Answer:
The mass of the mercury remaining in the bottle is 497.57 grams.
Explanation:
The mass of the mercury remaining in the bottle is found by subtracting the mass expeled due to heating from initial mass inside the bottle. That is:
[tex]m_{f} = m_{o}-\Delta m[/tex] (1)
Where:
[tex]m_{o}[/tex] - Initial mass, in grams.
[tex]\Delta m[/tex] - Mass expelled due to heating, in grams.
[tex]m_{f}[/tex] - Final mass, in grams.
If we know that [tex]m_{o} = 500\,g[/tex] and [tex]\Delta m = 2.43\,g[/tex], then the mass of the mercury remaining in the bottle is:
[tex]m_{f} = m_{o}-\Delta m[/tex]
[tex]m_{f} = 497.57\,g[/tex]
The mass of the mercury remaining in the bottle is 497.57 grams.
A 64.0 cm long cord is vibrating in such a manner that it forms a standing wave with two antinodes. (The cord is fixed at both ends.) Which harmonic does this wave represent
Answer:
the wave represents the second harmonic.
Explanation:
Given;
length of the cord, L = 64 cm
The first harmonic of a cord fixed at both ends is given as;
[tex]f_o = \frac{V}{2L}[/tex]
The wavelength of a standing wave with two antinodes is calculated as follows;
L = N---> A -----> N + N ----> A -----> N
Where;
N is node
A is antinode
L = N---> A -----> N + N ----> A -----> N = λ/2 + λ/2
L = λ
The harmonic is calculated as;
[tex]f = \frac{V}{\lambda} \\\\f = \frac{V}{L} = 2(\frac{V}{2L} ) = 2(f_o) = 2^{nd} \ harmonic[/tex]
Therefore, the wave represents the second harmonic.
L = λ
Please answer this for 15 points please don’t put in a link.
Answer:
c. Double Replacement
Explanation:
As in Double Replacement reaction exchanges the cations (or the anions) of two ionic compounds.
Here, in BaCl2 , Ba has replaced with NO3 to form Ba(NO3)2
and in 2AgNo3 , Ag has replaced with Cl to form 2AgCl.
Which one the answer to this question
If our atmosphere had a uniform density of 1.25 kg/m3 all the way up to a border with empty space above, that border would be Answer km above sea level. The pressure at sea level is 1 atm = 105 N/m2 and g = 10 m/s2. Enter your answer as an integer.
Answer:
The border is 8km above sea level.
Explanation:
We know that:
Density = 1.25 kg/m^3
Pressure = 10^5 N/m^2
g = 10m/s^2
Now, suppose that we have a virtual rectangle, such that its bases have an area of 1m^2 and the rectangle has a height equal to H.
This virtual figure has a volume V = 1m^2*H, and it is filled with air (which we know that has a density 1.25 kg/m^3)
Then the total mass inside that volume is:
M = (1.25 kg/m^3)*V = (1.25 kg/m^3)*(1m^2*H)
The weight of this mass is:
W = g*M = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)
And if we divide the weight in a given surface, let's say 1 m^2, we get the pressure per square meter, which we know is equal to 10^5 N/m^2
then:
P = 10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)
Whit this equation we can find the value of H.
10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)
10^5 N = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)
(10^5 N)/(10 m/s^2) = (1.25 kg/m^3)*(1m^2*H)
(10^4 kg) = (1.25 kg/m^3)*(1m^2*H)
(10^4 kg)/( 1.25 kg/m^3) = 1m^2*H
8,000 m^3 = 1m^2*H
(8,000 m^3)/(1m^2) =H
8,000 m = H
And we want this answer in km, knowing that 1,000m = 1km
8,000m = 8km = H
The border is 8km above sea level.
Height of boundaries is 8.2 km
Given that:Normal density = 1.25 kg/m³
1 atm = 101325 N/m²
Find:Height of boundaries
Computation:Pressure = Height × Density × Gravitational acceleration
101325 = Height × 1.25 × 9.8
101325 = Height × 12.25
Height of boundaries = 101325 / 12.25
Height of boundaries = 8271.42 m
Height of boundaries = 8.2 km
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A go-cart is traveling at 15 mi/hr. How long does it take the go-cart to travel 3 miles?
Answer:
12 min
Explanation:every 4 minutes is 1 mile
if the density of a napthalene ball is 0.02kg.what is the mass of the napthalene ball if it has a volume of 100m³
a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel during 30 seconds of rotation?
Answer:
239 rpm
Explanation: So the distance covered in one minute is 75,000 centimeters. The diameter of the wheel is 100 cm, so the radius is 50 cm, and the circumference is 100π cm. How many of these circumferences (or wheel revolutions) fit inside the 75,000 cm? In other words, if I were to peel this wheel's tread from the cart and lay it out flat, it would measure a distance of 100π cm. How many of these lengths fit into the entire distance covered in one minute? To find out how many of (this) fit into so many of (that), I must divide (that) by (this), so:
100πcm/rev
75,000cm/min
750 min rev≈238.7324146RPM
A 0.70-kg disk with a rotational inertia given by MR 2/2 is free to rotate on a fixed horizontal axis suspended from the ceiling. A string is wrapped around the disk and a 2.0-kg mass hangs from the free end. If the string does not slip then as the mass falls and the cylinder rotates the suspension holding the cylinder pulls up on the mass with a force of______
Answer:
The force will be "9.8 N".
Explanation:
The given values are:
mass,
m = 0.7 kg
M = 2
g = 9.8
Now,
⇒ [tex]\tau = T \alpha[/tex]
then,
⇒ [tex]\frac{1}{2}mR^2(\frac{1}{R}\frac{dv}{dt}) =M(g-a_t)R[/tex]
⇒ [tex]\frac{1}{2}m \ a_t=m(g-a_t)[/tex]
⇒ [tex]a_t=\frac{2g}{(\frac{m}{M} +2)}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{2\times 9.8}{\frac{0.7}{2} +2}[/tex]
⇒ [tex]=8.34 \ m/s[/tex]
hence,
⇒ [tex]T=mg+M(g-a_t)[/tex]
On substituting the values, we get
⇒ [tex]=0.7\times 9.8+2(9.8-8.34)[/tex]
⇒ [tex]=6.86+2(1.46)[/tex]
⇒ [tex]=6.86+2.92[/tex]
⇒ [tex]=9.8 \ N[/tex]
An object, with mass 64 kg and speed 14 m/s relative to an observer, explodes into two pieces, one 2 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame
Answer:
K_f = 1881.6 J
Explanation:
To solve this exercise, let's start by finding the velocities of the bodies.
We define a system formed by the initial object and its parts, with this the forces during the explosion are internal and the moment is conserved
initial instant. Before the explosion
p₀ = M v₀
final instant. After the explosion
p_f = m₁ v + m₂ 0
the moeoto is preserved
p₀ = p_f
M v₀ = m₁ v
v = [tex]\frac{m_1}{M}[/tex] v₀
in the exercise they indicate that the most massive part has twice the other part
M = m₁ + m₂
M = 2m₂ + m₂ = 3 m₂
m₂ = M / 3
so the most massive part is worth
m₁ = 2 M / 3
we substitute
v = ⅔ v₀
with the speed of each element we can look for the kinetic energy
initial
K₀ = ½ M v₀²
Final
K_f = ½ m₁ v² + 0
K_f = ½ (⅔ M) (⅔ v₀)²
K_f = [tex]\frac{8}{27}[/tex] (½ M v₀²)
K_f = [tex]\frac{8}{27}[/tex] K₀
the energy added to the system is
ΔK = Kf -K₀
ΔK = (8/27 - 1) K₀
ΔK = -0.7 K₀
K_f = K₀ + ΔK
K_f = K₀ (1 -0.7)
K_f = 0.3 K₀
let's calculate
K_f = 0.3 (½ 64 14²)
K_f = 1881.6 J
A person applies a constant force of 100 N on an object of mass 20 kg. The object moves horizontally at a constant speed of 0.20 m/s through a distance of 0.80 m. What is the net work done on the object
Answer:
the net work done on the object is 80 J.
Explanation:
Given;
force applied on the object, F = 100 N
mass of the object, m = 20 kg
speed of the object, v = 0.2 m/s
distance moved by the object, d = 0.8 m
The net work done on the object is calculated as follows;
W = F x d
W = 100 N x 0.8 m
W = 80 J
Therefore, the net work done on the object is 80 J.
Which runner finished the 100 m race in the least amount of time?
Ming
Which runner stopped running for a few seconds during the race?
At what distance did Anastasia overtake Chloe in the race?
1: Ming
2: Chloe
3: 40m
Light of wavelength 656 nm and 410 nm emitted from a hot gas of hydrogen atoms strikes a grating with 5300 lines per centimeter. a) Determine the angular deflection of both wavelengths in the 1st and 2nd order.
Answer:
[tex]20.32^{\circ}[/tex] and [tex]44.08^{\circ}[/tex]
[tex]12.56^{\circ}[/tex] and [tex]25.77^{\circ}[/tex]
Explanation:
[tex]\lambda[/tex] = Wavelength
[tex]\theta[/tex] = Angle
m = Order
Distance between grating is given by
[tex]d=\dfrac{1}{5300}\\\Rightarrow d=0.0001886\ \text{cm}[/tex]
[tex]\lambda=656\ \text{nm}[/tex]
We have the relation
[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}[/tex]
m = 1
[tex]\theta=\sin^{-1}\dfrac{1\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=20.35^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=44.08^{\circ}[/tex]
The first and second order angular deflection is [tex]20.32^{\circ}[/tex] and [tex]44.08^{\circ}[/tex]
[tex]\lambda=410\ \text{nm}[/tex]
m = 1
[tex]\theta=\sin^{-1}\dfrac{1\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=12.56^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=25.77^{\circ}[/tex]
The first and second order angular deflection is [tex]12.56^{\circ}[/tex] and [tex]25.77^{\circ}[/tex].
the number of perpandicular components of a force is
I WILL GIVE BRAINLIEST TO CORRECT ANSWER
Two boxes are being pulled to the right by a force F across a frictionless table. The force is acting directly on M 2 which is connected to M 1 by a massless cord. Which of the following expressions best represents the acceleration of M 1?
Daryl ties a rope to a brick and lifts the brick straight up. The free-body
diagram below shows the brick when it is suspended above the ground.
Force 1
Force 2
What is force 1 in this diagram?
O A. Friction
OB. Tension
O C. Normal force
O D. Weight
The force 1 is tension force.
To find the correct statement among all the options, we need to know more about friction, tension, normal force and weight.
What is friction?Friction force is found between two surfaces when one is kept or moved on another surface.It is directed opposite to the direction of motion.What is tension force?When any object is hanged by an thread or rope, that object exerts a force on that rope. This force is called as tensional force.It's directed from along the rope towards the point of hanging.What is normal force?When an object is kept on a surface, the surface exerts a force on the object to oppose the weight of the object which is the normal force.It's perpendicular to the surface that an object contacts.What is weight?Weight is the gravitational force exerted by earth on that object. It's always directed towards the center of the earth.Thus, we can conclude that the correct option is (B).
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A bird travels at a speed of 14.2 m/s for 514 meters. How many seconds did it
fly?
Answer:
0.54 sec
Explanation:
Answer:
Time = 36.19 secondsExplanation:
Speed = 14.2 m/s
Distance = 514 m
Time = Distance / Speed
Time = 514 / 14.2
Time = 36.19 seconds
A hand dryer blows heated air downwards out of the exit duct at a velocity of 4 m/s. The temperature and density of the ambient air at the inlet are 15 C and 1.23 kg/m3, while at the outlet it has temperature 35 C and density 1.15 kg/m3 The blower power is 10.0 W and the heater power is 715 W. Consider the inlet to be at the large mass of ambient air which has negligible velocity.
a) What is the pressure at the outlet? 4 m/s, 35 C
b) You will be applying the energy equation. Why can you ignore any height differences in this situation?
c) If the specific heat of air C-1000 J/(kg K), where Δυ-C Δ T, find the change in internal energy per unit mass from the inlet to outlet.
d) Find the mass flow rate through the dryer.
e) What is the power loss in the system?
f) What is the loss in the system?
g) What is the head loss in the system?
h) What is the total loss coefficient of the system, referred to the outlet velocity?
i) If there were no heater, would the temperature of gas at the outlet be higher, the same, or lower than the inlet? Explain why.
Answer:
nzkdjdksishdjsdjjdjnzkskejeoueeuieeoyrie ryrhdhcksodopdncndnszdkhfoeosheiwuef wokxkzdkjdoeehxjbxbdkeiehdhdhddjjddjdhhdhdhhhjdhfjdjjfjddhdheudiwiehdbejwowud
Explanation:
isos
Light containing two different wavelengths passes through a diffraction grating with 1,250 slits/cm. On a screen 17.5 cm from the grating, the third-order maximum of the shorter wavelength falls midway between the central maximum and the first side maximum for the longer wavelength. If the neighboring maxima of the longer wavelength are 8.44 mm apart on the screen, what are the wavelengths in the light
Answer:
[tex]\lambda_s =6.43*10^-4m[/tex]
Explanation:
From the question we are told that:
Diffraction grating [tex]N=1250slits/cm[/tex]
Distance b/w Screen and grating length [tex]d_{sg}=17.5 cm[/tex]
Distance b/w neighboring maxima and Screen [tex]d_{ms}=8.44[/tex]
Generally the equation for grating space is mathematically given by
[tex]d(g)=\frac{1}{N}[/tex]
[tex]d(g)=\frac{100}{1250}[/tex]
[tex]d(g)=0.08[/tex]
Generally the equation for small angle approximation is mathematically given by
[tex]\triangle y=\frac{\lambda d}{L}[/tex]
Therefore for longest wavelength
[tex]\lambda _l=\frac{8.44*10^{-3}*(0.08)}{0.175m}[/tex]
[tex]\lambda _l=3.858*10^{-3}[/tex]
Therefore the third order maximum equation for the shorter wavelength as
[tex]\lambda_s =\frac{1}{6} \lambda_l[/tex]
[tex]\lambda_s =\frac{1}{6} (3.858*10^-^3)[/tex]
[tex]\lambda_s =6.43*10^-4m[/tex]
The wavelengths in the light is given as
[tex]\lambda_s =6.43*10^-4m[/tex]
need help ASAP!!!!!!!!!!!
Answer:
The equation says that due to variation in temperature is
delt T = .59 m/s / C = 16 C * .59 m/s = 9.44 m/s
So v = 332 m/s + 9.44 m/s = 341 m/s (to three significant figures)
Which layer of the atmosphere is the least dense?
Question 5 options:
exosphere
thermosphere
mesosphere
troposphere
the exosphere is the least dense
Answer:
exosphere
Explanation: Just took the quiz and got an A, 100%
Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A person stands 5.0 m away from one speaker and 6.2 m away from the other. Calculate the second lowest frequency that results in destructive interference at the point where the person is standing. Assume the speed of sound to be 330 m/ s. A) 183 Hz B) 275 Hz C) 413 Hz D) 137 Hz E) 550 Hz
Answer:
C) 413 Hz
Explanation:
For destructive interference, the path difference ΔL = (n + 1/2)λ where ΔL = L₂ - L₁ where L₁ = person's distance from one speaker (the closer one) = 5.0m and L₂ = person's distance from other speaker (the farther one) = 6.2 m and λ = wavelength = v/f where v = speed of sound = 330 m/s and f = frequency
So, ΔL = (n + 1/2)λ
L₂ - L₁ = (n + 1/2)v/f
f = (n + 1/2)v/(L₂ - L₁)
At the second lowest frequency that results in destructive interference at the point where the person is standing, n = 1.
So,
f = (1 + 1/2)v/(L₂ - L₁)
f = 3v/2(L₂ - L₁)
Substituting the values of the variables into the equation, we have
f = 3v/2(L₂ - L₁)
f = 3(330 m/s)/2(6.2 m - 5.0 m)
f = 3(330 m/s)/2(1.2 m)
f = 990 m/s ÷ 2.4 m)
f = 412.5 Hz
f ≅ 413 Hz
Water can form large dewdrops in nature how would droplets made of isopropyl alcohol instead of water be different
Answer:
isopropyl alcohol would form smaller droplets, because it has lower surface tension than water has
Explanation:
Ap3x
The droplets made of isopropyl alcohol instead of water be smaller due to surface tension.
What is droplets?The single drop of a liquid in the form of sphere is called droplet.
Water can form large dewdrops in nature. Isopropyl alcohol would form smaller droplets, because it has lower surface tension than water.
Surface tension is the property of the liquid to acquire minimum surface area.
Thus, droplets made of isopropyl alcohol instead of water be smaller.
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What is a transfer of energy called?
A. Displacement
B. Acceleration
C. Work
D. Torque
At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM
Answer:
[tex]4.47\ \text{km/h}[/tex]
Explanation:
[tex]\dfrac{da}{dt}[/tex] = Rate at which the distance between A and starting point of B is changing = -20 km/h
[tex]\dfrac{db}{dt}[/tex] = Rate at which the distance of B is changing = 15 km/h
[tex]\dfrac{dc}{dt}[/tex] = Rate at which the distance between A and B is changing
Time after which the rate at which the distance between A and B is changing is 4 hours
Distance covered by A in 4 hours = [tex]20\times 4=80\ \text{km}[/tex]
a = Distance remaining to the start point of B = [tex]110-80=30\ \text{km}[/tex]
b = Distance covered by B in 4 hours = [tex]15\times 4=60\ \text{km}[/tex]
Distance between A and B after 4 hours
[tex]c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}[/tex]
[tex]c^2=a^2+b^2[/tex]
Differentiating with respect to time we get
[tex]c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}[/tex]
The rate at which the distance between the ships is changing at 4 PM is [tex]4.47\ \text{km/h}[/tex].
a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel during 30 seconds of rotation?
A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1490 Hz. The bird-watcher, however, hears a frequency of 1505 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound
Answer:
The speed of the bird is 1.00% of the speed of sound.
Explanation:
The speed of the bird can be found by using the Doppler equation:
[tex] f = f_{0}(\frac{v - v_{r}}{v - v_{s}}) [/tex]
Where:
v: is the speed of sound = 343 m/s
f₀: is the frequency emitted = 1490 Hz
f: is the frequency observed = 1505 Hz
[tex]v_{r}[/tex]: is the speed of the receiver = 0 (it is stationary)
[tex]v_{s}[/tex]: is the speed of the source =?
The minus sign of [tex]v_{s}[/tex] is because the source is moving towards the receiver.
By solving the above equation for [tex]v_{s}[/tex] we have:
[tex] v_{s} = v - \frac{f_{0}*v}{f} = 343 - \frac{1490*343}{1505} = 3.42 m/s [/tex]
The above speed in terms of the speed of sound is:
[tex]\% v_{s} = \frac{3.42}{343}\times 100 = 1.00 \%[/tex]
Therefore, the speed of the bird is 1.00% of the speed of sound.
I hope it helps you!
Find the wavelength of light which is capable of ionizing a hydrogen atom?
Answer:
The correct answer is - 91.4 nm
Explanation:
According to Bohr's model, the minimum wavelength to ionize Hydrogen atom from n= 1 state is expressed as:
(h×c)/λ=13.6eV
here,
h - Planck constant
c - the speed of light
λ - wavelength
Placing the value in the formula for the wavelength
(6.626×10^−34J.s × 3×10^8 m/s)/λ = 13.6 ×1.6 × 10^−19 J
λ≈91.4nm
Thus, the correct answer would be = 91.4 nm
explain the refraction of light on a glass slab
Answer:
refraction is the change in direction of a wave passing from one medium to another or from a gradual change in the medium.
types of wave interactions include
Light rays enter a transparent material. Which description best describes what happens to the light rays
When light encounters transparent materials, almost all of it passes directly through them. Glass, for example, is transparent to all visible light. ... Most of the light is either reflected by the object or absorbed and converted to thermal energy. Materials such as wood, stone, and metals are opaque to visible light.