Answer:
75%
Explanation:
We'll begin by calculating the final volume of the gas. This can be obtained as follow:
Initial volume (V₁) = V
Initial temperature (T₁) = T
Initial pressure (P₁) = P
Final pressure (P₂) = 2P
Final temperature (T₂) = ½T
Final volume (V₂) =?
P₁V₁/T₁ = P₂V₂/T₂
PV/T = 2P × V₂ / ½T
Cross multiply
T × 2P × V₂ = PV × ½T
T × 2P × V₂ = PV × ½T
Divide both side by T × 2P
V₂ = (PV × ½T) / T × 2P
V₂ = ¼V
Next, we shall determine the absolute change in the volume of the gas. This can be obtained as follow:
Initial volume (V₁) = V
Final volume (V₂) = ¼V
Absolute change in volume |ΔV| =?
|ΔV| = |V₂ – V₁|
|ΔV| = |¼V – V|
|ΔV| = |0.25V – V|
|ΔV| = 0.75V
Finally, we shall determine percentage change in the volume of the gas. This can be obtained as follow:
Initial volume (V₁) = V
Absolute change in volume |ΔV| = 0.75V
Percentage change =?
Percentage change = |ΔV| / V × 100
Percentage change = 0.75V / V × 100
Percentage change = 0.75 × 100
Percentage change = 75%
Thus, the percentage change in the volume of the gas is 75%
The Earth is a small, rocky planet. It is known as one of the terrestrial planets.
A planet that is similar to the Earth is
A. Uranus
B. Saturn
C. Jupiter
D. Venus
Answer:
D Venus
Explanation:
in terms of size, average, density, mass and surface gravity venus is similar to earth
Q11:A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5500 Hz. When the siren is receding away from you, the frequency of the sound is 4500 Hz. Use the Doppler formula to determine the velocity of the police car. Use vsound
Answer:
v_s = 34.269 m / s
Explanation:
This is a Doppler effect exercise, in this case the observer is fixed and the sound source is moving.
f ’= f [tex]\frac{v}{v \mp v_s }[/tex]
where the negative sign is used for when the source approaches the observer and the positive sign for when the source moves away from the observer
In this case when f ’= 5500 Hz approaches and when f’ = 4500 Hz moves away, let's write the two expressions together
5500 = f ([tex]\frac{v}{v - v_s }[/tex])
4500 = f ( [tex]\frac{v}{v + vs}[/tex])
let's solve these two equations
[tex]\frac{5500}{4500} = \frac{v+v_s}{v-v_s}[/tex]
1.222 (v-v_s) = v + v_s
v_s (1+ 1.22) = v (1.222 -1)
v_s = v [tex]\frac{0.222}{2.223}[/tex]
the speed of sound in air is v = 343 m / s
v_s = 343 0.09990
v_s = 34.269 m / s
A 0.20-kg block and a 0.25-kg block are connected to each other by a string draped over a pulley that is a solid disk of inertia 0.50 kg and radius 0.10 m. When released, the 0.25-kg block is 0.21 m off the ground. What speed does this block have when it hits the ground?
Answer:
The answer is "0.2711 m/s".
Explanation:
Potential energy = Kinetic energy + Potential energy
[tex]m_1 gh =\frac{1}{2} m_1v^2 +\frac{1}{2} m_2v^2 + \frac{1}{2} I\omega^2 + m_1gh\\\\[/tex]
[tex](m_1- m_2)gh =\frac{1}{2} m_1v^2 +\frac{1}{2} m_2v^2 +\frac{1}{2} I\omega^2\\\\2(m_1 - m_2)gh = m_1v^2 + m_1v^2 + I\omega^2\\\\solid \ disk (I) = \frac{1}{2} \ \ M r^2 \\\\[/tex]
When there is no slipping, \omega =\frac{ v]{r}\\\\
[tex]2(m_1 - m_2)gh = m_1v^2 + m_2v^2 + (\frac{1}{2} Mr^2) (\frac{v}{r})^2\\\\2(m_1 -m_2)gh = m_1v^2 + m_2v^2 + \frac{1}{2} Mv^2\\\\4(m_1 -m_2)gh = 2m_1v^2 + 2m_2v^2 + Mv^2\\\\4(m_1 - m_2)gh = (2m_1 + 2m_2 + M) v^2\\\\[/tex]
[tex]v^2 = \frac{4(m_1 - m_2)gh}{(2m_1 + 2m_2 + M)}v[/tex]
[tex]v^2 = \frac{4 (0.25 \ kg - 0.20 \ kg) (9.8 \frac{m}{s^2}) (0.21 m)}{ (2 \times 0.25 kg + 2 \times 0.20 kg + 0.50 kg)}[/tex]
[tex]=\frac{0.1029}{1.4} \ \ \frac{m^2}{s^2}\\\\=0.0735\ \ \frac{m^2}{s^2}\\\\= 0.2711 \ \frac{m}{s}[/tex]
A person pushes a 15.7-kg shopping cart at a constant velocity for a distance of 25.9 m on a flat horizontal surface. She pushes in a direction 23.7 ° below the horizontal. A 32.7-N frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.
Answer:
a) F = 35.7 N, b) W = 846.7 J, c) W = - 846.9 J, d) W=0
Explanation:
a) For this exercise let's use Newton's second law, let's set a reference frame with the x-axis horizontally
let's break down the pushing force.
cos (-23,7) = Fₓ / F
sin (-237) = F_y / F
Fₓ = F cos 23.7 = F 0.916
F_y = F sin (-23.7) = - F 0.402
Y axis
N- W - F_y = 0
N = W + F 0.402
X axis
Fₓ - fr = 0
F 0.916 = fr
F = fr / 0.916
F = 32.7 / 0.916
F = 35.7 N
It is asked to calculate several jobs
b) the work of the pushing force
W = fx x
W = 35.7 cos 23.7 25.9
W = 846.7 J
c) friction force work
W = F x cos tea
friction force opposes movement
W = - fr x
W = - 32.7 25.9
W = - 846.9 J
d) The work of the force would gravitate, as the displacement and the force of gravity are at 90º, the work is zero
W = 0
A truck is driving over a scale at a weight station. When the front wheels drive over the scale, the scale reads 5800 N. When the rear wheels drive over the scale, it reads 6500 N. The distance between the front and rear wheels is 3.20 m Determine the distance between the front wheels and the truck's center of gravity.
Answer:
[tex]x_2=1.60m[/tex]
Explanation:
From the Question We are told that
Initial Force [tex]F_1=5800N[/tex]
Final Force [tex]F_2=6500N[/tex]
Distance between the front and rear wheels \triangle x=3.20 m
Since
[tex]\triangle x=3.20 m[/tex]
Therefore
[tex]x_1+x_2=3.20[/tex]
[tex]x_1=3.20-x_2[/tex]
Generally the equation for The center of mass is at x_2 is mathematically
given by
[tex]x_2 =\frac{(F_1x_1+F_2x_2)}{(F_1+F_2)}[/tex]
[tex]x_2=3.20F_1-\frac{x_2F_1+F_2x_2}{(F_1+F_2)}[/tex]
[tex]2*F_1*x_2 =3.20F_1[/tex]
[tex]x_2=1.60m[/tex]
Center of gravity of a body is the sum of its moments divided by the overall weight of the object. The distance between the front wheels and the truck's center of gravity is 1.6 meters.
Given-
Scale reading value when the front wheels drive over the scale [tex]m_{1}[/tex] is 5800 N.
Scale reading value when the rear wheels drive over the scale [tex]m_{2}[/tex] is 6500 N
Distance between the front and rear wheel [tex]\bigtriangleup x[/tex] is 3.20 meters.
Let, the distance between the front wheels and the truck's center of gravity is [tex]x_{2}[/tex].
Since sum of the distance between front wheel to truck's center of gravity [tex]x_{1}[/tex], and rear wheel to truck's center of gravity [tex]x_{2}[/tex], is equal to the distance between the front and rear wheel [tex]\bigtriangleup x[/tex]. Therefore,
[tex]\bigtriangleup x=x_{1} +x_{2}[/tex]
[tex]3.20=x_{1} +x_{2}[/tex]
[tex]x_{1} =3.20-x_{2}[/tex]
For the distance between the front wheels and the truck's center of gravity is the formula of center of gravity can be written as,
[tex]x_{2} =\dfrac{m_{1}x_{1}+m_{2} x_{2} }{m_{1} +m_{2} }[/tex]
[tex]x_{2} =\dfrac{5800\times (3.20- x_{2})+6500\times x_{2} }{5800 +6500 }[/tex]
[tex]1230 x_{2} ={18560-5800 x_{2}+6500 x_{2} }[/tex]
[tex]x_{2}= 1.6[/tex]
Hence, the distance between the front wheels and the truck's center of gravity is 1.6 meters.
For more about the center of gravity, follow the link below-
https://brainly.com/question/20662119
Suppose that the electric field in the Earth's atmosphere is E = 131 N/C, pointing downward. Determine the electric charge (in C) in the Earth. (Assume the electric field is measured at the surface of the Earth. Include the sign of the value in your answer.) Hint
Answer:
the electric charge in the Earth is -5.91 × 10⁵ C
Explanation:
Given the data in the question;
Electric field E = 131 N/C
we know that; radius of the earth r = 6,371 km = 6371000 m
and Coulomb's constant k = 8.99 × 10⁹ Nm²/c²
Now, using the following formula to calculate the charge;
E = k × Q/r²
we make Q the subject of the formula
Q = Er² / k
so we substitute
Q = [ 131 N/C × ( 6371000 m )² ] / 8.99 × 10⁹ Nm²/c²
Q = [ 5.317242971 × 10¹⁵ ] / [ 8.99 × 10⁹ ]
Q = 5.91 × 10⁵ C
Since the electric field pointing downward
Q = -5.91 × 10⁵ C
Therefore, the electric charge in the Earth is -5.91 × 10⁵ C
An object accelerates 3 m/s2 , when a force of 6 N acts on it. What is the object’s mass
Answer:
2 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{6}{3} \\ = 2[/tex]
We have the final answer as
2 kgHope this helps you
15) If a magnet produces a force on a current-carrying wire, the wire
A) produces a force on the magnet.
B) may or may not produce a force on the magnet.
C) neither of these
Answer:
The answer is option no. A
A 0.390 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 75.0 pC charge on its surface. What is the potential (in V) near its surface
Answer:
V = 346.15 Volts
Explanation:
Given that,
Diameter of the sphere, d = 0.390 cm
Radius, r = 0.195 cm
Charge, [tex]q=75\ pC =75\times 10^{12}\ C[/tex]
The electric potential near its surface is given by :
[tex]V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 75\times 10^{-12}}{0.195\times 10^{-2}}\\\\V=346.15\ V[/tex]
So, the potential near its surface is equal to 346.15 V.
name 3 properties of solids
-A solid has a definite shape and volume.
-Solids in general have higher density.
-In solids, intermolecular forces are strong.
If you left a glass fiber-optic cable unshielded by any plastic covering, should the light still be able to travel through the cable?
1. Yes
2. No
Answer:
Yes
Explanation:
As part of an exercise program, a woman walks south at a speed of 2.8 m/s for 46 minutes. She then turns around and walks north a distance 3,132 m in 54 minutes . What is the woman's average speed in m/s during her entire motion
Answer:
[tex]A_[avg}=1.81m/s[/tex]
Explanation:
From the question we are told that:
North Movement
Velocity[tex]V=2.8[/tex]
Time [tex]t=46 minuites[/tex]
South movement
Distance [tex]d=3,132[/tex]
Time [tex]t'= 54 minutes[/tex]
Generally the equation for Average Velocity is mathematically given by
[tex]A_[avg}\]frac{Total distance}{Total time}[/tex]
Where
Total distance d_t
[tex]d_t=Souther\ distance\ traveled\ +Northern\ distance\ traveled[/tex]
[tex]d_t=(2.8*46*60)+(3132)[/tex]
[tex]d_t=10860[/tex]
An
Total Time
[tex]T=(46+54)60[/tex]
[tex]T=6000[/tex]
Therefore
[tex]A_[avg}=\frac{d_t}{T}[/tex]
[tex]A_[avg}=\frac{10860}{6000}[/tex]
[tex]A_[avg}=1.81m/s[/tex]
As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force
acting on it is equal to (41 - 9j) N. If the speed of the object at the initial position is 4.0 m/s,
what is its kinetic energy at its final position?
Answer:
Answer:
v_f = 10.38 m / s
Explanation:
For this exercise we can use the relationship between work and kinetic energy
W = ΔK
note that the two quantities are scalars
Work is defined by the relation
W = F. Δx
the bold are vectors. The displacement is
Δx = r_f -r₀
Δx = (11.6 i - 2j) - (4.4 i + 5j)
Δx = (7.2 i - 7 j) m
W = (4 i - 9j). (7.2 i - 7 j)
remember that the dot product
i.i = j.j = 1
i.j = 0
W = 4 7.2 + 9 7
W = 91.8 J
the initial kinetic energy is
Ko = ½ m vo²
Ko = ½ 2.0 4.0²
Ko = 16 J
we substitute in the initial equation
W = K_f - K₀
K_f = W + K₀
½ m v_f² = W + K₀
v_f² = 2 / m (W + K₀)
v_f² = 2/2 (91.8 + 16)
v_f = √107.8
v_f = 10.38 m / s
The temperature of 10 kg of a substance rises by 55oC when heated. Calculate the
temperature rise when 22 kg of the substance is heated with the same quantity of heat.
Given :
The temperature of 10 kg of a substance rises by 55oC when heated.
To Find :
The temperature rise when 22 kg of the substance is heated with the same quantity of heat.
Solution :
We know, change in temperature when heat is given to object is given by :
[tex]\Delta H = m S\Delta T[/tex]
It is given that same amount of heat is given in both the cases also the substance is same.
So,
[tex]M_1 S \Delta T_1 = M_2 S \Delta T_2\\\\10\times 55 = 22 \times \Delta T_2\\\\\Delta T_2 = 25^o\ C[/tex]
Hence, this is the required solution.
a 2kg block of wood starts at rest and slides down a ramp. Its initail height is 12m. if the final velocity of the block is 13m/s, determine the energy of this system that has been turned into heat
Answer:
E = 66.44 J
Explanation:
From the law of conservation of energy:
Total Mechanical Energy at Start = Total Mechanical Energy at the End
Potential Energy at Start = Kinetic Energy at End + Energy Lost
[tex]mgh = \frac{1}{2} mv^2 + E\\\\E = mgh - \frac{1}{2} mv^2\\\\[/tex]
where,
E = Energy turned into heat = ?
m = mass of block = 2 kg
g = acceleration due to gravity = 9.81 m/s²
h = height = 12 m
v = final speed = 13 m/s
Therefore,
[tex]E = (2\ kg)(9.81\ m/s^2)(12\ m)-\frac{1}{2} (2\ kg)(13\ m/s)^2\\\\E = 235.44\ J - 169\ J\\[/tex]
E = 66.44 J
4. A 268 kg boulder rolls down a hill. If it had 50,000 J of energy to start with, how high was the hill? (Can anyone explain or give the equation for it? Thank you!)
Answer:
h = 18.65 m
Explanation:
Given that,
The mass of a boulder, m = 268 kg
The potential energy at the hill, E = 50,000 J
We need to find the height of the hill.
We know that, the potential energy of an object is given by :
E = mgh
Where
h is the height of the hill
So,
[tex]h=\dfrac{E}{mg}\\\\h=\dfrac{50000}{268\times 10}\\\\h=18.65\ m[/tex]
So, the height of the hill is 18.65 m.
[tex]\\ \sf\longmapsto PE=mgh[/tex]
[tex]\\ \sf\longmapsto 50000=268(10)h[/tex]
[tex]\\ \sf\longmapsto 50000=2680h[/tex]
[tex]\\ \sf\longmapsto h=\dfrac{50000}{2680}[/tex]
[tex]\\ \sf\longmapsto h=18.65m[/tex]
in a Mercury thermometer the level of Mercury Rises when its bulb comes in contact with a hot object what is the reason for this rise in the level of Mercury
Answer:
because thermometric liquid readily expands on heating or contracts on cooling even for a small difference in the temperature of the body.
Light energy from the Sun reaches an ocean beach, where people are
walking. Which transfer of thermal energy involved in this scenario is an
example of radiation?
Answer:
the correct answers is D
Explanation:
Thermal energy can be transferred by three methods: conduction, convention, and radiation.
Radiation transfer occurs when there is no movement of matter for the exchange of energy.
In this case, checking the correct answers is D
since in this case the transfer is between light and sand without matter exchange
Work of 2 Joules is done in stretching a spring from its natural length to 14 c m beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (14 c m )
Answer:
F =28.57 N
Explanation:
Assuming that the spring is ideally followed Hooke's law, This means
F=kx
Here F is the force required to stretch or compress the spring through x distance.
the amount of the work done in the displacement of the spring would be:
W = ∫F.dx
=∫kx dx
=1/2kx^2
=1/2k×0.14^2
⇒ 0.0098k =2
⇒ k = 204.081
Therefore, the force (in Newtons) that holds the spring stretched at the same distance.
F = 204.081×0.14
F =28.57 N
A man is pushing a box of weight W with a forward force of magnitude F. The box
moves forward with a constant speed. What is the magnitude f of the friction force?
Answer:
The magnitude of the friction force is also F.
Explanation:
By the second Newton's law, we know that:
F = m*a
Net force is equal to mass times acceleration.
Here, we know that the box moves with constant speed, thus, the box has no acceleration, then the net force applied to the box is zero.
Also, remember that the friction force is given by:
[tex]F_f = -\mu*W[/tex]
Where mu is the coefficient of friction, and this force opposes to the direction of motion (that coincides with the direction of our forward force, that is why this has a negative sign)
The net force will be equal to the sum of our two horizontal forces (as the weight is already canceled by the normal force)
[tex]F_{total} = F + F_f[/tex]
And this is equal to zero, because we know that the box is non-accelerated.
Then we must have that:
[tex]F_f = -F[/tex]
Then we can conclude that the magnitude of the friction force is F.
1. What did you observe about the magnitudes of the forces on the two charges? Were they the same or different? Does your answer depend on whether the charges were of the same magnitude or different? How does this relate to Newton’s 3rd law?
Answer:
Following are the solution to the given question:
Explanation:
Its strength from both charges is equivalent or identical. The power is equal. And it is passed down
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Therefore, the extent doesn't rely on the fact that charges are the same or different. Newton's third law complies with Electrostatic Charges due to a couple of charges. They are similar in magnitude, and they're in the other way.
[tex]|F_{12}| = |F_{21}|[/tex]
If v = 4.00 meters/second and makes an angle of 60° with the positive direction of the y–axis, what is the magnitude of vx?
An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and the drag coefficient is 0.065. The mass of the aircraft is 900 kg. Calculate the effective lift area for the aircraft and the required engine thrust and power to maintain level flight.
Answer:
- the effective lift area for the aircraft is 8.30 m²
- the required engine thrust is 1275 N
- required power is 79.7 kW
Explanation:
Given the data in the question;
Speed V = 225 km/hr = 62.5 m/s
The lift coefficient CL = 0.45
drag coefficient CD = 0.065
mass = 900 kg
g = 9.81 m/s²
a) the effective lift area for the aircraft
we know that for a steady level flight, weight = lift and thrust = drag
Using the equation for the lift force
F[tex]_L[/tex] = C[tex]_L[/tex][tex]\frac{1}{2}[/tex]ρV²A = W
we substitute
0.45 × [tex]\frac{1}{2}[/tex] × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )
1081.05 × A = 8829
A = 8829 / 1081.05
A = 8.30 m²
Therefore, the effective lift area for the aircraft is 8.30 m²
b) the required engine thrust and power to maintain level flight.
we use the expression for drag force
F[tex]_D[/tex] = T = C[tex]_D[/tex][tex]\frac{1}{2}[/tex]ρV²A
we substitute
= 0.065 × [tex]\frac{1}{2}[/tex] × 1.21 × ( 62.5 )² × 8.30
T = 1275 N
Since drag and thrust force are the same,
Therefore, the required engine thrust is 1275 N
Power required;
P = TV
p = 1275 × 62.5
p = 79687.5 W
p = ( 79687.5 / 1000 )kW
p = 79.7 kW
Therefore, required power is 79.7 kW
A rocket moving around the earth at height "H", If the gravitational acceleration "g1" at height
His of gravitational acceleration 'g at earth surface. If Earth radius is "R", find "H"
using R
Answer:
At the earth's surface g = G M / R^2 where G is the gravitational constant
at H g1 = G M / (R + H)^2 using Gauss' theorem for enclosed mass
g1 = G M / (R^2 + 2 R H) ignoring H^2 as it is small compared to R^2
g / g1 = (R^2 + 2 R H) / R^2 = 1 + 2 R H
g = g1 + 2 R H g1
g1 - g = - 2 R H or H = (g1 - g) / 2 R
(A)
(B)
(C)
(D)
Which graph is a quadratic graph?
OA graph A
OB. graph B
OC graph
OD. graph D
A compound consisting of Cr3+ ions and OH ions would be named chromium (III) hydroxide
True or false
Two long, straight wires are fixed parallel to one another a distance do apart. The wires carry equal constant currents 1, in the same direction. The attractive magnetic force per unit length between them if f = F/L. What is the force per unit length between the wires if their separation is 2d, and each carries current 210? 10 da lo A. f/4 B. f/2 C. 3f/2 D. 2f
Answer:
The option (B) is correct.
Explanation:
The magnetic force between the two current carrying wires is given by
[tex]F =\frac{\mu o}{4\pi}\times \frac{2 I I'}{r}[/tex]
where, I and I' be the currents in the wires and r is the distance between the two wires.
Here, we observe that the force between the wires is inversely proportional to the distance between them.
So, when the distance is doubled , let the new force is F'.
[tex]\frac{F'}{F}=\frac{r}{r'}\\\\\frac{F'}{F}=\frac{d}{2d}\\\\F'=\frac{F}{2}[/tex]
So, option (B) is correct.
Angles t and v are complementary.angles T has a mesure of (2X+10). Angle v has a measure of 48 what is the value of x
16 because complementary angles equal up to 90. 2x+58 = 90 x= 16
Rest and motion are relative terms. write in brief
Answer:
Rest and motion are the relative terms because they depend on the observer's frame of reference. So if two different observers are not at rest with respect to each other, then they too get different results when they observe the motion or rest of a body.
Explanation With Example:
Imagine you are sitting inside a moving bus. When you look outside you will observe that you are moving. ... With respect to the roof of bus, you are at rest. Hence it is concluded that rest and motion are relative terms.
Darcy suffers from farsightedness equally severely in both eyes. The focal length of either of Darcy's eyes is 19.8 mm in its most accommodated state (i.e, when the eye is focusing on the closest object that it can clearly see) What lens strength (a.k.a., lens power) of contact lenses should be prescribed to correct the farsightedness in Darcy's eyes? (Assume the lens-to retina distance of Darcy's eyes is 2.00 cm, and the contact lenses are placed a negligibly small distance from the front of Darcy's eyes)
i. 1.64D
ii. 2.98 D
iii. 2.19 D
iv. 3.49 D
v. 1.37 D
Answer:
Explanation:
Power of defective eye = (1 / near point )+ (1 / lens-retina distance )
1 / .0198 m = (1 / near point )+ (1 / .02 m )
= 50.50 = (1 / near point )+ 50
0.50 = 1 / near point
near point = 1 / .5 = 2m
= 200 cm .
for near point at 25 cm , convex lens will be required.
u = - 25 cm , v = - 200 cm ,
1 / v - 1 / u = 1 /f for lens
- 1 / 200 + 1 / 25 = 1 / f
= .04 - .005 = 1 /f
.035 = 1 / f
f = 28.57 cm = 0.2857 m
power = 1 /f
= 1 / 0.2857
= 3.5 D .
i v ) option is correct .