tex]
[tex]a = \frac{4\pi^{2}\cdot (93,000,000\,mi)}{(8,765.76\,h)^{2}}[/tex]
[tex]a\approx 47.782\,\frac{mi}{h^{2}}[/tex]
The magnitudes of the speed and acceleration of the Earth are approximately 66,661.217 miles per hour and 47.782 miles per square hour, respectively.
An asteroid has acquired a net negative charge of 149 C from being bombarded by the solar wind over the years, and is currently in equilibrium whereby it expels electrons at the same rate as it acquires them. How many more electrons does it have than protons
Answer:
93.125 × 10^(19)
Explanation:
We are told the asteroid has acquired a net negative charge of 149 C.
Thus;
Q = -149 C
charge on electron has a value of:
e = -1.6 × 10^(-19) C
Now, for us to determine the excess electrons on the asteroid, we will just divide the net charge in excess on the asteroid by the charge of a single electron.
Thus;
n = Q/e
n = -149/(-1.6 × 10^(-19))
n = 93.125 × 10^(19)
Thus, it has 93.125 × 10^(19) more electrons than protons
Which of the following is the best example of Newton’s third law? A. A person riding a bike at a constant speed B. A car accelerating C. Rowing a boat D. An orange rolling off a tray
Answer:
C; Rowing a boat
Explanation:
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how many molecules of o2 are in 8.0 g of oxygen?if the o2 molecules were completely split into 9,how many mole of atoms of oxygen would be obtained?
A child pulls a sled up a snow covered hill. If the child does 504J of work on the sled while pulling the sled 23m up the hill then how much force did they exert?
Explanation:
ans is equal to 504j* 23 m* 10 ms
How would taping a coin to the balloon affect the overall motion of the balloon?
Answer:
the motion of the coin taping the balloon is the balloon squshing down
In order for convection to transfer heat, particles need to
Select one:
a.
absorb solar and fossil energy
b.
circulate and move within a liquid or gas
c.
make contact with the heat source
d.
transmit electromagnetic waves
Answer:
the answer is b mark me as the brqinlist
A spanner is an example of:
i. screw ii. wheel and axle
iii. pulley
iv. wedge
formula
Answer:
The answer is "wedge"
Explanation:
A wedge is an item that tightens to a meager edge. Pushing the wedge one direction makes a force in a sideways direction. It is normally made of metal or wood and is used for parting, tightening, lifting, or fixing, as in making sure about a hammer head onto its handle.
The wedge was used in ancient occasions to part logs and shakes; an ax is also a wedge, similar to the teeth on a saw. As far as its mechanical capacity, the screw might be considered as a wedge folded over a cylinder.
Compare the weight of a 60 kg person on the earth with the weight of the same person on
the moon. Then, describe a quick (but very costly) way for dieters
at NASA to lose weight.
Answer:
Explanation:
The formula for weight is
W = mg, where
W = the weight of the object or person
m = mass of the object or person
g = acceleration due to gravity
Now, we're given the mass of the person to be 60 kd, and thus, the weight of that person would be
W = 60 * 9.81
W = 588.6 N
On the surface of the moon, the weight of the person would be
W = 60 * 1.625
W = 97.5 N
Therefore, the weight of the person on both surfaces are 588.6 and 97.5 respectively
Question 4(Multiple Choice Worth 2 points)
Satellite technology has allowed us to monitor areas that were once so remote that we had no way of monitoring them. Review the statements and choose the one that best describes how we can use this technology to benefit the environment.
Large ocean liners use satellites to aid in navigation. This type of navigation aid is also used on cruise ships. Using satellites can help these large ships navigate waters during bad storms.
Satellite surveillance can be used to aid in port security. Satellite data can be monitored and evaluated by complex computer systems. These systems can alert officials when something unusual occurs.
Satellite technology allows us to monitor and track the changes in the polar ice sheets. This can help us plan and direct conservation efforts for the animals that depend on polar ice in order to survive.
Satellite surveillance has allowed us to map areas that were previously too remote to observe. This helps developers to locate and choose sites for large resorts.
Answer:
satellite technology helped large ocean liners guide you to our space mission.
Explanation:
Large ocean liners use satellites to aid in navigation. This type of navigation aid is also used on cruise ships. Using satellites can help these large ships navigate waters during bad storms.
Satellite surveillance can be used to aid in port security. Satellite data can be monitored and evaluated by complex computer systems. These systems can alert officials when something unusual occurs.
Satellite technology allows us to monitor and track the changes in the polar ice sheets. This can help us plan and direct conservation efforts for the animals that depend on polar ice in order to survive.
Satellite surveillance has allowed us to map areas that were previously too remote to observe. This helps developers to locate and choose sites for large resorts.
Nuclear power plants produce energy using:________
a. nuclear fission.
b. nuclear fusion.
c. combustion.
d. critical mass.
Answer: a. nuclear fission.
Explanation:
Nuclear power plants use heavy materials, like uranium 238, which is bombarded with neutrons. When the neutrons interact with the uranium, there is fission (a division of one atom into smaller ones and other subatomic particles), and in this process, more neutrons are produced. These new neutrons may impact other atoms of uranium and continue with this cycle, generating heat in the process.
Then we can conclude that nuclear power plants produce energy using nuclear fission.
You might think that an ultraviolet light shining on an initially uncharged electroscope would cause the electroscope to become positively charged as photoelectrons are emitted. In fact, ultraviolet light has no noticeable effect on an uncharged electroscope. Why not
Answer:
Positively Charge formation on the electroscope pulls the electrons so no emission happens.
Explanation:
Photoelectric effect is the process by which the electrons get ejected when light of certain frequency falls on the material. The energy from the electromagnetic radiation excites the electron by providing it enough energy as a result it gets ejected.
The Ultraviolet(UV) light is a form of electromagnetic radiation, If the UV radiation strikes an uncharged electroscope, it is predicted that it will cause the electrons in the electroscope to excite and to be emitted. But on the contrary no noticeable effect is observed on the electroscope this happened because when UV light makes the electroscope positively(+) charged this creates a pull for the electrons, as a result rather emitting the electrons gets pulled by the positively charged electroscope.
Therefore the positively charge formation on the electroscope hinder the electrons emission and there is no noticeable effect.
A body travels 10 meters during the first 5 seconds of its travel, and it travels a total of 30 meters over the first 10 seconds of its travel. What is
its average speed during the time from t 5 seconds to t - 10 seconds?
A. 2 meters/second
В. 3 meters/second
С. 4 meters/second
D. 5 meters/second
E. 6 meters/second
A 0.3 g mosquito is flying toward a girl with a speed of 4.5 mph. Just before landing on the girl, the fly is swatted straight back at a speed of 12 mph. If the fly swatter and the fly were in contact for 0.2 s, what is the force that was exerted on the fly
Answer:
1.1x10^-2N
Explanation:
We have the change in momentum as
P = 0.3(4.5+12)g.mph
= 0.3x0.447x(4.5+12)x10^-3
Then the force that is exerted will be
F = p/∆t
∆t = 0.2
= 0.3x0.447x(4.5+12)x10^-3/0.2
= 0.1341x16.5x10^-3/0.2
= 1.1x10^-2
Therefore the force that was exerted is equal to 1.1x10^-2
The required magnitude of the force exerted on the fly is of [tex]5.025 \times 10^{-3} \;\rm N[/tex].
Given data:
The mass of mosquito is, [tex]m =0.3 \;\rm g =3 \times 10^{-4} \;\rm kg[/tex]
The speed of flying is, u = 4.5 mph = 4.5 ( 0.447) = 2.01 m/s.
The swatting speed of mosquito is, v = 12 mph = 12 (0.447 ) = 5.36 m/s.
The time of contact is, t = 0.2 s.
In this problem, we will first calculate the change in momentum, and the change in momentum is given as,
p = m ( v - u)
Solving as,
[tex]p = 3 \times 10^{-4} (5.36 - 2.01)\\\\p = 1.005 \times 10^{-3} \;\rm kg.m/s[/tex]
Now as per the Newton's second law,
[tex]F = p/t\\\\F = 1.005 \times 10^{-3} / 0.2\\\\F= 5.025 \times 10^{-3} \;\rm N[/tex]
Thus, the required magnitude of the force exerted on the fly is of [tex]5.025 \times 10^{-3} \;\rm N[/tex].
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