Answer:
A) 8,368 J
B) ) 0.893 J/gºC
Explanation:
A)
The heat gained by the water can be obtained solving the following equation:[tex]q_{g} = c_{w} * m * \Delta T (1)[/tex]
where cw = specific heat of water = 4.184 J/gºCm= mass of water = 1,000 gΔT = 2ºC Replacing these values in (1) we get:[tex]q_{g} = c_{w} * m * \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)[/tex]
B)
Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative: -8,368 J.Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:[tex]q_{l} = c_{Al} * m_{Al} * \Delta T (3)[/tex]
⇒ [tex]-8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)[/tex]
[tex]c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)[/tex]
which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.
A rectangular reflecting pool is 85.0 ft wide and 120 ft long. What is the area of the pool in square meters?
true or false A person's speed around the Earth is faster at the poles than it is at the equator.
Answer:False
Explanation:The Earth rotates faster at the equator than at the poles.
Can a single atom be considered a molecule?
A:only if the atom is found in water
B:no, it takes two or more atoms bonded to create a molecule
C:only if it is an oxygen atom floating in the air
D:yes, all atoms are made up of many different molecules
The speed limit on some segments of interstate 4 is 70 mph. What is this in km/h?
Answer:
112.63km/hr
Explanation:
The given dimension is :
70mph
We are to convert this to km/hr
1 mile = 1.609km
so;
70mph x 1.609 = 112.63km/hr
So,
The solution is 112.63km/hr
The second law of thermodynamics imposes what limit on the efficiency of a heat engine?
A. The energy a heat engine must deposit in a cold reservoir is greater than or equal to the energy it extracts from a hot reservoir.
B. The energy a heat engine must deposit in a cold reservoir is greater than or equal to the energy extracted as useful work.
C. A heat engine must deposit some energy in a cold reservoir.
Answer:
C. A heat engine must deposit some energy in a cold reservoir.
Explanation:
The second law of thermodynamics says that "It is impossible to extract an amount of heat Q from a hot reservoir and use it all to do work W. Some amount of heat q must be exhausted to a cold reservoir."
This means that if we extract an amount of heat Q from the hot reservoir, the work W can never be exactly equal to Q, then there is a surplus of heat q that must be deposited in a cold reservoir.
Then we have the equation:
Q = W + q
From this we can conclude that the correct option is:
C. A heat engine must deposit some energy in a cold reservoir.
There will be always some energy that is not transformed into work, and is deposited in a cold reservoir.
C. A heat engine must deposit some energy in a cold reservoir.
The second law of thermodynamics says that "It is impossible to extract an amount of heat Q from a hot reservoir and use it all to do work W. Some amount of heat q must be exhausted to a cold reservoir". This means that if we extract an amount of heat Q from the hot reservoir, the work W can never be exactly equal to Q, then there is a surplus of heat q that must be deposited in a cold reservoir. Then we have the equation: Q = W + q There will be always some energy that is not transformed into work, and is deposited in a cold reservoir.Therefore, option C is correct.
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at what speed does the kg ball move ?
Answer: Choice A) 2 meters per second
=======================================================
Explanation:
The smaller ball has momentum of
p = m*v
p = (1 kg)*(4 m/s)
p = 4 kg*m/s
All of this momentum transfers into the larger ball because the smaller ball comes to a complete stop.
For the larger ball, we have p = 4 and m = 2. Let's find v.
p = m*v
4 = 2*v
4/2 = v
2 = v
v = 2 m/s which is why the answer is choice A
The larger ball moves at a speed of 2 meters per second. The speed is cut in half compared to the smaller ball because the larger ball has more inertia (aka more mass), and therefore it takes more energy to move it. If you apply the same energy to each, then the smaller object moves faster.
A 2.6 kg ball is accelerated at 4.5 m/s2.
Calculate the force needed to achieve this feat.
Show all work including formula and units!
Answer:
[tex]12\:\mathrm{N}[/tex]
Explanation:
Force is given by the equation [tex]F=ma[/tex].
Plugging in given values, we have:
[tex]F=ma=2.6\cdot 4.5=11.7=\fbox{$12\:\mathrm{N}$}[/tex] (two significant figures).
WHat does that mean?
Which ray diagram demonstrates the phenomenon of absorption?
An illustration with a vector pointed right going through an opening in a boundary and splitting into 3 vectors. One up and to the right, one straight and one down to the right.
An illustration with a vector pointed right going through an opening in a boundary and turning into a transverse wave on the other side.
An illustration with a vector striking a boundary at an angle and a second vector coming off the boundary at the exact same angle.
Answer:
B
Explanation:
on edge
The illustration with a vector pointed right going through an opening in a boundary and turning into a transverse wave on the other side demonstrates the phenomenon of absorption, so, option B is correct.
What is absorption?Absorption, in wave motion, is the process by which a wave's energy is transferred to matter when the wave travels through it. The energy of an electromagnetic, acoustic, or other wave is related to the square of its amplitude, which is the maximum displacement or movement of a point on the wave.
The amplitude of a wave continuously diminishes as it travels through a substance. The medium is described as being transparent to a specific type of radiation if just a tiny portion of the energy is absorbed, whereas it is described as opaque if all the energy is lost.
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A bicycle racer rides from a starting marker to a turnaround marker at 10 m/s. She then rides back along the same route from the turnaround marker to the starting marker at 16 m/s. What is her average speed for the whole race?
Answer:
12.31 m/s
Explanation:
If we recall from the previous knowledge we had about speed,
we will know that:
speed = distance/ time.
As such:
The average speed of the rider bicycle is
average speed = total distance/ total time
Mathematically, it can be computed as:
[tex]v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}[/tex]
[tex]v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}[/tex]
[tex]v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}[/tex]
[tex]v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}[/tex]
[tex]\mathbf{v_{avg} =12.31 \ m/s}[/tex]
A student asks the following question:
"If all things with mass have a gravitational field, why doesn't this glue bottle and
stapler, sitting on the counter, stick together because of gravitational forces?"
Which classmate answers correctly?
Ashton says that the gravitational fields between the bottle and the stapler
cancel out because of Newton's 3rd Law.
O Natalie says that all things with mass have a gravitational field, but the force is
very weak and cannot be perceived around small objects.
Xavier says the bottle and the stapler are way too small to have a gravitational
field.
Katherine says the bottle and the stapler have a strong gravitational field, and
would move towards each other quickly if there were no friction on the counter.
Answer:
Natalie says that all things with mass have a gravitational field, but the force is very weak and cannot be perceived around small objects.
Explanation:
The force due to gravity is proportional to the mass of the object and inversely proportional to the square of the distance between objects. The Earth is so massive that the force due to its gravity is much greater than the force between objects on the counter.
If there were no friction, the objects might move toward each other, depending on what other masses were near them tending to cause them to move in other directions.
Natalie's explanation is about the best.
__
Additional comment
The universal gravitational constant was determined by Henry Cavendish in the late 18th century using lead balls weighing 1.6 pounds and 348 pounds. His experiment was enclosed in a large wooden box to minimize outside effects. While these masses are somewhat greater than those of a glue bottle and stapler, the experiment shows the force of gravity between "small" objects can be measured.
A ray of monochromatic light is incident on a plane mirror at and angle of 30. The angle of reflection for the light is
1)15
2)30
3)60
4)90
Answer:
30 degrees
Explanation: reflection, same angle
For a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.
Reflection occurs when radiation bounces off from a surface. Light is an electromagnetic wave and it can be reflected. According to the laws of reflection, the angle of incidence is equal to the law of reflection.
Hence, for a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.
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A girl walks 20.0 m east then 70.0 meters north. What is the girl’s displacement (mag. And direction)? What is the girl’s distance?
Explanation:
Given that,
A girl walks 20.0 m east then 70.0 meters north.
Displacement is the shortest path covered by an object. Let it is d. It is calculated as :
[tex]d=\sqrt{20^2+70^2} \\\\=72.80\ m[/tex]
For direction,
[tex]\theta=\tan^{-1}(\dfrac{70}{20})\\\\\theta=74.05^{\circ}[/tex]
Girl's distance = 20 m + 70 m
= 90 m
Hence, this is the required solution.
a 1 mole of an ideal gas is kept at 0°C during expansion from 30l to 10l .How much work is done on the gas during expansion
Answer:
20 J
Explanation:
Work done is given force by distance .
W= F * d where F is force given by the product of pressure and area
W= P* Δv where Δv is change in volume.
Given that ;
1 mole of an ideal gas is kept at 0°C, the pressure of the gas is : 1 atm.
Δv is change in volume , 30 l - 10l = 20 l
W= 1 * 20 = 20 J
Two spherical objects have masses of 100 kg and 200 kg. Their centers are
separated by a distance of 40 cm. Find the gravitational attraction between
them.
Answer:
8.34 x 10⁻⁶N
Explanation:
Given parameters:
Mass 1 = 100kg
Mass 2 = 200kg
Distance of separation = 40cm = 0.4m
Unknown:
Gravitational force of attraction between them = ?
Solution:
To solve this problem, we use the expression below which is derived from the Newton's law of universal gravitation:
Fg = [tex]\frac{G x mass 1 x mass 2}{d^{2} }[/tex]
G is the universal gravitation constant = 6.67 x 10⁻¹¹
d is the separation
Now;
Fg = [tex]\frac{6.67 x 10^{-11} x 100 x 200}{0.4^{2} }[/tex] = 8.34 x 10⁻⁶N
Many scientific studies have found that colds are caused by viruses. What is this? *
Fact
Interpretation
Analysis
Opinion
Answer:
Analysis
Explanation:
Because you must Analysis each and every cold too find out which virus caused this.
It’s weird because Interpretation and Analysis have the meaning of examination
i’ll mark you brainlist :)
A racecar makes 24 revolutions around a circular track of radius 2 meters in
162 seconds. Find the racecar's frequency
Answer:
[tex]0.15\: \mathrm{Hz}[/tex]
Explanation:
The frequency is of an object is given by [tex]f=\frac{1}{T}[/tex], where [tex]T[/tex] is the orbital period of the object.
Since the racecar makes 24 revolutions around a circular track in 162 seconds, it will take the racecar [tex]\frac{162}{24}=6.75\:\mathrm{s}[/tex] per revolution.
Therefore, the frequency of the racecar is [tex]\frac{1}{6.75}=\fbox{$0.15\:\mathrm{Hz}$}[/tex] (two significant figures).
The radius of the track is irrelevant in this problem.
which changes will increase the rate of reaction during combustion
Answer:
reducing temperature of the surrounding
Explanation:
combustion reactions are exothermic so they give off heat. reducing the temperature of the surrounding will enable more efficient energy transfer
A car is traveling at a constant speed of 20 m/s for 3 seconds. Then the driver puts on the brakes. The total distance the car travels is 100 m. What is the total time the car was moving?
Answer:
15 seconds
Explanation:
If car was moving at 20m/s for 3 sec.
if car traveled 100m = 15 sec total
A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from it at 21.8 m/s, what is the received sound frequency?
f= ? Hz
Answer:
f" = 40779.61 Hz
Explanation:
From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;
from the Doppler effect equation, we can calculate the initial observed frequency as:
f' = f(1 - (v_o/v))
We are given;
f = 46.2 kHz = 46200 Hz
v_o = 21.8 m/s
v is speed of sound = 343 m/s
Thus;
f' = 46200(1 - (21/343))
f' = 43371.4285 Hz
In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;
Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;
f" = f'/(1 + (v_o/v))
f" = 43371.4285/(1 + (21.8/343))
f" = 40779.61 Hz
Our Sun’s mass is 1.0 and our Earth’s mass is 2.0. The distance is standard as given on the simulation. Describe the path of the Earth.
Answer:
Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).
Explanation:
PLS HELP ME!
A motorist is traveling 40ms-¹ and applies brakes and slow down at a rate of 2ms-² the available distance for the the motorist to stop is 400m will the motorist be able to stop?
Answer:
[tex] \underline{ \boxed{ yes}}\\[/tex]
Explanation:
[tex]given : initial \: velocity \: (u )= 40 {ms}^{ - 1} \\ given : final \: velocity \: (u )= 0 {ms}^{ - 1} \\ given : - (acceleration) \: (a_r) = 2 {ms}^{ - 2} \\ given : distance \: (s) \: = \: ? : \\ but \: {v}^{2} = {u}^{2} + 2( a)s\\ {0}^{2} = {40}^{2} + 2( - 2)s \\ - {40}^{2} = - 4s \\ s = \frac{ - {40}^{2} }{ - 4} \\ s = \frac{1600}{4} \\s = 400 \: m[/tex]
A charge of 7.1 x 10-4 C is placed at the origin of a Cartesian coordinate system. A second charge of 6.5 x 10-4 C lies 20 cm above the origin, and a third charge of 8.9 x 10-4 C lies 20 cm to the right of the origin. Determine the direction of the total force on the first charge at the origin. Express your answer as a positive angle in degrees measured counter clockwise from the positive x-axis.
Answer:
α = 36.21 °
β = 143.79°
Explanation:
To do this, we need to know the expression to calculate the angle.
In this case:
α₁ = tan⁻¹ (Fy₁/Fx₁) (1)
Now, let's analize the given data.
We have a charge q₁ at the origin of the cartesian coordinate system, so, it's at the 0. The charge q₂ is 20 cm above q₁, meaning is on the y-axis. Finally q₃ it's 20 cm to the right, meaning it's on the x-axis.
Knowing this,we can calculate the force that q₂ and q₃ are exerting over q₁. As these forces are in the x and y-axis respectively, we also are calculating the value of the forces in the x and y axis, that are needed to calculate the direction.
The expression to calculate the force would be Coulomb's law so:
F = K q₁q₂ / r² (2)
The value of K is 9x10⁹ N m² / C². Let's calculate the forces:
F₁₂ = Fy = 9x10⁹ * (7.1x10⁻⁴) * (6.5x10⁻⁴) / (0.020)²
Fy = 1.04x10⁷ N
F₁₃ = Fx = 9x10⁹ * (7.1x10⁻⁴) * (8.9x10⁻⁴) / (0.020)²
Fx = 1.42x10⁷ N
Now that we have both forces, we can calculate the magnitude of the force:
F = √(Fx)² + (Fy)²
F = √(1.04x10⁷)² + (1.42x10⁷)²
F = 1.76x10⁷ N
Finally, the direction would be applying (1):
α = tan⁻¹ (1.04x10⁷/1.42x10⁷)
α = 36.21 °
And counter clockwise it would be:
β = 180 - 36.21 = 143.79°
Hope this helps
Two 0.60-kilogram objects are connected by a thread that passes over a light, frictionless pulley. The objects are initially held at rest. If a third object with a mass of 0.30 kilogram is added on top of one of the 0.60-kilogram objects and the objects are released, the magnitude of the acceleration of the 0.30-kilogram object is most nearly:______
Answer:
2 m/s²
Explanation:
From the given information:
The first mass m_1 = 0.6 kg
The second mass m_2 = 0.3 kg
The magnitude for the acceleration of 0.3 kg is:
a = net force/ effective mass
Mathematically, it can be computed as follows:
[tex]a = \dfrac{F}{m}[/tex]
[tex]a = \dfrac{(m_2 +m_1 -m_1) }{(m_2+m_1+m_1)}(g)[/tex]
[tex]a = \dfrac{0.3 +0.6 -0.6}{(0.3 +0.6+0.6)}(9.8)[/tex]
a ≅ 2 m/s²
To understand and apply the formula τ=Iα to rigid objects rotating about a fixed axis. To find the acceleration a of a particle of mass m, we use Newton's second law: F net=ma , where F net is the net force acting on the particle.To find the angular acceleration α of a rigid object rotating about a fixed axis, we can use a similar formula: τnet=Iα, where τnet=∑τ is the net torque acting on the object and I is its moment of inertia.
Part A:
Assume that the mass of the swing bar, is negligible. Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, l, as well as the acceleration due to gravity g.
Part B:
Now consider a similar situation, except that now the swing bar itself has mass mbar.Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, mbar, l, as well as the acceleration due to gravity g.
Answer:
Hello your question is incomplete attached below is the missing part of the question
In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.
You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I
answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L
part A = attached below
Explanation:
Part A :
Assuming that mass of swing is negligible
α = T/I
where ; T = torque, I = inertia,
hence T = L/2*9*(M1 - M2)
also; I = [tex]M1*(L/2)^2 + M2*(L/2)^2[/tex]= ( M1 + M2) * (L/2)^2
Finally the magnitude of the angular acceleration α
α = 2*[(M1 - M2)/(M1 + M2)]*g/L
Part B attached below
A pingpong ball has 2 kg/s of momentum when
thrown 8 m/s. Find the mass of the ball.
Answer:
0.25 kg
Explanation:
p = mv
2 = m(8)
2/8 = m(8)/8 *cancels
m = 1/4 OR 0.25 kg
In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars, sending data back to Earth. Assume a circular orbit with a period of 7.08 × 103 s and orbital speed of 3.40 × 103 m/s . The mass of the GS is 930 kg and the radius of Mars is 3.43 × 106 m. Calculate the mass of Mars.
Answer: [tex]5.944\times 10^{23}\ kg[/tex]
Explanation:
Given
Time period [tex]T=7.08\times 10^3\ s[/tex]
Orbital speed [tex]v=3.40\times 10^3\ m/s[/tex]
mass of GS [tex]m_{GS}=930\ kg[/tex]
Radius of Mars [tex]r=3.43\times 10^6\ m[/tex]
Consider the mass of mars is M
Here, Gravitational pull will provide the centripetal force
[tex]F_G=F_c[/tex]
[tex]\dfrac{GMm_{GS}}{r^2}=\dfrac{m_{GS}v^2}{r}\\M=\dfrac{v^2\cdot r}{G}\\M=\dfrac{(3.43\times 10^3)^2\cdot 3.43\times 10^6}{6.67\times 10^{-11}}[/tex]
[tex]M=5.944\times 10^{23}\ kg[/tex]
In March 1999 the Mars Global Surveyor (GS) entered its final orbit on Mars, sending data back to Earth. The mass of Mars is approximately 6.419 × 10²³ kg.
Kepler's Third Law states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (a) of the orbit:
T² = (4π² / GM) × a³
In a circular orbit, the semi-major axis is equal to the radius of the orbit (r).
Given:
Orbital period (T) = 7.08 × 10³ s
Orbital speed (v) = 3.40 × 10³ m/s
Mass of GS (m) = 930 kg
Radius of Mars (r) = 3.43 × 10⁶ m
The orbital speed (v) is related to the radius (r) and the gravitational constant (G) by:
v = √(GM / r)
v² = GM / r
G = (v² × r) / M
T² = (4π² / [(v² × r) / M]) × r³
T² = (4π² × M × r²) / v²
M = (T² × v²) / (4π² × r²)
M = ( (7.08 × 10³)² × (3.40 × 10³)² ) / (4π² × (3.43 × 10⁶)²)
M = 6.419 × 10²³ kg
Therefore, the mass of Mars is approximately 6.419 × 10²³ kg.
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1.How much work does it take to get a 2Kg ball moving 15m/s if it starts from rest?
2. If a force of 235N was added to the ball, through what distance would this force have to act to give the ball a velocity of 15m/s
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.69 times a second. A tack is stuck in the tire at a distance of 0.331 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed.
Answer:
the tack's tangential speed is 5.59 m/s
Explanation:
Given that;
R = 0.331 m
wheel rotates 2.69 times a second which means, the wheel complete 2.69 revolutions in a second, so
ω = 2.69 rev/s × 2π/1s = 16.9 rad/s
using the relation of angular speed with tangential speed
tangential speed v of the tack is expressed as;
v = R × ω
so we substitute
v = 0.331 m × 16.9 rad/s
v = 5.59 m/s
Therefore, the tack's tangential speed is 5.59 m/s