Answer: we divide this equation by pV and use {C}_{p}={C}_{V}+ . ... The temperature, pressure, and volume of the resulting gas-air mixture
Explanation:
The new pressure of the gas relative to its initial pressure P₁ is P₁ × 1.09
From the question given above, the following data were obtained:
Initial temperature (T₁) = 20 °C = 20 + 273 = 293 K
Initial pressure (P₁) = P₁
New temperature (T₂) = 46 °C = 46 + 273 = 319 K
New pressure (P₂) =?The new pressure of the gas can be obtained as follow:
P₁ / T₁ = P₂ / T₂
P₁ / 293 = P₂ / 319
Cross multiply
293 × P₂ = P₁ × 319
Divide both side by 293
P₂ = (P₁ × 319) / 293
P₂ = P₁ × 1.09Thus, the new pressure relative to it's initial pressure is P₁ × 1.09
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what is the acceleration of a satellite moving in a circular orbit around the earth of radius 2r
Explanation:
You do the radius times the circumference of the earth
If the kinetic energy of the 40kg box is 784 J, what is the velocity before it strikes the ground?
Answer:
Explanation:
[tex]KE=\frac{1}{2}mv^2[/tex]
[tex]784=\frac{1}{2}(40)v^2[/tex]
[tex]784=20v^2[/tex]
[tex]39.2=v^2[/tex]
[tex]v=6.26m/s[/tex]
If the magnetic force is 3.5 × 10–2 N, how fast is the charge moving?
Answer:
D
Explanation:
Took it on edg
The speed of the charge at the given magnetic force and field is determined as 1.1 x 10⁴ m/s.
Speed of the chargeThe speed of the charge is calculated as follows;
F = qvBsinθ
v = F/qBsinθ
where;
F is the magnetic forceB is magnetic fieldv is speed of the chargev = (3.5 x 10⁻²)/(8.4 x 10⁻⁴ x 6.7x 10⁻³ x sin35)
v = 10,842.33 m/s
v ≅ 1.1 x 10⁴ m/s
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How can you describe the orbital period (planetary year) of each planet?
Answer:
How can you describe the orbital period (planetary year) of each planet?
A year is defined as the time it takes a planet to complete one revolution of the Sun, for Earth this is just over 365 days. This is also known as the orbital period. Unsurprisingly the the length of each planet's year correlates with its distance from the Sun.
Explanation:
Answer: The orbital period is how long it takes a plane to fulfill its "revolution"(365 days)
Explanation:
A stone of mass 1kg is thrown at 10m/s upwards making an angle of 37°with the horizontal from a building that is 20m high. Using the law of conservation of energy calculate the speed wjen the stone hits the ground.
Answer:
31.68 m/s
Explanation:
The law of conservation of energy states that energy is not lost or gained it is just converted, in this example, since it is not given any resistance from the wind, you'd have two variables Speed on the Y-axis and on the X-axis, since both of them would result in the same decrease and increase with against gravity, it doesn't matter the value of both.
As the stone continues to go upwards it will continue to lose speed due to de-acceleration from the gravity acting on it, similarly, it will continue to gain Potential energy, instead of kinetic energy, when it reaches its highest point the speed on Y will be "0" and the free fall will start, since the up and down movement will be equal in time the and acceleration would be equal -9.81 m/s and 9.81 m/s because the only acceleration you have is gravity, you only need to calculate how much speed will gain a rock accelerating at 9.81 m/s falling 20 m:
[tex]H=\frac{1}{2}g*t^{2} \\\frac{20}{1/2*9.81} =t^{2} \\\\t^{2} =4.08\\t=\sqrt{4.08} \\t=2.21[/tex]
Now we just add the time accelerating:
[tex]Vf=Vo+at\\Vf=10 m/s+ 2.21*9.81\\Vf=10 m/s+21.68\\Vf=31.68 m/s[/tex]
Using Figure 2, what is the momentum of Train Car A before the collision?
A
180,000 kg*m/s
B
0 kg*m/s
C
11,250 kg*m/s
D
4 kg*m/s
Answer:
Option A. 180000 Kgm/s.
Explanation:
From the question given above, the following data were obtained:
For Train Car A:
Mass of train car A = 45000 Kg
Velocity of train car A = 4 m/s
Momentum of train car A =?
For Train Car B:
Mass of train car B = 45000 Kg
Velocity of train car B = 0 m/s
Momentum is simply defined as the product of mass and velocity. Mathematically, it can be expressed as:
Momentum = mass × velocity
With the above formula, the momentum of train car A before collision can be obtained as follow:
Mass of train car A = 45000 Kg
Velocity of train car A = 4 m/s
Momentum of train car A =?
Momentum = mass × velocity
Momentum = 45000 × 4
Momentum of train car A = 180000 Kgm/s
A skateboarder travels on a horizontal surface with an initial velocity of 4.2 m/s toward the south and a constant acceleration of 2.6 m/s^2 toward the east. Let the x direction be eastward and the y direction be northward, and let the skateboarder be at the origin at t=0.
Required:
a. What is her x position at t=0.60s?
b. What is her y position at t=0.60s?
c. What is her x velocity component at t=0.60s?
d. What is her y velocity component at t=0.60s?
Answer:
Explanation:
Let the skateboarder's movement in x -direction be taken into consideration .
a )
initial velocity in x direction u = 0
acceleration a = 2.6 m /s²
time t = .6 s
displacement in x direction
s = ut + 1/2 a t²
= 0 + .5 x 2.6 x .6²
= .468 m
= 46.80 cm
c )
velocity after .6 s
v = u + at
= 0 + 2.6 x .6
= 1.56 m /s
Let the skateboarder's movement in y -direction be taken into consideration .
b ) initial velocity in y direction u = - 4.2 m /s ( velocity is towards south or - y direction )
acceleration a = 0
time t = .6 s
displacement in y direction
s = ut + 1/2 a t²
= - 4.2 x .6 + 0
= - 2.52 m
2.52 m towards south .
d )
velocity after .6 s
v = u + at
= - 4.2 + 0
= - 4.2 m /s
or 4.2 m /s towards south .
Solids have a definite shape and volume this is because
High-voltage power lines are a familiar sight throughout the country. The aluminum wire used for some of these lines has a cross-sectional area of 4.8 x 10-4 m2. What is the resistance of 14 kilometers of this wire
Answer:
Explanation:
For resistance of a wire , the formula is as follows
R = ρ L / S
where ρ is specific resistance , L is length and S is cross sectional area
Given L = 14 000 m ,
S = 4.8 x 10⁻⁴ m²
specific resistance of aluminum = 2.8 x 10⁻⁸ ohm-meter
Putting the values in the formula
R = 2.8 x 10⁻⁸ x 14 x 10³ / (4.8 x 10⁻⁴ )
R = 0.8167 ohm .
= .82 ohm .
Question 2 of 10 How many oxygen (O) atoms are in a molecule of C3 H4 03? A. 3 OB. 4 C. 10 ОО D. 1
Answer:
a. 3
Explanation:
Answer:
A. 3
Explanation:
AP3X
The all-digital touch-tone phones use the summation of two sine waves for signaling. Frequencies of these sine waves are defined as 697, 770, 852, 941, 1209, 1336, 1477, and 1633 Hz. Since the sampling rate used by the telecommunications is 8000 Hz, convert those eight analog frequencies into digital frequencies of radians and cycles.
How do we know that an object has accelerated?
Two objects are located on an airtrack (you may assume there is no friction). The magnitude of the charge on object A is 5 times higher than on object B, and object A also has a 6 times higher mass than object B (the picture is not scale or necessarily in the correct direction). Each object is accelerating in the direction of the arrows.
a. Draw a system schema.
b. Draw the force body diagrams for each charge and identify all newton 3d law pairs.
c. Write (in symbolic form) expressions for the net force on each object.
d. Find how many times faster/slower the acceleration of object B is compared to object A?
Answer:
Explanation:
From the information given:
The schematic diagram for the system and the force body diagrams for each charge can be seen in the image attached below
(c)
The symbolic expressions for the net force on each object is as follows:
[tex]N_A = m_Ag \\ \\ F_{net}, A= F_A = \dfrac{kQ_AQ_B}{r^2}[/tex]
[tex]N_B = m_Bg \\ \\ F_{net}, B= F_B = \dfrac{kQ_AQ_B}{r^2}[/tex]
(d) From above
[tex]F_A =F_B[/tex]
[tex]m_Aa_A = m_Ba_B \\ \\ (6m_B)a_A = m_B a_B[/tex]
[tex]a_B = 6a_A[/tex]
Question: How did NASA use Newton's Laws to land the Perseverance lander,
safely
on Mars?
Please help asap.
Answer:
If the thrust is increased, the aircraft accelerates and the velocity increases. This is the second part sited in Newton's first law; a net external force changes the velocity of the object. The drag of the aircraft depends on the square of the velocity. So the drag increases with increased velocity.
Write any two uses of plane mirrors?
Answer:
Uses of plane mirrorsThey are used in periscopes, for signalling, in kaleidoscopes, to see round dangerous bends, in meters, as mirror tiles, in a sextant, in an overhead projector, an SLR camera, car wing mirrors, in microscopes and as reflecting number plates to mention only some!
Explanation:
Hope it is helpful....
Answer:
two uses are:
they are using for looking glassthey are used to make periscopeA constant electric field of 5.00 N/C points along the positive x-direction. An electron, initially at rest, moves a distance of 2.00 m in this space. How fast is the electron moving after its 2.00 m journey
Answer:
1.875 x 10⁶ m /s .
Explanation:
Force on electron = E e where E is electric field and e is charge on electron
acceleration generated = Ee / m where m is mass of the electron .
Putting the values
acceleration generated = 5 x 1.6 x 10⁻¹⁹ / 9.1 x 10⁻³¹
= .879 x 10¹² m /s²
v² = u² + 2 as , initial velocity u = 0 , displacement s = 2 m
v² = 0 + 2 x .879 x 10¹² x 2
v = 1.875 x 10⁶ m /s .
Astronomers discover a planet orbiting around a star similar to our sun that is 35 light years away. How fast must a rocket ship go if the round trip is to take no longer than 70 years in time for the astronauts aboard
Answer:
[tex]v = 0.7071c[/tex]
Explanation:
Given
Distance to the planet = 35 light years. So, the entire distance is: 2 * 35 = 70.
[tex]\triangle{x'} = 70[/tex]
[tex]T_0 = 70\ years[/tex] i.e time of travel of the ship.
For the observer on earth, the time is:
[tex]T' = \gamma T_0[/tex]
The required speed so that it does not take more than 70 years is then calculated using:
[tex]\triangle x' = vT'[/tex]
Substitute [tex]T' = \gamma T_0[/tex]
[tex]\triangle x' = v\gamma T_0[/tex]
[tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]
So, we have:
[tex]\triangle x' = \frac{vT_0}{\sqrt{1 - v^2/c^2}}[/tex]
Make v the subject of formula.
Square both sides
[tex]\triangle x'^2 = \frac{v^2T^2_0}{1 - v^2/c^2}[/tex]
Cross Multiply
[tex](1 - \frac{v^2}{c^2}) *\triangle x'^2 = v^2T^2_0[/tex]
Divide both sides by [tex]\triangle x'^2[/tex]
[tex](1 - \frac{v^2}{c^2}) = \frac{v^2T^2_0}{\triangle x'^2}[/tex]
Divide through by [tex]v^2[/tex]
[tex](\frac{1}{v^2} - \frac{v^2}{v^2*c^2}) = \frac{v^2T^2_0}{v^2\triangle x'^2}[/tex]
[tex]\frac{1}{v^2} - \frac{1}{c^2} = \frac{T^2_0}{\triangle x'^2}[/tex]
Make [tex]\frac{1}{v^2}[/tex] the subject
[tex]\frac{1}{v^2} = \frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}[/tex]
Inverse both sides
[tex]v^2 = \frac{1}{\frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}}[/tex]
Take square root of both sides
[tex]v = \sqrt{\frac{1}{\frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}}}[/tex]
Substitute values for [tex]T_0[/tex] and [tex]\triangle x[/tex]
[tex]v = \sqrt{\frac{1}{\frac{70^2}{(70c)^2} + \frac{1}{c^2}}}[/tex]
[tex]v = \sqrt{\frac{1}{\frac{70^2}{70^2*c^2} + \frac{1}{c^2}}}[/tex]
[tex]v = \sqrt{\frac{1}{\frac{1}{c^2} + \frac{1}{c^2}}}[/tex]
[tex]v = \sqrt{\frac{1}{\frac{2}{c^2}}}[/tex]
[tex]v = \sqrt{\frac{c^2}{2}}[/tex]
[tex]v = c\sqrt{\frac{1}{2}}[/tex]
[tex]v = c * 0.7071[/tex]
[tex]v = 0.7071c[/tex]
list 5 types of food that should be consumed daily in a healthy diet.Give an example of each type.
Answer:
vegetables and legumes or beans
fruit
lean meats and poultry, fish, eggs, tofu, nuts and seeds, legumes or beans
grain (cereal) foods, mostly wholegrain or high cereal fibre varieties
milk, yoghurt, cheese or alternatives, mostly reduced fat.
Explanation:
Foods are grouped together because they provide similar amounts of key nutrients. For example, key nutrients of the milk, yoghurt, cheese and alternatives group include calcium and protein, while the fruit group is a good source of vitamins, especially vitamin C.
The resolution of a lens can be estimated by treating the lens as a circular aperture. The resolution is the smallest distance between two point sources that produce distinct images. This is similar to the resolution of a single slit, related to the distance from the middle of the central bright band to the firstorder dark band; however, the aperture is circular instead of a rectangular slit which introduces a scale factor. Suppose the Hubble Space Telescope, 2.4 m in diameter, is in orbit 90.4 km above Earth and is turned to look at Earth. If you ignore the effect of the atmosphere, what is the resolution of this telescope for light of wavelength 557 nm?
Answer:
y = 2.56 10⁻² m
Explanation:
The resolution of this telescope is given by the Rayleigh criterion, for the phenomenal diffraction the first minimum for a linear slit is in
a sin θ = λ
in general the angles are very small, so we approximate
sin θ = θ
we substitute
θ = λ / a
in the case of circular slits we must use polar coordinates, which introduces a numerical factor, leaving the equation
θ = 1.22 [tex]\frac{\lambda }{D}[/tex]
where D is the diameter of the circular opening
In this case they indicate the lens diameter D = 2.4 m, the observation distance r = 90.4 km = 90.4 10³ m
how angles are measured in radians
θ = y / r
we substitute
y / r = 1.22\frac{\lambda }{D}
y = 1.22 \frac{\lambda r }{D}
let's calculate
y = [tex]1.22 \frac{ 557 \ 10^{-9} \ \ 90.4 \ 10^{3} }{2.4}[/tex]
y = 2.56 10⁻² m
this is the minimum distance that can differentiate two objects on Earth
Giving 50 points exactly 50 point and making brainles and if answered wrong I will report and get you kicked out
Draw a picture showing how Heat is added, released, and transferred from one object to another. Also, draw a picture explaining how sublimation or deposition works. please you can draw in your book and take a picture and post it
Answer:
i hope this helps i did'nt quiet understand the secound one. I hope youcan see my picture well .
Explanation:
sincerily, MEMC3891
If the gravitational constant is extremely weak, how is the force of gravity on earth so strong?
When a wave hits an object,energy from the wave is both absorbed and reflected off the object
A. True
B. False
P.S pls help
The nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons).
(a) What is the force between the two alpha particles when they are 6.60 ✕ 10−15 m apart? N
(b) What is the initial magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u. m/s2
Answer:
a) F = 21.16 N, b) a = 3.17 10²⁸ m / s
Explanation:
a) The outside between the alpha particles is the electric force, given by Coulomb's law
F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]
in that case the two charges are of equal magnitude
q₁ = q₂ = 2q
let's calculate
F = [tex]9 \ 10^9 \ \frac{ (2 \ 1.6 \ 10^{-19} )^2 }{ (6.60 \ 10^{-15} )^2 }[/tex]
F = 21.16 N
this force is repulsive because the charges are of the same sign
b) what is the initial acceleration
F = ma
a = F / m
a = [tex]\frac{21.16}{4.0026 \ 1.67 \ 10^{-27} }[/tex]21.16 / 4.0025 1.67 10-27
a = 3.17 10²⁸ m / s
this acceleration is in the direction of moving away the alpha particles
I need help please will mark brainliest
Answer:
200
Explanation:
20m/s*10sec=200
What is the acceleration when a force of 2 N is applied to a ball that has a mass of 0.60 kg?
Answer:
3.333
Explanation:
Acceleration is force divided by mass. So divide the force, 2, by the mass, 0.60, and you will get 3.333. I hope this helps :)
The acceleration of an object is the force divided by time. The acceleration of the ball of 0.60 kg when a force of 2N is applied, is 3.33 m/s².
What is acceleration ?Acceleration of an object is the measure of rate of change in its velocity. Like velocity acceleration is a vector quantity having both magnitude and direction.
Acceleration is defined as the ratio of change in velocity to the change in time. However, according to Newton's second law of motion fore applied on an object is the product of its mass and acceleration.
Hence, F = m a .
Given that, force applied on the ball = 2 N
mass of the ball = 0.60 Kg.
Then acceleration a = force/mass
a = 2 N/ 0.60 Kg = 3.33 m/s²
Therefore, the acceleration of the ball is 3.33 m/s².
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pls help everything is in the pic
Answer:
c
Explanation:
A girl standing on a bridge throws a stone vertically downward with an initial velocity of 15.0 m/s into the river below. If the stone hits the water 2.00 seconds later, what is the height of the bridge above the water
Answer:
the height of the bridge above the water is 49.6 m.
Explanation:
Given;
initial velocity of the stone, u = 15 m/s
time of motion of the stone, t = 2 s
The height of the bridge above the water is calculated from the following kinematic equation as follows;
h = ut + ¹/₂gt²
h = (15 x 2) + ¹/₂(9.8)(2²)
h = 30 + 19.6
h = 49.6 m
Therefore, the height of the bridge above the water is 49.6 m.
5. Friction has two components __________ and ___________.
Answer:
static friction and kinetic friction.
Explanation:
Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 3 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Find the tension in each wire and the magnitude of each tension. (Use g = 9.8 m/s2 for the acceleration due to gravity. Round your answers to two decimal places.)
Answer:
Explanation:
Let the tension in ropes be T₁ and T₂ . Ropes are making angle of 52 and 40 degree with the horizontal . The vertical component of tension will add up together to balance the weight of the decoration
T₁ sin52 + T₂ sin40 = 3 x 9.8
.788 T₁ + .643 T₂ = 29.4
The horizontal component of tension will add up to zero because the decoration piece is at rest .
T₁ cos52 + T₂ cos40 = 0
.615 T₁ - .766 T₂ = 0
T₁ = 1.24 T₂
Substituting this value of T₁ in earlier equation , we have
.788 x 1.24 T₂ + .643 T₂ = 29.4
1.62 T₂ = 29.4
T₂ = 18.15 N
T₁ = 1.24 x 18.15 = 22.51 N .
The current supplied by a battery in a portable device is typically about 0.122 A. Find the number of electrons passing through the device in two hours.