Answer:
Approximately [tex]13\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] (assuming that the boiling point of water in this experiment is [tex]100\; \rm ^\circ C\![/tex].)
Explanation:
Latent heat of condensation/evaporation of water: [tex]2260\; \rm J \cdot g^{-1}[/tex].
Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to [tex]\rm J \cdot g^{-1}[/tex].
Specific heat of water: [tex]4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}[/tex].
Specific heat of copper: [tex]0.39\; \rm J \cdot g^{-1}\cdot K^{-1}[/tex].
The temperature of this calorimeter and the [tex]250\; \rm g[/tex] of water that it initially contains increased from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex]. Calculate the amount of energy that would be absorbed:
[tex]\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J \end{aligned}[/tex].
[tex]\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J \end{aligned}[/tex].
Hence, it would take an extra [tex]585\; \rm J + 31500\; \rm J = 32085\; \rm J[/tex] of energy to increase the temperature of the calorimeter and the [tex]250\; \rm g[/tex] of water that it initially contains from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex].
Assume that it would take [tex]x[/tex] grams of steam at [tex]100\; \rm ^\circ C[/tex] ensure that the equilibrium temperature of the system is [tex]50\; \rm ^\circ C[/tex].
In other words, [tex]x\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] would need to release [tex]32085\; \rm J[/tex] as it condenses (releases latent heat) and cools down to [tex]50\; \rm ^\circ C[/tex].
Latent heat of condensation from [tex]x\; \rm g[/tex] of steam: [tex]2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J[/tex].
Energy released when that [tex]x\; {\rm g}[/tex] of water from the steam cools down from [tex]100\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex]:
[tex]\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J \end{aligned}[/tex].
These two parts of energy should add up to [tex]32085\; \rm J[/tex]. That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex].
[tex](2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J[/tex].
Solve for [tex]x[/tex]:
[tex]x \approx 13[/tex].
Hence, it would take approximately [tex]13\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] for the equilibrium temperature of the system to be [tex]50\; \rm ^\circ C[/tex].
The sound from a trumpet radiates uniformly in all directions in 20C air. At a distance of 5.00 m from the trumpet the sound intensity level is 52.0 dB. The frequency is 587 Hz. (a) What is the pressure amplitude at this distance
Answer:
The answer is below
Explanation:
The intensity level (B) of a sound wave is given by:
B = 10log(I/I₀);
where I₀ is the threshold intensity = 1 * 10⁻¹² W/m², I is the intensity at distance 5 m, B is the intensity level = 52 dB
Substituting gives:
[tex]52=10log(\frac{I}{10^{-12}} )\\\\log(\frac{I}{10^{-12}} )=5.2\\\\I=1.58*10^{-7}\ W/m^2[/tex]
The pressure is given by:
[tex]I=\frac{p_{max}^2}{2\rho v} \\\\\rho=air\ density=1.2\ kg/m^3,v=speed\ of\ sound\ in\ air=344\ m/s,p_{max}=pressure:\\\\p_{max}=\sqrt{2\rho vI}=\sqrt{2*1.58*10^{-7}*1.2*344} =1.14*10^{-2}Pa[/tex]
Which one is the dependent variable in distance, force, or work
Answer:
Distance
Explanation:
Work can be defined as the energy transferred to a physical object by exertion of a force on the object to cause a displacement of the object. Thus, work is typically done when a person or simple machine move an object over a distance through the application of a force.
Mathematically, work done is given by the formula;
[tex] W = F * d[/tex]
Where,
W is the work done
F represents the force acting on a body.
d represents the distance covered by the body.
A dependent variable is the event expected to change when an independent variable is manipulated.
Hence, distance is the dependent variable because its value changes with respect to the amount of force exerted on an object.
What is the internal resistance of a current source with an EMF of 12 V if, when a resistor with an unknown resistance is connected to it, a current of 2 A flows through the circuit? A voltmeter connected to the source terminals shows 8 V.
How much time will it take for a person to walk the length of a football field (100 yards)
at a constant speed of 5 ft/s ?
The speed is in feet per seconds so change the length of the field from yards to feet.
1 yard = 3 feet
100 yards x 3 = 300 feet
The field is 300 feet long
Time = distance / speed
Time = 300 feet / 5 feet per second = 60 seconds = 1 minute
It will take 1 minute
Answer:
A person will take 60 Seconds to walk the distance of 100 yards.
Explanation:
Data Given ;
Distance ( d ) = 100 yards = 300 Ft
Speed ( v ) = 5 Ft/s
Time ( t ) = ?
What is speed ?The distance travelled in unit time is called speed.
formula ; [tex]v = \frac{d}{t} \\[/tex]
On putting values,
[tex]5 = \frac{300}{t}[/tex]
[tex]t =\frac{300}{5}[/tex]
[tex]t = 60 sec[/tex]
Hence the time taken by the person is 60 sec.
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A electron gains electric potential energy as it moves from point 1 to point 2. Which of the following is true regarding the electric potential at points 1 and 2?
a. V1 = V2
b. V1 > V2
c. V1 < V2
Answer:
We know that the change in electric potential energy is defined as:
q*ΔV = ΔP
So, the change in the electric potential energy is the charge times the change in the electric potential.
For the case of an electron gas, we have:
q = -e
where -e is the charge of an electron (remember that is negative).
So, if the electron gains electric potential then:
ΔP > 0
this means that the final potential energy is larger than the initial one, then we have:
-e*ΔV > 0
This means that ΔV must be negative.
V₂ = electric potential at point 2, so it is the final electric potential
V₁ = electric potential at point 1, so it is the final electric potential
Then we should get:
ΔV = V₂ - V₁ < 0.
This means that:
V₂ < V₁
The correct option is b.
Convert Rev/min to rad/s x 2pie/60?
Anyone knows this please?
Answer:
Thus, [tex]\frac{1 rev}{min} =\frac{2\pi}{60} rad/s[/tex]
Explanation:
The angular speed is defined as the rate of change of angular velocity.
Its SI unit is rad/s and other units are rev/min or rev/s.
[tex]\frac{1 rev}{min } = \frac{1 rev}{60 sec}\\\\1 rev = 2\pi rad\\\\So\\\\\frac{1 rev}{min} = \frac{2\pi}{60} rad/s[/tex]
What do we mean by a penetrative pass in the game of football?
Answer:
Penetration means forward passes can go through the opposition lines. Once these penetrative passes get through each line, it eliminates the line of players it broke through and leaves the player in possession closer to the opposition goal.
Explanation:
On Ramesh’s13th birthday, his father invited all his friends and their relatives. It was a big party with lots of food and DJs. Ramesh didn’t like the loud sound of DJs and asked his father to play it in a low volume so that their neighbours do not get much disturbed and people at the party can also enjoy the music. Ramesh’s father felt good for his wisdom and did as he said.
→Do you think when loud music is played at a party is acceptable to all the people living in, neighbourhood? Give a reason for your answer
→How can you control noise pollution at your end?
A horizontal force is applied to a 4.0 kg box. The box starts from rest, moves a horizontal distance of 10.0 meters, and obtains a velocity of 7.0 m/s. The change in the kinetic energy is:_____.
Answer:
98 J
Explanation:
Applying,
Change in kinetic energy = Final kinetic energy- initial kinetic energy
ΔK.E = mv²/2-mu²/2..............Equation 1
Where ΔK.E = Change in kinetic energy, m = mass of the box, u = initial velocity of the box, v = final velocity of the box.
From the question,
Given: m = 4.0 kg, u = 0 m/s, v = 7 ,0 m/s
Substitute these values into equation 1
ΔK.E = (4(7²)/2)-(4(0²)/2)
ΔK.E = (2×49)-0
ΔK.E = 98 J
Hence the change in kinetic energy 98 J
human activities that interfere with distribution of natural resources are contributing to the increase of earthquake risk.
Answer:
Changes in climate conditions, especially the warming of global temperatures increases the likelihood of weather-related natural disasters. ... This is most visible when seen through changes in the intensity and frequency of droughts, storms, floods, extreme temperatures and wildfires.
It is said that a gas fills all the space available to it. Why then doesn't the atmosphere go off into space?
Estimate the force a person must exert on a massless string attached to a 0.15 kg ball to make the ball revolve in horizontal circle of radius 0.6 m. The ball makes 2 revolutions per second.
Answer:
[tex]F = 14.2 N[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=0.15kg[/tex]
Radius [tex]r=0.6[/tex]
Angular Velocity [tex]\omega=2rev/s[/tex]
[tex]\omega= =2x2 \pi rad/s=>4 \pi rad/s[/tex]
Generally the equation for Force applied is mathematically given by
[tex]F =mrw2[/tex]
[tex]F=0.15*0.6* (4*x3.14^)2[/tex]
[tex]F = 14.2 N[/tex]
A charged particle accelerates as it moves from location A to location B. If VA = 260 V and VB = 210 V, what is the sign of the charged particle? positive negative (b) A electron loses electric potential energy as it moves from point 1 to point 2. Which of the following is true regarding the electric potential at points 1 and 2?
Answer:
(a) Positive
(b) Electron gains energy as it moves from A to B.
Explanation:
VA = 260 V
VB = 210 V
An electron moves from lower to higher potential which is negatively charged and a positively charged particle moves from higher to lower potential, so the charge particle is positive in nature.
(a) Positive
(b) No, electron gains energy as it moves from A to B.
explain how a lever can act as a force multiplier
Answer:
Example:Opening of a bottle cap by tool
when we hold a tool and open the bottle cap this is because , force x tool force .
The load arm is usually shorter than the effort arm in second order levers. Moving a large weight hence requires less effort. A force multiplier lever or effort multiplier lever is the name for this kind of lever. A boat's oars, for instance, can increase the force.
What is second order levers?Second-order levers are devices with the input force farthest from the fulcrum and the output force on the same side of the fulcrum. A wheelbarrow is an excellent illustration of a second-order lever.
A second-order lever will have an output force greater than an input force, similar to first-order levers. The output journey, however, will be shorter than the input length. Both the input and output forces in this situation will move in the same direction.
Learn more about lever here:
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A circular wire loop is placed near a long, straight, current-carrying wire in which the current is either increasing or decreasing. Draw the situations in which the induced current in the loop counterclockwise?
Answer:
The induced current is counter clockwise if the current is decreasing and towards right.
Explanation:
When the current is decreasing in the wire, the direction of magnetic field at the center of the loop is outwards to the plane of paper which is given by the Maxwell's right hand thumb rule. The magnetic field is decreasing in nature.
So according to the Lenz's law, the induced current is such that which opposes the cause of its production, so that the induced current is counter clockwise.
A man standing on a frictionless ice throws a 1.00kg mass at 20m/s at an angle elevation of 40.0 degrees. What was the magnitude of the mans momentum immediately after the the throl
Answer:
Explanation:
1.00kg×20m/s×cos40=15.3
Kinematics equations tells us the position of an object under constant acceleration increases linearly with time.
A. True
B. False
Answer:
False.
Explanation:
Suppose that we have an object that moves with constant acceleration A.
Then the acceleration of the object is defined by the equation:
a(t) = A
The acceleration is the rate of change of the velocity, then the velocity equation is given by the integration of the acceleration equation, we will get:
v(t) = A*t + V₀
Where V₀ is the velocity of the object at the time t = 0s.
Now, if we integrate it again, we will get the position equation:
p(t) = (1/2)*A*t^2 + V₀*t + P₀
Where P₀ is the initial position equation.
Here, we can see that the position equation is a quadratic equation (not a linear equation), then the statement is false.
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a potential difference of 1.43 V is needed to reduce the current to zero. What is the energy of a photon of this light in electron volts? energy of a photon: Find the work function of the irradiated metal in electron volts. work function:
Answer:
The right solution is:
(a) 2.87 eV
(b) 1.4375 eV
Explanation:
Given:
Wavelength,
= 433 nm
Potential difference,
= 1.43 V
Now,
(a)
The energy of photon will be:
E = [tex]\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}[/tex]
= [tex]4.59\times 10^{-19} \ J[/tex]
or,
= [tex]\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}[/tex]
= [tex]2.87 \ eV[/tex]
(b)
As we know,
⇒ [tex]Vq=\frac{hc}{\lambda}-\Phi_0[/tex]
By substituting the values, we get
⇒ [tex]1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0[/tex]
⇒ [tex]\Phi_0=2.3\times 10^{-19} \ J[/tex]
or,
⇒ [tex]=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}[/tex]
⇒ [tex]=1.4375 \ eV[/tex]
A 10,000J battery is depleted in 2h. What power consumption is this? *
A) 5000W
B) 3W
C) 1.4W
D) 20000W
show your work please
Answer:
P = 1.4 W
Explanation:
Given that,
The work done or the energy of the battery, E = 10,000 J
Time, t = 2 h
We need to find the power consumption. Let it is P. Power is the rate of doing work. So,
[tex]P=\dfrac{W}{t}\\\\P=\dfrac{10,000}{2\times 3600}\\\\P=1.38\ W[/tex]
or
P = 1.4 W
So, the power of the battery is 1.4 W.
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.26.
Required:
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?
Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
A rock is thrown from the edge of the top of a 51 m tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 74 m from the base of the building 8 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.
Answer:
The speed of projection is 34 m/s.
Explanation:
Height of building, h = 51 m
horizontal distance, d = 74 m
time, t = 8 s
Let the angle is A and the speed is u.
d = u cos A x t
74 = u cos A x 8
u cos A = 9.25 .... (1)
Use second equation of motion
[tex]h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)[/tex]
Squaring and adding both the equations
[tex]u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s[/tex]
We say that evaporation is a cooling process. (a) What cools and what warms? We say that condensation is a warming process. (b) What warms and what cools?
Answer:
a liquid warms gas cools
b solid piece warms and liquid cools
An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the point of origin of the flight. The plane flies with an airspeed of 120 m/s. If a constant wind blows at 10 m/s toward the west during the flight, what direction must the plane fly relative to the air to arrive at the destination
Answer:
The right solution is "4.8° east of north".
Explanation:
Given:
Distance,
= 500 km
Speed,
[tex]\vec{v}=120 \ m/s[/tex]
Wind (towards west),
[tex]v_0=10 \ m/s[/tex]
According to the question, we get
The angle will be:
⇒ [tex]\Theta=Cos^{-1}(\frac{v_0}{v_1} )[/tex]
[tex]=Cos^{-1}(\frac{10}{120} )[/tex]
[tex]=85.21[/tex] (north of east)
hence,
The direction must be:
⇒ [tex]\Theta'=90-85.21[/tex]
[tex]=4.79^{\circ}[/tex]
or,
[tex]=4.8^{\circ}[/tex] (east of north)
calculate the electric potential 3mm from a point charge of 16Nc
[tex]4.8 \times 10^8[/tex] volts
Explanation:
The electric potential due to a point charge is given by
[tex]V= \dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{Q}{r}[/tex]
where Q = charge = [tex]16 \times 10^{-9}[/tex] C
r = distance from a point = [tex]3 \times 10^{-3}[/tex] m
[tex]\varepsilon_{0}[/tex] = permitivity of free space
= 8.85×10^-12 C^2/N-m^2
Plugging in the numbers,
[tex]V = \dfrac{1}{4 \pi (8.85 \times 10^{-12})} \dfrac{16 \times 10{-9}}{3 \times 10^{-3}}[/tex]
[tex]= 4.8 \times 10^8[/tex] volts
How does the density of water change when: (a) it is heated from 0o
C to
4o
C; (b) it is heated from 4o
C to 10o
C ?
Answer:
[b] it id heated from 4o
Explanation:
A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?
Answer:
First of all the formula is F= uR,( force= static friction× reaction)
mass= 5+25=30
F= 50
R= mg(30×10)=300
u= ?
F=UR
u= F/R
u= 50/300=0.17N
If you wanted to know how much the temperature of a particular piece of material would rise when a known amount of heat was added to it, which of the following quantities would be most helpful to know?
a. coefficient of linear expansion
b. specific heat
c. initial temperature
d. thermal conductivity
e. density
Answer:
Option (b) is correct.
Explanation:
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C, is called specific heat of the substance.
The formula of the specific heat is
H = m c (T' - T)
where, m is the mass, c is the specific heat and T' - T is the change in temperature.
So, to know the rise in temperature, by adding the known amount of heat, the specific heat is required.
So, option (b) is correct.
At a playground, Maryam a 3-year old girl and Zahirah a 6-year old girl are playing with the swings. Maryam is sitting while Zahirah is standing on the swing. Both of them were given the same push by their mother. Choose the CORRECT statements:
A. Maryam is swinging faster than Zahirah.
B. Zahirah is swinging faster than Maryam.
C. Both swings at the same pace.
D. Maryam is swinging faster since she is younger.
E. Zahirah is swing faster since she is older.
Answer:
both swings at the same place
Explanation:
because there mother is giving same amount of force to both.
A 16 kg science book is dropped of a 120 meter high cliff. Assuming a closed system:
a) how fast is book traveling the instant before it impacts the ground below the cliff?
b) how far above the bottom of the cliff is the object moving at 12 m/s?
Answer:
Explanation:
The mass of that science book...wow. In pounds that would be 35.2! Yikes!
Anyway, we need final velocity here, and the mass of the book has nothing to do with how fast it falls. Everything is pulled by the same gravity. A feather falls at 9.8 m/s/s and so does an elephant. Mass is useless information. The equation we will use is
[tex]v^2=v_0^2+2a[/tex]Δx where
v is the final velocity, our unknown,
v₀ is the initial velocity which is 0 since someone had to be holding the book before dropping it,
a is the pull of gravity which is always -9.8 m/s/s, and
Δx = -120 which is the displacement (it's negative because the book falls below the point from which it was dropped). Filling in:
[tex]v^2=0^2+2(-9.8)(-120)[/tex] so
[tex]v=\sqrt{2(-9.8)(-120)}[/tex] and
v = 48 m/s
As far as how far above the bottom of the cliff the object is when it is moving at 12 m/s we will use the same equation, but the velocity will be 12:
[tex]12^2=0^2+2(-9.8)[/tex]Δx and
144 = -19.6Δx so
Δx = -7.3 m. That's how far from the top of the cliff it is. We subtract then t find out how far it is from the bottom:
120 - 7.3 = 112.7 m off the ground.
How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?
Answer:
(a) v₁ = 51.96 km/h
(b) v₁ = 178 km/h
Explanation:
(a)
For having the same momentum:
m₁v₁ = m₂v₂
where,
m₁ = mass of Volkswagen = 816 kg
v₁ = speed of Volkswagen = ?
m₂ = mass of Cadillac = 2650 kg
v₂ = speed of Cadillac = 16 km/h
Therefore, using these values in the equation, we get:
[tex](816\ kg)v_1 = (2650\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{2650\ kg}{816\ kg})[/tex]
v₁ = 51.96 km/h
(b)
For having the same momentum:
m₁v₁ = m₂v₂
where,
m₁ = mass of Volkswagen = 816 kg
v₁ = speed of Volkswagen = ?
m₂ = mass of Truck = 9080 kg
v₂ = speed of Truck = 16 km/h
Therefore, using these values in the equation, we get:
[tex](816\ kg)v_1 = (9080\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{9080\ kg}{816\ kg})[/tex]
v₁ = 178 km/h
