What determines the type of air mass that forms in an area?
Question 11 options:
The amount of oxygen present
The amount of air present
The direction of air flow
The location where it forms
Answer: no because you have left the number
Explanation:
A toy car weighing 3.2 N gets pushed a distance of 10 m in 1.3 s. What is the momentum of the car?
Answer:
2.51 kg * m/s
Explanation:
In order to find momentum, use the equation below:
momentum = mass * velocity.
Since neither mass nor velocity was given, you must solve for both variables.
In order to solve for mass, use the force equation for its weight / gravitational force.
Fg (gravitational force) = 3.2 N = ma = 9.8m
mass = 3.2 N / 9.8 m/s^2 = 0.326531 kg
In order to solve for velocity, use the equation:
velocity = displacement / time
velocity = 10m / 1.3 s = 7.69231 m/s
Momentum = mass * velocity = 0.326531 kg * 7.69231 m/s = 2.51177 kg * m/s = 2.51 kg * m/s
Define each of the three heat transfer methods:
Conduction
Convection
Radiation
The temperature of the surface of the Sun is 5500°C.
a. What is the average kinetic energy, in joules, of hydrogen atoms on the surface of the Sun?
b. What is the average kinetic energy, in joules, of helium atoms in a region of the solar corona where the temperature is 6.00 times 10^5 K?
Answer:
a. the average kinetic energy of hydrogen atoms is 1.20 × 10^-19J
b. the average kinetic energy of helium atoms is 1.24 × 10^-17J
Explanation:
The computation is shown below;
As we know that
Kinetic energy = 3 ÷ 2 kT
where,
K = Boltzmann constant
And, T = Temperature
a. Now the temperature in kelvin is
T = (5,500 × (°C ÷ K) + 273.15 K)
= 5773.15 K
As
Kinetic energy = 3 ÷ 2 kT
So now 1.38 × 10^-23 J/K for K would be substituted and 5773.15 K for Temperature T
Now Kinetic energy is
= 3 ÷ 2 (1.38 × 10^-23 J/K) ( 5773.15 K)
= 1.20 × 10^-19J
hence, the average kinetic energy of hydrogen atoms is 1.20 × 10^-19J
b. As
Kinetic energy = 3 ÷ 2 kT
now 1.38 × 10^-23 J/K for K would be substituted and 6 × 10^5K for Temperature T
Now Kinetic energy is
= 3 ÷ 2 (1.38 × 10^-23 J/K) (6 × 10^5K )
= 1.24 × 10^-17J
hence, the average kinetic energy of helium atoms is 1.24 × 10^-17J
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]
In order to model the motion of an extinct ape, scientists measure its hand and arm bones. From shoulder to wrist, the arm bones are 0.60 m long and their mass is 4.0 kg. From wrist to the tip of the fingers, the hand bones are 0.10 m long and their mass is 1.0 kg. In the model above, each bone is assumed to have a uniform density. When the arm and hand hang straight down, the distance from the shoulder to the center of mass of the arm-hand system is most nearly
Answer:
0.37 m
Explanation:
Let the shoulder be the origin.
The center of mass of the arm bones is 0.60 m/2 = 0.30 m and the center of mass of the hand bones is 0.10 m/2 = 0.05 m since they are modeled as straight rods with uniform density and the center of mass of a rod is x = L/2 where L is the length of the rod.
The center of mass y = (m₁y₁ + m₂y₂)/(m₁ + m₂) where m₁ = mass of arm bones = 4.0 kg, y₁ = distance center of mass of arm bones from shoulder = 0.30 m, m₂ = mass of hand bones = 1.0 kg and y₂ = distance of center of mass hand bones from shoulder = x₁ + distance of center of hand bones from wrist = 0.60 m + 0.05 m = 0.65 m
Substituting these into the equation for the center of mass, we have
y = (m₁y₁ + m₂y₂)/(m₁ + m₂)
y = (4.0 kg × 0.30 m + 1.0 kg × 0.65 m)/(4.0 kg + 1.0 kg)
y = (1.20 kgm + 0.65 kgm)/5.0 kg
y = 1.85 kgm/5.0 kg
y = 0.37 m
The distance of the center of mass from the shoulder is thus y = 0.37 m
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.
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 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
g Estimate the number of photons emitted by the Sun in a second. The power output from the Sun is 4 × 1026 W and assume that the average wavelength of each photon is 550 nm.
Answer:
The value is [tex]N = 1.107 *10^{45 } \ photons[/tex]
Explanation:
From the question we are told that
The power output from the sun is [tex]P_o = 4 * 10^{26} \ W[/tex]
The average wavelength of each photon is [tex]\lambda = 550 \ nm = 550 *10^{-9} \ m[/tex]
Generally the energy of each photon emitted is mathematically represented as
[tex]E_c = \frac{h * c }{ \lambda }[/tex]
Here h is the Plank's constant with value [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]E_c = \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 550 *10^{-9} }[/tex]
=> [tex]E_c = 3.614 *10^{-19} \ J[/tex]
Generally the number of photons emitted by the Sun in a second is mathematically represented as
[tex]N = \frac{P }{E_c}[/tex]
=> [tex]N = \frac{4 * 10^{26} }{3.614 *10^{-19}}[/tex]
=> [tex]N = 1.107 *10^{45 } \ photons[/tex]
diffine thermodynamics
Answer:
The science of the conversions between heat and other forms of energy is thermodynamics.
Hope it helps :)
Answer:
thermodynamic is the branch of physics that deals with the relationship between heat and other forms of energy .it describes how thermal energy is converted to other forms and how it will affect matter.
Explanation:
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:
A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of 1.72 kg. The spring is released and the object moves along a frictionless surface when it reaches a small embankment. If the speed of the object is 2.45 m/s at location A, what is the embankment height h
Given :
A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of 1.72 kg.
To Find :
The embankment in the height.
Solution :
Since no external force is acting in the system, therefore total energy will be conserved.
Initial kinetic energy of the object = Energy stored in spring
[tex]K.E _i = \dfrac{kx^2}{2}\\\\K.E_i = \dfrac{1730\times 0.136^2}{2}\\\\K.E_i = 16\ J[/tex]
Also, initial potential energy is 0.
Now,
[tex]K.E_i + P.E_i = K.E_f + P.E_f\\\\16 + 0 = \dfrac{1.72\times 2.45^2}{2}+ mgh\\\\mgh =16 - 5.16\\\\h = \dfrac{16 - 5.16}{1.72 \times 9.8}\\\\h = 0.64\ m[/tex]
Therefore, the embankment height is 0.64 m.
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
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
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.
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 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.
a box is 30 cm wide, 40 cm long and 25 cm high calculate: what is the area of its base.
Answer:
1200cm²
Explanation:
Width= 30cm
Length= 40cm
Area of base(A) = Width×Length
= 30cm×40cm
= 1200cm²
Taking into account the definition of reactanguar prism and area of rectangle, the area of its base is 1200 cm².
A rectangular prism is a polyhedron whose surface is formed by two equal and parallel rectangles called bases and by four lateral faces that are also parallel and equal rectangles two by two.
So, the base being a rectangle, its area is calculated as the multiplication between the base and the height. In this case, these values correspond to the width and length of the figure.
Then, in this case, you know:
Width= 30cm Length= 40cm
Being:
Area of base (A) = Width×Length
Then, the are of base (A) is calculated as:
Area of base (A)= 30cm×40cm
Solving:
Area of base (A)= 1200cm²
Finally, the area of its base is 1200 cm².
Learn more:
https://brainly.com/question/20360303Including 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.
oo hi granger ru online here i have a doubt in physics .
Answer:
.
Explanation:
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?
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
If a man has
a mass of 115 pounds what what is his mass in gram
Answer:
52,163.1
Explanation:
It is difficult to lift a bigger stone than the smaller
stone. why?
plzzz give me short and brillient answer
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be when it is very far from the Earth
Answer:
The value is [tex]v = 2.3359 *10^{4} \ m/s[/tex]
Explanation:
From the question we are told that
The initial speed is [tex]u = 2.05 *10^{4} \ m/s[/tex]
Generally the total energy possessed by the space probe when on earth is mathematically represented as
[tex]T__{E}} = KE__{i}} + KE__{e}}[/tex]
Here [tex]KE_i[/tex] is the kinetic energy of the space probe due to its initial speed which is mathematically represented as
[tex]KE_i = \frac{1}{2} * m * u^2[/tex]
=> [tex]KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2[/tex]
=> [tex]KE_i = 2.101 *10^{8} \ \ m \ \ J[/tex]
And [tex]KE_e[/tex] is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as
[tex]KE_e = \frac{1}{2} * m * v_e^2[/tex]
Here [tex]v_e[/tex] is the escape velocity from earth which has a value [tex]v_e = 11.2 *10^{3} \ m/s[/tex]
=> [tex]KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2[/tex]
=> [tex]KE_e = 6.272 *10^{7} \ \ m \ \ J[/tex]
Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as
[tex]KE_p = \frac{1}{2} * m * v^2[/tex]
Generally from the law energy conservation we have that
[tex]T__{E}} = KE_p[/tex]
So
[tex]2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2[/tex]
=> [tex]5.4564 *10^{8} = v^2[/tex]
=> [tex]v = \sqrt{5.4564 *10^{8}}[/tex]
=> [tex]v = 2.3359 *10^{4} \ m/s[/tex]
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
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
Car 1 drives 20 mph to the south, and car 2 drives 30 mph to the north. From
the frame of reference of car 1, what is the velocity of car 2?
Answer:
10 mph faster than car 1 is going
Explanation:
what belongs in the center section
Answer:
The second one I think
Explanation:
B