Answer:
How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistors R1 ,R2 and R3 in series connected to a battery of V volt.
we can connect voltmeter across each resistor
V = V1 + V2 + V3
I1 = current across resistor R1
I1 = V1/R1
similarly
I2 = current across resistor R2
I2= V2/R2
I3 = current across resistor R3
I3 = V3/R3
we will see that
I1 = I2 = I3
also it will be equal to V/(R1 + R2+R3)
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
You and your buddy are scuba diving and notice that air bubbles triple in volume as they rise to the surface from where you are in the ocean. Ignoring any temperature changes, how far below the surface of the water (in m) are you when these bubbles are released
Answer:
You and your buddy are scuba diving and notice that air bubbles triple in volume as they rise to the surface from where you are in the ocean. Ignoring any temperature changes, how far below the surface of the water (in m) are you when these bubbles are released? The ocean has a density of 1034 kg/m3
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:
Which of these is considered a contact force?
a the force between two magnets.
b the gravitational pull of a planet.
c the friction between an object and air.
d The force between two charged particles.
Answer:
C -) The friction between an object and air.
Explanation:
The frictional force is the force that exists between two surfaces in contact, which is opposed to the movement.
That is, this type of force exists as long as there is physical contact between surfaces.
While the other types of force always act at a separation distance between bodies.
How does radiation transfer thermal energy from the Sun to Earth?
Answer:
The thermal energy is carried by electromagnetic waves
Explanation:
There are three types of transfer of heat (thermal energy):
- Conduction: conduction occurs when two objects/two substances are in contact with each other. The heat is transferred from the hotter object to the colder object by the collisions between the molecules of the two mediums.
- Convection: convection occurs when a fluid is heated by an external source of heat. The part of the fluid closer to the heat source gets warmer, therefore it becomes less dense and it rises, and it is replaced by the colder part of the fluid, which is colder. Then, this part of fluid is heated as well, so it gets warmer, it rises, etc.. in a cycle.
- Radiation: radiation occurs when thermal energy is carried by electromagnetic waves. Since electromagnetic waves do not need a medium to propagate, this is the only method of heat transfer that can occur through a vacuum (so, in space as well).
Indeed, the Sun emits a lot of electromagnetic radiation, which travels through space and eventually reaches the Earth, heating it.
Explanation:
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]
Mass of egg is 65g and volume is 60cc.The mass of salt dissolved in 100cc of water so that egg just float in water is
Answer:
108.33 grams.
Explanation:
Given that,
Mass of a egg, m = 65 g
Volume of an egg, V = 60 cc
We need to find the mass of the salt dissolved in 100cc of water so that egg just float in water.
The density will remains constant,
[tex]\dfrac{m_1}{V_1}=\dfrac{m_2}{V_2}\\\\m_2=\dfrac{m_1V_2}{V_1}\\\\m_2=\dfrac{65\times 100}{60}\\\\=108.33\ g[/tex]
So, the required mass is 108.33 grams.
Two workers are sliding 500 kg crate across the floor. One worker pushes forward on the crate with a force of 440 N while the other pulls in the same direction with a force of 340 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor
Answer:
0.159
Explanation:
Given that
F1 = 440 N
F2 = 340 N
m = 500 kg
acceleration, a = 0 m/s²
Remember that F = m.a
If F - f(f) = 0, then F = f(f)
N = mg
f(f) = μN, substituting for N, we have
f(f) = μmg
The total force, F = F1 + F2
F = 440 + 340
F = 780 N
Recall that we'd already proven that
F = f(f), so F = 780 N = f(f)
And again, f(f) = μmg, if we substitute for all the values, we have
780 = μ * 500 * 9.81
780 = μ * 4905
μ = 780 / 4905
μ = 0.159
Therefore, the coefficient of static friction of the crate on the floor is 0.159
Mark walked 2 miles east than 1 mile north how would you determine his total displacement
They could determine it by counting their total amount of miles mark went by the directions he went to walk those miles.
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]
Chemical reactions and nuclear reactions cause matter to change in different
ways. Which two statements describe how matter changes only in a chemical
reaction?
A. The nuclei of the atoms change.
B. Atoms and their electrons are rearranged.
O C. New atoms and elements are formed.
D. The elements stay the same.
Answer:
B. Atoms and their electrons are rearranged.
D. The elements stay the same.
Oxygen-15 undergoes beta decay. Which particle correctly completes the
equation to show that the numbers of nucleons on the two sides of the
equation are equal?
150 - 15N+ ?
Answer: [tex]_8^{15}\textrm{O}\rightarrow _{7}^{15}\textrm{N}+_{+1}^0{e}[/tex]
Explanation:
Positron emission: In this process, in which a proton inside a radionuclide nucleus is converted into a neutron while releasing a positron and an electron neutrino. The beta positive particle released is basically a positron with +1 charge and no mass.
General representation of an element is given as:
[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{+1}^0{e}[/tex]
where,
Z represents Atomic number
A represents Mass number
X and Y represents the symbol of an element
For [tex]_8^{15}\textrm{O}\rightarrow _{7}^{15}\textrm{N}+_{+1}^0{e}[/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:
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:
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
A person who weighs 614 N is riding a 85-N mountain bike. Suppose the entire weight of the rider plus bike is supported equally by the two tires. If the gauge pressure in each tire is 9.00 x 10^5 Pa. What is the area of contact between each tire and the ground?
Answer:
3.88 × 10^-4 m
Explanation:
Given that a person who weighs 614 N is riding a 85-N mountain bike. Suppose the entire weight of the rider plus bike is supported equally by the two tires. If the gauge pressure in each tire is 9.00 x 10^5 Pa. What is the area of contact between each tire and the ground?
The total weight = 614 + 85
The total weight = 699N
Let the total area of contact = A
Pressure = Force / A
Substitute all the parameters into the formula
900000 = 699 /A
A = 699 / 900000
A = 7.77 × 10^-4 m
The area of contact between each tire and the ground will be = A/2
That is, 7.77 / 2 = 3.88 × 10^-4 m
Therefore, the area of contact between each tire and the ground is 3.88 × 10^-4 m approximately.
A person pushes a box filled with books with a force of 120 newtons a distance of 30 meters within 3 minutes Calculate the work the person did and the amount of power required
120/30-4
I will give brainliest
Ist Law: a object continues in a state of
rest or motion unless an external force
applied to it
This image shows an example of
the first law because:
Answer:
Explanation:
This image is the example of Newton's first law of motion because:
As per Newton's first law, football will remain in the state of rest until a player applies an external force by kicking the ball.
And the ball will keep on moving until unless the net of a goal post exerts the external force to stop the ball.
What is the magnitude of the velocity after it hits the
ground?
9.3 m/s
12 m/s
41 m/s
73 m/s
Answer: 9.3m/s
Explanation:
Your question isn't complete but let me help out:
A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball. What is the magnitude of the velocity after it hits the ground?
We would use Newton's law of motion to solve this which goes thus:
F = ma
f = m(v-u)/t
Cross multiply
ft = m(v-u)
where,
f = 55
t = 45/1000 = 0.045
m = 0.0060
u = -32
v = Unknown
Therefore,
55 × 45/1000 = 0.060(v - -32)
55 × 0.045 = 0.060(v + 32)
2.475 = 0.06(v + 32)
2.475 = 0.06v + 1.92
0.06v = 2.475 - 1.92
0.06v = 0.555
v = 0.555/0.06
v = 9.25m/s
v = 9.3m/s Approximately
Answer:
A.9.3 m/s
Explanation:
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
Please help
4. What are the lowest points on a transverse wave called?
a. Crests
b. troughs
C. compressions
d. rarefractions
5. Any substance that a wave moves through is called a
a. medium
b. vibrate
C. crest
d. frequency
Answer:
Explanation:
the lowest points of a transverse wave are called the troughs what is the wavelength of a wave traveling through a rope if the distance from one crest to the next is 1 meter a. 2 m
4. Low points are called troughs
5. Medium: substance through which a wave can travel
oo hi granger ru online here i have a doubt in physics .
Answer:
.
Explanation:
the tighter the threads of the screw the greater
Two ice skater are at rest, andy and brenda. andy has a mass of 62.5 kg. they push off each other. after, andy moves 1.59 m/s east, while brenda moves 2.22 m/s west. what is brenda's mass?
Unit=kg
Answer: 44.76 kg
Explanation: There are two options for formulas to use here.
0=(m1)(V1f)+(m2)(V2f)
Or
(m1)(V1f) = —(m2)(V2f)
Both formulas should give you an answer of 44.763514, which can be rounded to the specifications of whatever your curriculum or teacher wants. Most likely 44.76.
Blessings to all of you in the Name of Jesus Christ our Savior!
A rotating space station simulates artificial gravity by means of centripetal acceleration at the rim. The radius of the station is 1156 m and the apparent acceleration of gravity at the rim is 6.13 m/s2. What is the rotation rate of the station in rpm (revolutions per minute).
a= 1156× 6.13= 7086.28
Explanation:
a = m× m/ s^2
It is difficult to lift a bigger stone than the smaller
stone. why?
plzzz give me short and brillient answer
18
AX
A bicycle rider travels 15 km in 1.25 hours. What is the rider's average speed? v=
At
rage speed [v-ante
10.5 km/h
13.75 km/h
12 km/h
22.5 km/h
The required average speed of bicycle rider is 12 km/h.
Given data:
The distance travelled by the bicycle rider is, d = 15 km.
Time taken to cover the distance is, t = 1.25 h.
Here, we need to use the simple relation between the speed, time and distance. The distance covered covered by any object per unit of time is known as average speed of object. And the mathematical expression for the average speed is given as,
[tex]v = \dfrac{d}{t}[/tex]
Here, v is the average speed.
Solving as,
[tex]v = \dfrac{15}{1.25}\\\\v = 12\;\rm km/h[/tex]
Thus, we can conclude that the required average speed of bicycle rider is 12 km/h.
Learn more about the average speed here:
https://brainly.com/question/12322912
In seismology, the P wave is a longitudinal wave. As a P wave travels through the Earth, the relative motion between the P wave and the particles is
Answer:
A longitudinal wave is a wave where the displacement of the medium is in the same direction than the propagation of the wave.
This means that as a P wave travels through the Earth, the relative motion between the p wave and the particles is near zero, as the motion of the particles is parallel to the motion of the wave.
An example of this would be the waves generated when you throw a rock in water, you can see how the water particles move along the waves in the water.
The relative motion between the p wave and the particles is almost negligible.
In case of a longitudinal wave, the displacement of the medium is in the same direction as the direction of the propagation of the wave. This property suggests that as a P wave travels through the Earth, the particles also move along in the same direction. The relative motion between the p wave and the particles is almost negligible. In case of transverse waves, the displacement of particles of the medium and the direction of the propagation of the wave is perpendicular to each otherLearn more:
https://brainly.com/question/26095273
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/20360303A 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.