Answer:
Answer:- Volume of the balloon is 5.78 L.
Solution:- There are 0.24 moles of hydrogen gas in a balloon at 35 degree C and 1.05 atm pressure. It asks to calculate the volume of the balloon.
This problem is based on ideal gas law equation:
P = 1.05 atm, n = 0.24 mole, T = 35 + 273 = 308 K
R =
V = ?
The equation could be rearranged for the volume as:
Let's plug in the values and do the calculations to get the volume of the balloon:
V = 5.78 L
So, the volume of the gas balloon is 5.78 L.
what weight is recorded by a scale when it is placed inside a lift which is in free fall? Enplain.
Answer:
Explanation:
There is no pressure of your feet on the scales, and no pressure of the floor on the scales, so the scales will read zero*. Hence, your weight, in a freely falling lift is zero
Question in the picture please help me...
Show all steps please....
Answer: i)A to B : (ice) freezing
ii) B to C (water) boiling
C to D (steam) evaporating
explanation: 0° is the freezing point of water when temperature increases from 0° the water starts melting. As 100° is the boiling point of water so at 100° the water completely melts and it starts boiling during boiling water changes into steam(water vapour) and it evaporates
Two rams run toward each other. One ram has a mass of 44 kg and runs south with a speed of 6 m/s, while the other has a mass of 50 kg and runs north with a speed of 3 m/s. What will the momentum of the system made up of the two rams be after they collide? Assume the total momentum of the system is conserved.
A. 114 kg-m/s south
B. 414 kg-m/s south
C. 414 kg m/s north
D. 114 kg-m/s north
Question 12 of 19 Lions can run at speeds up to approximately 80.0 km/h. A hungry 109 kg lion running northward at top speed attacks and holds onto a 36.0 kg Thomson's gazelle running eastward at 78.0 km/h. Find the final speed vf of the lion‑gazelle system just after the lion attacks.
Answer:
The final velocity of the lion-gazelle system is approximately 62.321 kilometers per hour.
Explanation:
Let suppose that both lion and the Thomson's gazelle collide each other inelastically, the use of the Principle of Linear Momentum Conservation suffices to describe the entire phenomenon:
[tex]m_{L}\cdot \vec v_{L} + m_{T}\cdot \vec v_{T} = (m_{L}+m_{T})\cdot \vec v[/tex] (1)
Where:
[tex]m_{L}[/tex] - Mass of the lion, in kilograms.
[tex]m_{T}[/tex] - Mass of the Thomson's gazelle, in kilograms.
[tex]\vec v_{L}[/tex] - Initial velocity of the lion, in meters per second.
[tex]\vec v_{T}[/tex] - Initial velocity of the Thomson's gazelle, in meters per second.
[tex]\vec v[/tex] - Final velocity of the lion-gazelle system, in meters per second.
Let suppose that both northward velocity and eastward velocity are positive.
If we know that [tex]m_{L} = 109\,kg[/tex], [tex]\vec v_{L} = (0, 22.222)\left[\frac{m}{s} \right][/tex], [tex]m_{T} = 36\,kg[/tex] and [tex]\vec v_{T} = (21.667, 0)\,\left[\frac{m}{s} \right][/tex], then the final velocity of the lion-gazelle system is:
[tex]109\cdot (0,22.222)+36\cdot (21.667,0) = 147\cdot (v_{x},v_{y})[/tex]
[tex](v_{x}, v_{y}) = (0, 16.478) + (5.306, 0) \,\left[\frac{m}{s} \right][/tex]
[tex](v_{x}, v_{y}) = (5.306, 16.478)\,\left[\frac{m}{s} \right][/tex]
[tex](v_{x}, v_{y}) = (19.102, 59.321)\,\left[\frac{km}{h} \right][/tex]
And the final speed of the lion-gazelle system is calculated by the Pythagorean Theorem:
[tex]v = \sqrt{19.102^{2}+59.321^{2}}\,\left[\frac{km}{h} \right][/tex]
[tex]v \approx 62.321\,\frac{km}{h}[/tex]
The final velocity of the lion-gazelle system is approximately 62.321 kilometers per hour.
If a fan draws 220 watts on a 120- volt circuit what is the current being drawn?
22. amperes
O 5.2 amperes
O 1.8 ampmes
15. amperes
Explanation:
here's the answer to your question
Which of the following is an example of a molecule?
A. A sodium atom forms a metallic bond with another sodium atom.
B. A hydrogen atom forms a covalent bond with a chlorine atom.
C. A sodium atom forms an ionic bond with a chlorine atom.
D. A sodium atom forms a metallic bond with another sodium atom.
Answer:
C
Explanation:
state the advantages and disadvantages of alkaline battery over a lead acid accumulator
Answer:
Alkaline batteries have longer shelf life than other batteries of the chloride type electrolyte batteries. They have a much higher energy density compared with other batteries. This allows the battery to produce the same energy while lasting longer than other batteries.
Advantage
1) Alkaline battery lasts long than other batteries
2) It have a much higher energy density
Disadvantage :
1) They are heavier and bulkier
2) They have high internal resistance
What is alkaline battery ?An alkaline battery is a type of primary battery that derives its energy from the reaction between zinc metal and manganese dioxide.
Advantage
Alkaline battery can produce same amount of energy by lasting longer than other batteries as it have longer shelf life than lead acid accumulator
Alkaline battery have a much higher energy density compared to lead acid accumulator
Disadvantage :
Alkaline battery are heavier and bulkier
Alkaline battery have high internal resistance
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in california the pacific plate slides past the north american plate. If the pcific plate is moving at a speed of 5 centimeters per year, how long will take for the plate to travel 100 meters
Answer:
6.34×10⁹ seconds
Explanation:
Applying,
V = d/t............. Equation 1
Where V = speed plate, d = distance covered by the plate, t = time it takes the plate to travel
make t the subject of the equation
t = d/V............ Equation 2
From the question,
Given: d = 5 cm/yr = (5×3.154×10⁻⁹) = 1.577×10⁻⁸ m/s, d = 100 meter
Substitute these values into equation 2
t = 100/1.577×10⁻⁸
t = 6.34×10⁹ seconds
Hence the time it takes the the plate to travel is 6.34×10⁹ seconds
Which law states that absolute zero cannot be reached?
A CROW BAR WITH LENGTH 200 CM IS USED TO LIFT A LOAD OF 600N . IF THE DISTANCE BETWEEN FULCRUM AND LOAD IS 0.75. CALCULATE ; a, effort b, MA c, VR
Answer:
a. Effort = 960 Newton
b. Mechanical advantage (M.A) = 0.625
c. Velocity ratio (V.R) = 1.67
Explanation:
Given the following data;
Load = 600 NLength of crowbar = 200 cmLength of load arm = 0.75 mConversion:
100 cm = 1 m
X cm = 0.75 m
Cross-multiplying, we have;
X = 0.75 * 100 = 75 cm
First of all, we would find the effort arm;
Effort arm = length of crow bar - length of load arm
Effort arm = 200 - 75
Effort arm = 125 cm
Next, we would determine the mechanical advantage (M.A) of the crow bar;
[tex] M.A = \frac {Effort \; arm}{Load \; arm} [/tex]
Substituting the values into the formula, we have;
[tex] M.A = \frac {125}{200} [/tex]
M.A = 0.625
To find the effort of the crow bar;
[tex] M.A = \frac {Load}{Effort} [/tex]
Making "effort" the subject of formula, we have;
[tex] Effort = \frac {Load}{M.A} [/tex]
[tex] Effort = \frac {600}{0.625} [/tex]
Effort = 960 Newton
Lastly, we would determine the velocity ratio (V.R);
[tex] V.R = \frac {length \; of \; effort \; arm}{length \; of \; load \; arm} [/tex]
[tex] V.R = \frac {125}{75} [/tex]
V.R = 1.67
The heat required to raise 3 kg of copper from 0°C to 10°C raises 1kg of lead from 10°C to 100°C. If the specific heat capacity of copper is 0.095 cal/g°C,find the specific heat capacity of lead.
The specific heat capacity of lead has been determined to be 0.032 cal/g°C.
The specific heat capacity of a substance is the amount of heat required to increase the temperature of its unit mass by 1 Kelvin rise in temperature. There is no phase change during the process. This can be expressed as:
Q = mcT
c = [tex]\frac{Q}{mT}[/tex]
Where c is the specific heat capacity of the substance, Q is the quantity of heat required, m is the mass of the substance and T is its change in temperature.
In the given question, the heat emitted by the copper mass would be absorbed by the lead mass.
So that;
heat emitted by copper = heat absorbed by lead
[tex]m_{c}[/tex] [tex]c_{c}[/tex]([tex]T_{2}[/tex] - [tex]T_{1}[/tex]) = [tex]m_{l}[/tex][tex]c_{l}[/tex]([tex]T_{2}[/tex] - [tex]T_{1}[/tex])
With the given information in the question, the above equation is expressed as:
3 * 0.095 * (10 - 0) = 1 * [tex]c_{l}[/tex]* (100 - 10)
3*0.095*10 = [tex]c_{l}[/tex] * 90
2.85 = [tex]c_{l}[/tex]*90
[tex]c_{l}[/tex] = [tex]\frac{2.85}{90}[/tex]
= 0.031667
[tex]c_{l}[/tex] = 0.032 cal/g°C
Therefore, the specific heat capacity of lead is 0.032 cal/g°C.
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A rocket travels vertically at a speed of 1200 km/h. The rocket is tracked through a telescope by an observer located 16km from the launching pad. Find the rate at which the angle between the telescope and the ground is increasing 3 min after lift-off. Round your answer to the nearest hundredth rad/hr
Answer:
w = 0.01 t⁻² rad/h
Explanation:
Let's use trigonometry to find the elevation angle of the telescope, we assume that the rocket remains fixed in the telescope.
tan θ = CO / CA
the adjacent leg is CA = 16 km = 16 10³ m
Let's use kinematics to find the height of the rocket
v = y / t
y = v t
this height is equal to the opposite leg
CO = v t
we substitute
tan θ = vt / 16
tan θ = 1200/16 t
tan θ = 75 t
θ = tan⁻¹ ( 75 t )
speed is defined by
w = dθ/dt
w = [tex]\frac{1}{1 + (75 t)^2} \ 75[/tex]
as time increases we can neglect the 1 of the denominator
w = 1/75 t²
w = 0.0133 t⁻²
Using three significant figures
w = 0.01 t⁻² rad/h
for this answer the time must be given in hours
Tính công của dòng điện
Answer:
CG gh sure er go b vh pxuh FPI OO c AM h kh
state physical quantity represented by the gradient of a displacement - time graph
Answer:
velocity
Explanation:
this is because velocity has both magnitude and direction
The speed of sound where a tuning fork of frequency 262 Hz produces the third resonance position above a closed air column that is 1.59 m in length is ___m/s.
Answer:
v = 333.26 m/s
Explanation:
Given that,
The frequency of the tuning fork, f = 262 Hz
It produces the third resonance position above a closed air column that is 1.59 m in length.
We need to find the speed of sound in tuning fork. Let it is f. The third resonance position corresponds to the fifth harmonic of the closed air column. The wavelength in third resonance is given by :
[tex]\lambda=\dfrac{4}{5}L\\\\=\dfrac{4}{5}\times 1.59\\\\=1.272\ m[/tex]
The speed of sound is :
[tex]v=f\lambda\\\\=262\times 1.272\\\\=333.26\ m/s[/tex]
So, the speed of sound is 333.26 m/s.
Answer:
The speed of sound is 555.44 m/s.
Explanation:
frequency of third resonance position, f''' = 262 Hz
length of pipe, L = 1.59 m (closed)
Let the speed is v.
The frequency of third resonance in closed organ pipe is
[tex]f''' =\frac{3 v}{4 L}\\\\262 = \frac{3 v}{4\times 1.59}\\\\v = 555.44 m/s[/tex]
You toss an apple across the room to a friend. Which of the following statements is true about the apple at the top of its trajectory?
A. Its acceleration is zero.
B. The horizontal component of its velocity is zero.
C. The vertical component of its velocity is 9.8 m/s down.
D. Its acceleration is 9.8 m/s2 down.
write any two uses of a wedge.
Answer: two uses of a wedge can be to tighten things as well as to lift things.
Explanation: because that is some things wedges (a mechanical tool) can do.
is there any machine that is 100% efficient? why?why not
Answer:
No, it's not there.
Explanation:
For a machine to be 100% efficient, it has to be with an output which is equal to its input. But machines have an out put less than an input, hence efficiency below 100%.
y’all know what to do… please help
Answer:
A. It increased gradually as the reaction progressed.
Answer: A it increased gradually
as the reaction progressed
What is the apparent deflection of a light beam in an elevator which is 2.7 m wide if the elevator is accelerating downward at 9.8 m/s^2?
Answer:
[tex]X=3.37561*10^{-16}[/tex]
Explanation:
From the question we are told that:
Width [tex]w=2.7m[/tex]
Acceleration [tex]a=9.8m/s^2[/tex]
Generally the equation for The apparent deflection is mathematically given by
X= 0.5 *g*t^2
Where
[tex]t=\frac{width}{Velocity of light}[/tex]
[tex]t=\frac{2.5}{3*10^8}[/tex]
[tex]t=8.3*10^{-9}[/tex]
Therefore
[tex]X= 0.5 *9.8*(8.3*10^{-9})^2[/tex]
[tex]X=3.37561*10^{-16}[/tex]
En un rio una Onda viaja con una velocidad de propagación de 50 m/s con una longitud de Onda de 40 metros. Hallar la frecuencia de la Onda.
Answer:
Frequencia = 1.25 Hz
Explanation:
Dados los siguientes datos;
Velocidad = 50 m/s Longitud de onda = 40 metrosPara encontrar la frecuencia de la onda;
Matemáticamente, la velocidad de una onda viene dada por la fórmula;
[tex] Velocidad = Longitud \; de \; onda * Frequencia [/tex]
Haciendo de la frecuencia el tema de la fórmula, tenemos;
[tex] Frequencia = \frac {Velocidad}{Longitud \; de \; onda} [/tex]
Sustituyendo en la fórmula, tenemos;
[tex] Frequencia = \frac {50}{40} [/tex]
Frequencia = 1.25 Hz
A sound frequency 100Hz and wavelength 3.34m is travelling through air, calculate the Velocity of sound in air.
Answer:
334m/s
Explanation:
If you take the data you gave and insert it into the equation f=v/λ the frequency would come out to be 3.34m
334 m/s is the Velocity of sound in air.
What is the velocity of a wave?Wave velocity in common usage refers to speed, although, properly, velocity implies both speed and direction.
The velocity of a wave is equal to the product of its wavelength and frequency (number of vibrations per second) and is independent of its intensity.
Mathematically , v=fλ
According to the question,
Sound frequency (f) = 100Hz
Wavelength (λ)= 3.34m
Computing the values in the formula,
v=fλ
v = 100Hz x 3.34m
v= 334m/s
Therefore,
Velocity of sound in air is 334 m/s.
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(1) A string of constant thickness and length I cm is stretched by a force of T Newton. A
tuning fork stamped 256 Hz sets the string vibrating at its fundamental frequency.
Find the frequency of the string when:
(i) Its length is doubled and the tension constant;
(ii) the tension is doubled and the length constant.
2) State the effect of increase in the tension on a wire on its frequency.
Answer:
Explanation:
i) frequency is the square root of the quotient of Tension over linear density/ all divided by two times the length.
linear density and tension are constant, so the numerator remains constant
The length is doubled, so the frequency is halved.
f = 128 Hz
ii) Doubling the tension increases the frequency by √2
f = 256√2 = 362 Hz
2) If other factors remain constant, increasing tension increases frequency.
Why does a small piece of steel rod sinks while a large steel oil tanker floats?
Plss help thanks! pls include EXPLANATION and EXAMPLES that I can proof to people that has no knowledge
Answer:
Too what i know its because of the air
Explanation:
The steel rod is more dense and doesnt have air whil'st the oil tanker is less dense with lots of air
Hai điện tích đặt cách nhau một khoảng R trong không khí thì lực tương tác
giữa chúng là 2.10−3N. Nếu khoảng cách đó mà đặt trong môi trường điện môi thì
lực tương tác giữa chúng là 10−3N. Để lực tương tác giữa hai điện tích đó khi đặt
trong môi trường điện môi bằng lực tương tác giữa hai điện tích đó khi đặt trong
không khí thì khoảng cách giữa 2 điện tích là bao nhiêu?
how does the siphon work
Answer:
A simple siphon raises water over a crest and discharges it at a lower level.It works coninueasly due to pull of gravity.
Explanation:
Siphons are tubes which draw fluid over the rim of a tank to a lower point. After an initial pressure change to initiate the flow, siphons operate continuously due to the pull of gravity.
Working:
A simple siphon raises water over a crest and discharges it at a lower level. As water flows through a siphon, energy due to pressure and elevation is either lost to pipe friction or converted to velocity energy.
The coefficient of kinetic friction between an 100-kg desk and the wood floor is 0.4. What force (in Newtons) must be applied to move the desk at a constant speed? help now
Assuming the applied force is exerted parallel to the floor, by Newton's second law both the net vertical and net horizontal forces would be zero:
∑ F (horizontal) = p - f = 0 … … … because the desk is pushed at a constant speed
∑ F (vertical) = n - mg = 0 … … … because the desk doesn't move up or down
where p is the magnitude of the applied force, f is the mag. of kinetic friction, n is the mag. of the normal force due to contact between the floor and desk, and mg is the weight of the desk. We have
n = mg = (100 kg) g = 980 N
and the mag. of friction is proportional to n according to
f = 0.4 n = 392 N
Then the applied force p has magnitude
p = f = 392 N
calculate the potential energy stored in a meta ball of mass of 80 kg kept at a height of 15m from the earth surface . What will be the potential energy when the metal ball is kept on the earth surface Take (g- 9.8 m/s]
Answer:
11760 J
Explanation:
cuz potential engery is PE = MHG
SO 80×15×9.8= 11760 J
Answer:
11769j
Explanation:
here,
mass(m)=80 kg
height(h)=15m
acceleration due to gravity(g)=9.8m/s^2
now,
potential energy = m×g×h
= 80×9.8×15
= 11760j
Two Carnot air conditioners, A and B, are removing heat from different rooms. The outside temperature is the same for both rooms, 314 K. The room serviced by unit A is kept at a temperature of 292 K, while the room serviced by unit B is kept at 298 K. The heat removed from either room is 4430 J. For both units, find (a), (b) the magnitude of the work required and (c), (d) the magnitude of the heat deposited outside for A and B conditioners respectively
Answer:
(a) 333.77 J
(b) 237.85 J
(c) 4763.77 J
(d) 4667.85 J
Explanation:
Temperature of source, TH = 314 K
Temperature of A, Tc = 292 K
Temperature of B, Tc' = 298 K
heat taken out, Qc = 4430 J
Let the heat deposited outside is QH and QH' by A and B respectively.
[tex]\frac{Q_H}{Q_c}=\frac{T_H}{T_c}\\\\Q_H = \frac{4430\times314}{292}=4763.77 J[/tex]
Now
[tex]\frac{Q_H'}{Q_c}=\frac{T_H}{T_c'}\\\\Q_H' = \frac{4430\times314}{298}=4667.85 J[/tex]
(a) Work done for A
W = QH - QC = 4763.77 - 4430 = 333.77 J
(b) Work done for B
W' = QH' - Qc = 4667.85 - 4430 = 237.85 J
(c) QH = 4763.77 J
(d) QH' = 4667.85 J
why does the satellite not fall while revolving the earth
Answer:
Satellites don't fall from the sky because they are orbiting Earth. Even when satellites are thousands of miles away, Earth's gravity still tugs on them. Gravity--combined with the satellite's momentum from its launch into space--cause the satellite go into orbit above Earth, instead of falling back down to the ground.
