Answer:
the angular velocity increase by a factor of 2
Explanation:
using the law of the conservation of the angular momentum
[tex]L_i = L_f[/tex]
where [tex]L_i[/tex] is the inicial angular momentum and [tex]L_f[/tex] is the final angular momentum.
also, the angular momentum can be calculated by:
L = IW
where I is the inertia momentum and the W is the angular velocity.
so:
[tex]I_i W_i = I_fW_f[/tex]
we know that [tex]I_f = \frac{1}{2}I_i[/tex] then,
[tex]I_iW_i = \frac{1}{2}I_iW_f[/tex]
solving for [tex]W_f[/tex]:
[tex]W_f = 2W_i[/tex]
When a figure skater pulls her arms in while spinning, her moment of inertia decreases and her angular speed increases. In the provided case, with the moment of inertia decreasing by half, the angular speed will double.
Explanation:When a figure skater is spinning with her arms outstretched, and she pulls her arms in close to her body, her moment of inertia decreases. According to the law of conservation of angular momentum, if the moment of inertia of a spinning object decreases, its angular speed must increase to keep the angular momentum constant. In this case, since the skater's moment of inertia decreases by half, her angular speed will correspondingly double, or increase by a factor of 2.
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Two ropes have equal length and are stretched the same way. The speed of a pulse on rope 1 is 1.4 times the speed on rope 2. Part A Determine the ratio of the masses of the two ropes (m1m2).
Answer:
m1/m2 = 0.51
Explanation:
First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:
V = √F/u
This is the equation that describes the relation between speed of a pulse and a force exerted on it.
the value of "u" is:
u = m/L
Where m is the mass of the rod, and L the length.
Now, for the rod 1:
V1 = √F/u1 (1)
rod 2:
V2 = √F/u2 (2)
Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:
1.4V2 = √F/u1 (3)
Replacing (2) in (3):
1.4(√F/u2) = √F/u1 (4)
Now, let's solve the equation 4:
[1.4(√F/u2)]² = F/u1
1.96(F/u2) =F/u1
1.96F = F*u2/u1
1.96 = u2/u1 (5)
Now, replacing the expression of u into (5) we have the following:
1.96 = m2/L / m1/L
1.96 = m2/m1 (6)
But we need m1/m2 so:
1.96m1 = m2
m1/m2 = 1/1.96
m1/m2 = 0.51
The ratio of the masses of the two ropes, given that the speed of a pulse on rope 1 is 1.4 times the speed on rope 2, is approximately 1:2. This conclusion is attained by applying the concept of pulse velocity in a string or rope, which is governed by the tension and the linear mass density of the material.
Explanation:The principle guiding the speed of a pulse on a rope or string is tied to the tension and linear density of the rope. The velocity of a pulse in a string or rope is given as v = √(T/μ), where T represents tension, and μ represents linear mass density (mass/length).
In the context of this question, if the speeds of pulses on two ropes are different but the tensions are the same (since the ropes are stretched the same way), the difference in speed must be due to a difference in linear mass density. Specifically, the speed of a pulse is inversely proportional to the square root of the rope's mass per unit length.
Given that the speed of the pulse on rope 1 is 1.4 times the speed on rope 2, the ratio of the masses of the two ropes (m1/m2) will be the square of the inverse ratio of the speeds. Hence, m1/m2 = (1/1.4)² = 0.51 or approximately 1:2.
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Oil explorers set off explosives to make loud sounds, then listen for the echoes from underground oil deposits. Geologists suspect that there is oil under 485-m-deep Lake Physics. It's known that Lake Physics is carved out of a granite basin. Explorers detect a weak echo 0.920 s after exploding dynamite at the lake surface.
The geologists use echo sounding for oil exploration, where they analyze the time taken by sound waves (created by explosives) to travel from the surface, reflect off oil deposits, and return. Given the speed of sound in granite (5000 m/s) and considering the travel time of the sound wave is two-way, we find the oil deposits to be approximately 2300m under the surface of the Lake Physics.
Explanation:The question revolves around the concept of echo sounding used for oil exploration. Exploders use explosives to create sounds waves that travel down into the earth and reflect back when they hit a boundary between different materials like rock or oil. In this case, the echo was detected 0.920 seconds after the explosion.
To find out the depth of the oil deposit, we need to know the speed of sound in the medium. Sound travels at different speeds in different materials, and in granite, it's typically about 5000 m/s.
We can use the formula for distance in relation to speed and time: d = vt. The time we use in this equation should be the time for the sound to travel to the deposits and back, so we halve the given time to 0.460 s. Substituting the values into the equation, we get: d = 5000 m/s * 0.460 s = 2300 m.
This means that the oil deposits are approximately 2300 m below the surface of the lake.
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Imagine that the apparent weight of the crown in water is Wapparent=4.50N, and the actual weight is Wactual=5.00N. Is the crown made of pure (100%) gold? The density of water is rhow=1.00 grams per cubic centimeter. The density of gold is rhog=19.32 grams per cubic centimeter.
Answer:
Explanation:
Actual weight, Wo = 5 N
Apparent weight, W = 4.5 N
density of water = 1 g/cm^3 = 1000 kg/m^3
density of gold, = 19.32 g/cm^3 = 19.32 x 1000 kg/m^3
Buoyant force = Actual weight - Apparent weight
Volume x density of water x g = 5 - 4.5
V x 1000 x 9.8 = 0.5
V = 5.1 x 10^-6 m^3
Weight of gold = Volume of gold x density of gold x gravity
W' = 5.1 x 10^-6 x 19.32 x 1000 x 9.8 = 0.966 N
As W' is less than W so, it is not pure gold.
Using Archimedes' Principle, we determined the density of the crown to be 10.00 g/cm³, which is significantly lower than the density of pure gold (19.32 g/cm³). Therefore, the crown is not made of pure gold.
This problem involves determining whether a crown is made of pure gold using Archimedes' Principle.
According to the given information,
the actual weight of the crown is Wactual = 5.00N, and its apparent weight in water is Wapparent = 4.50N.Using Archimedes' Principle, we know that the buoyant force equals the weight of the water displaced:
Buoyant Force = Wactual - Wapparent = 5.00N - 4.50N Buoyant Force = 0.50NSince the density of water (ρw) is 1.00 g/cm³, and using the relation 1N = 0.10197 kg in gravitational terms, the weight of the water displaced in kilograms is
weight of the water displaced =0.50N / 9.81 m/s² weight of the water displaced = 0.05096 kg, which converts to 50.96 g.The volume of water displaced, which is also the volume of the crown (V), is
V = mass / density V = 50.96 g / 1.00 g/cm³ V = 50.96 cm³Density of the crown = mass / volume.
The mass of the crown (m) is calculated from its weight in air,
m = 5.00N / 9.81 m/s² m = 0.5096 kg m = 509.6 gThus, the density of the crown = 509.6 g / 50.96 cm³ = 10.00 g/cm³.
Since the density of pure gold is 19.32 g/cm³ and the calculated density is 10.00 g/cm³, the crown is not made of pure gold.
holiday lights are often connected in series and use special lamps that short out when the potential difference across a lamp increases to the line voltage. generate an explanation why and explain why these light sets might blow their fuses after many bulb
Explanation:
If there isn't the shorting mechanism, the whole set will be blown if anyhow one lamp burns out. Since having blown out several lamps and then shorted, the overall resistance of the remaining operating lamps will be decreased resulting in an increased working current that is adequate to blast the fuse.
During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuchet launches the stones with a velocity of +30.0 m/s at an angle of 50.0°, how long does it take the stone to hit the ground? What is the maximum distance that the trebuchet can be from the castle wall to be in range? How high will the stones go? Show all your work.
The projectile reaches a maximum height of 26.4 m.
The question is demanding us to find the;
i) Time of flight:
T = 2usinθ/g
u = initial velocity
θ = angle of projection
g = acceleration due to gravity
T = 2 × 30.0 × sin 50.0°/10
T = 4.6 s
ii) The range;
R = u^2sin2θ/g
R = (30.0)^2 × sin 2( sin 50.0°)/10
R = 88.6 m
ii) The maximum height;
H = u^2sin^2θ/2g
H = (30.0)^2 × sin^2(50.0°)/2 × 10
H = 26.4 m
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Magma forms when solid rock in the crust and upper mantle. True or False
Answer:
True
Explanation:
Magma is known as melted rock deep within the Earth, normally coming from the melting of the upper mantle or crust. Magma is formed by the partial melting of the mantle and crust and this can occur in different ways. One way can be called heat-transfer melting. Rising magma or rock will bring heat with it, and so can melt the surrounding mantle or crustal rock. For example, magmas generated in the mantle tend to be around 1200 degrees Celsius, whereas the more silicate minerals such as quartz and orthoclase feldspar (common in continental crustal rocks) begin to partially melt at around 650-850 degrees Celsius. Therefore, the crustal rock will begin to partially melt due to the introduction of heat from rising magma. A Another way of melting rock is known as decompression melting. During decompression melting, rock from within the mantle is brought to the surface adiabatically (no exchange of heat or energy with its surroundings) and so the lithostatic pressure decreases. This means that the parcel of rising rock crosses the solidus, and so at this point the thermal vibration of the molecules is no longer counteracted by the lithostatic pressure and the rock begins to partially melt.
Answer:
The answer is true
Explanation:
hope it helps
Space probes may be separated from their launchers by exploding bolts. (They bolt away from one another.) Suppose a 4800-kg satellite uses this method to separate from the 1500-kg remains of its launcher, and that 5000 J of kinetic energy is supplied to the two parts. What are their subsequent velocities using the frame of reference in which they were at rest before separation?
The subsequent velocities of the satellite and the remains of the launcher can be calculated using the principle of conservation of momentum. The resultant velocities are approximately 8.70×10−² m/s and 81.5 m/s.
Explanation:The subsequent velocities of the satellite and the remains of the launcher can be calculated using the principle of conservation of momentum. Since there are no external forces acting on the system, the initial momentum of the system is equal to the final momentum. Therefore, the final velocities of the satellite and the launcher can be calculated based on their masses and the kinetic energy supplied to them.
The mass of the satellite, m1 = 4800 kg, and the mass of the launcher, m2 = 1500 kg. The total initial kinetic energy supplied to the system is given as 5000 J. To calculate the velocities, we need to find the ratio of the kinetic energies of the satellite and the launcher.
Let v1 be the velocity of the satellite and v2 be the velocity of the launcher. According to the conservation of momentum, m1 * v1 + m2 * v2 = 0. Also, the total initial kinetic energy supplied to the system is given as 5000 J, where the kinetic energy of the satellite is (1/2) * m1 * v1^2 and the kinetic energy of the launcher is (1/2) * m2 * v2^2.
Using these equations, we can solve for v1 and v2. The subsequent velocities are approximately 8.70×10−² m/s in the direction of motion of the less massive satellite and 81.5 m/s, respectively.
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The satellite moves at approximately 0.704 m/s and the remains of the launcher move at approximately -2.253 m/s in the opposite direction.
To solve for the velocities of the satellite and the launcher remains after separation, we will use the principles of conservation of momentum and the given kinetic energy.
1. Conservation of Momentum
[tex]\[m_s v_s + m_l v_l = 0\][/tex]
Since the total momentum is zero:
[tex]\[4800 \, v_s + 1500 \, v_l = 0\][/tex]
Solving for one velocity in terms of the other:
[tex]\[v_l = - \frac{4800}{1500} v_s = -3.2 v_s\][/tex]
2. Conservation of Energy
The total kinetic energy provided to the system is 5000 J. The kinetic energy of the system is the sum of the kinetic energies of both parts:
[tex]\[\frac{1}{2} m_s v_s^2 + \frac{1}{2} m_l v_l^2 = 5000\][/tex]
Substituting [tex]\( v_l = -3.2 v_s \)[/tex]:
[tex]\[\frac{1}{2} (4800) v_s^2 + \frac{1}{2} (1500) (-3.2 v_s)^2 = 5000\][/tex]
[tex]\[2400 v_s^2 + 750 (10.24) v_s^2 = 5000\][/tex]
[tex]\[2400 v_s^2 + 7680 v_s^2 = 5000\][/tex]
[tex]\[10080 v_s^2 = 5000\][/tex]
Solving for [tex]\( v_s^2 \)[/tex]:
[tex]\[v_s^2 = \frac{5000}{10080} = 0.496 \, \text{m}^2/\text{s}^2\][/tex]
[tex]\[v_s = \sqrt{0.496} \approx 0.704 \, \text{m/s}\][/tex]
3. Determine [tex]\( v_l \)[/tex]
Using [tex]\( v_l = -3.2 v_s \)[/tex]:
[tex]\[v_l = -3.2 \times 0.704 \approx -2.253 \, \text{m/s}\][/tex]
The subsequent velocities of the satellite and the launcher remains after separation are:
- Velocity of the satellite: [tex]\( v_s \approx 0.704 \, \text{m/s} \)[/tex]
- Velocity of the launcher remains: [tex]\( v_l \approx -2.253 \, \text{m/s} \)[/tex]
A chunk of aluminum at 91.4°C was added to 200.0 g of water at 15.5°C. The specific heat of aluminum is 0.897 J/g°C, and the specific heat of water is 4.18 J/g°C. When the temperature stabilized, the temperature of the mixture was 18.9°C. Assuming no heat was lost to the surroundings, what was the mass of aluminum added?
The mass of the aluminum added is calculated through the principle of conservation of energy, specifically thermal energy. By considering the heat lost by the aluminum and gained by the water, we can rearrange the equation for heat transfer and find that the mass of the aluminum is approximately 37.9 grams.
Explanation:In this physics question, we're looking at a thermodynamic process involving a chunk of aluminum and water. Given the known values of their respective specific heats, the mass of water, and their final equilibrium temperature, we're aiming to find the mass of the aluminum.
We begin by understanding that in a closed system, the heat gained by one body is equal to the heat lost by another. In this case, the aluminum is losing heat, and the water is gaining it. The equation for heat transfer (Q = mcΔT), where m is mass, c is specific heat, and ΔT is change in temperature.
The heat gained by the water = mass of water * specific heat of water * change of temperature in water = 200g * 4.18J/g°C * (18.9°C - 15.5°C) = 2836.4J.
This is equal to the heat lost by the aluminum. Solving the analogous heat equation for the mass of the aluminum gives us the answer:
m = Q / (c * ΔT) = 2836.4J / (0.897J/g°C * (91.4°C - 18.9°C)) = 37.9g
So the mass of the aluminum is approximately 37.9 grams.
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Which type of indirect evidence can substitute for direct measurements of past climate change?
Answer:
Paleoclimate proxies
Explanation:
Paleoclimate proxies are materials which are preserved and which can be analysed and correlated with parameters that relates to the environment and climate change. These materials are physical, biological or chemical and are preserved in paleoclimate archives
Scientists always seek clues from paleoclimate archives to understand how climate varied over the recent past few millions of years
Region that contains the majority of molecules in the atmosphere
Answer:
Trophosphere
Explanation:
The troposphere is the atmospheric layer closest to the planet and is characterized because contains the largest percentage of the mass of the total atmosphere.
On special characteristic is that in this layer the temperature and water vapor content decrease a lot respect to the altitude. Also on this layer the Water vapor is important in order to regulate the air temperature since on this zone we have absorption of the solar energy.
The troposphere contains almost all the water vapor in the atmosphere. And specially on the tropics we have an accumulation of the water vapour.
All weather phenomena occur within the troposphere. Tropos means "change" and Troposphere means "region of mixing".
Above this layer, we have the tropopause, ranges in height from 5 miles near the poles up to 11 miles above the equator. And the height depends of the seasons, with an special characteristic: the is highest height occurs in the summer and lowest height occurs in the winter.
The troposphere contains almost 75% of the mass of the entire atmosphere. The air on this layer is composed by 78% nitrogen, 21% oxygen and 1% is made of argon, water vapor, and carbon dioxide.
So for this reason this is the Region that contains the majority of molecules in the atmosphere.
A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. Initially the magnetic field in the region is pointed out of the page and has a magnitude of 3.5 T, but it is decreasing at a rate of 21 G/s. Due to the changing magnetic field, an electric field will be induced in this space which causes the acceleration of charges in the region. What is the direction of acceleration of a proton placed in at the point P1, 1.5 cm from the center?
Answer:
Explanation:
Due to changing magnetic field , there will be emf induced in the region . EMF induced will create electric field which will be circular in shape and will be uniform along its circular path
The magnitude of circular electric field can be calculated as follows
We should apply Faraday law of electro magnetic induction
e = - dФ / dt = - ∫ E dl
Here Ф = π r² B
π r² dB / dt = - ∫ E dl
π r² dB / dt = E x 2π r
E = - r / 2 x dB / dt
For a circular electric field having a particular radius , magnitude of field will be constant .
The direction of electric field will be known by lenz's law
In the given case , magnetic field is upwards and it is reducing , therefore electric field induced will be such as to prevent this change of flux.
So electric field will be anticlock-wise . Hence direction of acceleration will also be anticlock-wise on proton at 1.5 cm from the centre.
A car and driver weighing 7130 N passes a sign stating...?
"Bridge Out 32 m Ahead." She slams on the brakes, and the car decelerates at a constant rate of 13.8 m/s^2. The acceleration of gravity is 9.8 m/s^2.
What is the magnitude of the work done stopping the car if the car just stops in time to avoid diving into the water?
Answer:
321 280 J
Explanation:
Work done = force * distance
The distance is 32 m
The force can be calculated using the second law of motion
F = ma = (7130 N ÷ 9.8 m/s²) * 13.8 m/s² = 10 040 N
Work done = force * distance
= 10 040 N * 32 m
= 321 280 J
Answer:
321 280
Explanation:
Work done = force * distanceThe distance is 32 mThe force can be calculated using the second law of motionF = ma = (7130 N ÷ 9.8 m/s²) * 13.8 m/s² = 10 040 NWork done = force * distance = 10 040 N * 32 m = 321 280 J
Aluminum Rod#1 has a length L and a diameter d. Aluminum Rod#2 has a length 2L and a diameter 2d. If Rod#1 is under tension T and Rod#2 is under tension 2T, how do the changes in length of the two rods compare?
A) Rod #2 has quadruple the change in length that Rod #1 has.
B) Rod #2 has double the change in length that Rod #1 has.
C) Rod #1 has double the change in length that Rod #2 has.
D) They are the same.
E) Rod #1 has quadruple the change in length that Rod #2 has.
Answer:D
Explanation:
For Rod 1
length,diameter and Tension is L, d and T
for Rod 2 length,diameter and Tension is 2L, 2d and 2T
Change in Length is given by
[tex]\Delta =\frac{PL}{AE}[/tex]
where P=load
L=length
A=area of cross-section
E=young's Modulus
[tex]\Delta _1=\frac{TL}{\frac{\pi d^2}{4}E}[/tex]
since both are aluminium rod therefore E is common
[tex]\Delta _2=\frac{2T\cdot 2L}{\frac{\pi (2d)^2}{4}E}[/tex]
[tex]\Delta _2=\frac{TL}{\frac{\pi d^2}{4}E}=\Delta _1[/tex]
Thus extension in both the rods are same
The change in length of a rod under tension or compression depends on several variables. The change in length is directly proportional to the force and the original length and inversely proportional to the cross-sectional area. Therefore, if Rod #1 is under tension T and Rod #2 is under tension 2T, the change in length of Rod #2 will be double the change in length of Rod #1.
Explanation:The change in length of a rod under tension or compression depends on several variables. These variables include the force applied (T or 2T), the original length of the rod (L or 2L), and the cross-sectional area of the rod (πd^2/4 or π(2d)^2/4). The change in length is directly proportional to the force and the original length and inversely proportional to the cross-sectional area. Therefore, if Rod #1 is under tension T and Rod #2 is under tension 2T, the change in length of Rod #2 will be double the change in length of Rod #1.
A student sees her physical science professor approaching on the sidewalk that runs by her dorm. She gets a water balloon and waits. When the professor is 2 s from being directly under her window 11 m above the idewalk, she drops the balloon. You finish the story. (Assume g # 10 m/s*)
Answer:Misses the Target
Explanation:
Given
Distance between window and Professor is h=11 m
she dropped the balloon 2 sec earlier
time taken by balloon to cover the 11 m
using
[tex]h=ut+\frac{at^2}{2}[/tex]
where h=height
u=initial velocity
t=time
a=acceleration
[tex]11=0+\frac{9.8t^2}{2}[/tex]
[tex]t=\sqrt{\frac{2\times 11}{10}}[/tex]
[tex]t=\sqrt{2.2}[/tex]
[tex]t=1.483 s[/tex]
Therefore she misses the Professor by 0.51 s
Final answer:
The water balloon will hit the ground before the professor is directly underneath the window since it takes approximately 1.48 seconds to fall 11 meters, which means the student is unlikely to hit the professor.
Explanation:
The student's question involves calculating the time it takes for a water balloon dropped from a height to hit a target. This is a problem in projectile motion and free fall within the context of physics, specifically under the topic of mechanics. As the professor is 2 seconds away from the point directly below the student's window, which is 11 meters above the sidewalk, we can determine if the balloon will hit the target by calculating how long it takes for the balloon to reach the ground.
Using the equation of motion for free fall s = 1/2gt^2, where g is the acceleration due to gravity (approximated to 10 m/s^2), and s is the distance (11 meters), we can solve for t (time): s = 1/2gt^2 implies t = sqrt(2s/g). Substituting the given values, we get t = sqrt(2*11/10), which gives t = sqrt(2.2), or approximately t = 1.48 seconds. Because the professor is 2 seconds away from being underneath the window, the water balloon will hit the ground before the professor reaches the point below the student's window, meaning the student's attempt to hit the professor with the balloon will likely fail unless the professor's speed changes.
Anissa slides down a playground slide sloped at 25o. The coefficient of kinetic friction between Anissa and the slide is 0.15. If Anissa’s mass is 55 kg, how fast is she going 2.0 s after starting from rest?
To find the speed of Anissa after 2.0 seconds, we need to consider the forces acting on her due to gravity and friction. By calculating the force parallel to the slide using the coefficient of kinetic friction and the normal force, we can determine the acceleration of Anissa. Finally, using the equations of motion, we can find her speed after 2.0 seconds.
Explanation:To find the speed of Anissa after 2.0 seconds, we need to consider the forces acting on her. The force of gravity pulling her down the slide can be divided into two components: a force parallel to the slide and a force perpendicular to the slide. The perpendicular force is canceled out by the normal force, leaving only the parallel force to consider. This force can be calculated using the coefficient of kinetic friction.
The force parallel to the slide is given by the equation F_parallel = μ_k * N, where F_parallel is the force parallel to the slide, μ_k is the coefficient of kinetic friction, and N is the normal force. The normal force can be calculated using the equation N = m * g * cos(θ), where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the slide.
Once we have the force parallel to the slide, we can use Newton's second law to find the acceleration of the object: F_parallel = m * a. Finally, we can use the equation v = u + a * t to find the speed of Anissa after 2.0 seconds, where v is the final velocity, u is the initial velocity (which is 0 since she starts from rest), a is the acceleration, and t is the time.
Plugging in the given values: μ_k = 0.15, m = 55 kg, g = 9.8 m/s^2, θ = 25 degrees, and t = 2.0 s, we can calculate the normal force, force parallel to the slide, acceleration, and final velocity.
The final velocity will depend on the length of the slide and other factors that are not mentioned in the question.
A rocket ship starts from rest in turns on it's forward booster rocket causing it to have a constant acceleration of 4m/s*rightward. After 3s, what will be the velocity of the rocket ship
The velocity after 3 s is 12 m/s rightward
Explanation:
The motion of the rocket is a motion at constant acceleration, therefore we can apply suvat equations:
[tex]v=u+at[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time interval
For the rocket in this problem:
u = 0 (the rocket starts from rest)
t = 3 s
[tex]a=4 m/s^2[/tex] rightward is the acceleration
Solving for v, we find the final velocity:
[tex]v=0+(4)(3)=12 m/s[/tex] (rightward)
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As sources of electrical power, windmills now account for only about 2,500 megawatts nationwide, but production is almost expected to double by the end of the year, which would provide enough electricity for 1.3 million households.
a. almost expected to double by the end of the year, which would provide
b. almost expected that it will double by the end of the year, thus providing
c. expected that it will almost double by the end of the year to provide
d. expected almost to double by the end of the year and thus to provide
e. expected almost to double by the end of the year, which would thus be providing
Answer: C
Explanation:
Production is expected that it will almost double by the end of the year is the answer because production in this context relates to a numerical quantity which is a function of time.
Therefore the probability there is on whether production will almost double to provide enough electricity and not if production will occur.
Windmills account for about 2,500 megawatts of electrical power nationwide, and the production is expected to almost double by the end of the year, providing enough electricity for 1.3 million households.
Windmills account for about 2,500 megawatts of electrical power nationwide, and the production is expected to almost double by the end of the year. This doubling of production would provide enough electricity for 1.3 million households.
The correct answer choice is c. expected that it will almost double by the end of the year to provide because it accurately represents the expectation of the doubling of production and the purpose of providing enough electricity for households.
1. A 0.250 kg baseball sits on the ledge of a window in Treadwell hall. If the ball has 18.5
Joules of potential energy, how high is the ball?
2. How long would it take to hit the ground?
Answer:
h = 7.54 m
t = 1.24 s
Explanation:
1.Let g = 9.81 m/s2 is the gravitational acceleration. Since the formula for potential energy is:
[tex]E_p = mgh[/tex]
where m = 0.25 is the ball mass and h is the height. We can solve for h
[tex]h = \frac{E_p}{mg} = \frac{18.5}{0.25*9.81} = 7.54m[/tex]
2. The time it take for the ball to reach a distance of 7.54m with a gravitational acceleration of 9.81m/s2:
[tex]h = \frac{gt^2}{2}[/tex]
[tex]t^2 = \frac{2h}{g} = \frac{2*7.54}{9.81} = 1.54[/tex]
[tex]t = \sqrt{1.54} = 1.24 s[/tex]
Deep in the interiors of the giant planets, water is still a liquid even though the temperatures are tens of thousands of degrees above the boiling point of water. This can happen because ________.
Answer:
High pressure inside the giant planet
Explanation:
As we move in the interior of the giant planet, the pressure and temperature in the interior of the planet increases. Since, the giant planets have hardly any solid surface and thus they are mostly constituted of atmosphere.
Also, the gravitational forces keep even the lightest of the matter bound in it contributing to the large mass of the planet.
If we look at the order of the magnitude of the temperature of these giant planets than nothing should be able to stay in liquid form but as the depth of the planet increases with the increase in temperature, pressure also increases which keeps the particle of the matter in compressed form.
Thus even at such high order of magnitude water is still found in liquid state in the interior of the planet.
At what displacement of a sho is the energy half kinetic and half potential?
Answer:
Displacement = 0.707A
Explanation:
To solve for the displacement we know that
Potential energy PE = 1/2Total energy (Etotal)
Therefore 1/2kx^2 = 1/2(1/2KA^2)
Solving for x we have
x^2 = √A^2/2
x = A/√2
x= 0.707A
Old-fashioned pendulum clocks are powered by masses that need to be wound back to the top of the clock about once a week to counteract energy lost due to friction and to the chimes. One particular clock has three masses: 4.0 kg, 4.0 kg, and 6.0 kg. They can drop 1.3 meters. How much energy does the clock use in a week?
Answer:
Energy, E = 178.36 J
Explanation:
It is given that,
Mass 1, [tex]m_1=4\ kg[/tex]
Mass 2, [tex]m_2=4\ kg[/tex]
Mass 3, [tex]m_3=6\ kg[/tex]
Height from which they are dropped, h = 1.3 m
Let m is the energy used by the clock in a week. The energy is equal to the gravitational potential energy. It is given by :
[tex]E=(m_1+m_2+m_3)gh[/tex]
[tex]E=(4+4+6)\times 9.8\times 1.3[/tex]
E = 178.36 J
So, the energy used by the clock in a week is 178.36 Joules. Hence, this is the required solution.
A long straight rod experiences several forces, each acting at a different location on the rod. All forces are perpendicular to the rod. The rod might be in translational equilibrium, rotational equilibrium, both, or neither.
(1) If a calculation reveals that the net torque about the left end is zero, then one can conclude that the rod
(A) is definitely in rotational equilibrium.
(B) is in rotational equilibrium only if the net force on the rod is also zero.
(C) might not be in rotational equilibrium even if the net force on the rod is also zero.
(D) might be in rotational equilibrium even if the net force is not zero.
(2) If a calculation reveals that the net force on the rod is zero, then one can conclude that the rod
(A) is definitely in rotational equilibrium.
(B) is in rotational equilibrium only if the net torque about every axis through anyone point is found to be zero.
(C) might be in rotational equilibrium if the net torque about every axis through anyone point is found to be zero.
(D) might be in rotational equilibrium even if the net
torque about any axis through anyone point is not zero.
Answer:
1. C
2.C
Explanation:
1. The rod is perpendicular to every axis and forces are acting on every location. If the torque on the left side is zero, this indicates that forces with respect to their distance on the left side is zero and doesn't account for the net force at a point.
2. If the net torque about every point on every axis is zero, the rod will be rotational because each axis will yield a magnitude of zero which obeys the principle of rotation at a point.
When the net torque on a rod is zero about the left end, the rod is in rotational equilibrium. If the net force is zero, the rod is in translational equilibrium, and rotational equilibrium depends on the net torque being zero about every axis.
Equilibrium Conditions in Physics
To determine whether a rod is in equilibrium under the influence of several forces, we need to examine the conditions for both translational and rotational equilibrium. The conditions necessary for equilibrium are quite straightforward:
The net external force on the system must be zero (net F = 0), ensuring there is no linear acceleration.
The net external torque must also be zero (net T = 0), preventing any angular acceleration.
Addressing the questions:
If the net force on the rod is zero, it indicates that the rod is in translational equilibrium. However, for the rod to be in rotational equilibrium, we must also confirm that the net torque about every axis through anyone point is zero (Option B).
Each static equilibrium condition is critical in itself. While the zero net force ensures no linear acceleration, the zero net torque ensures no rotational acceleration. Both conditions are required for complete static equilibrium of an object.
While driving down the road, a firefly strikes the windshield of a bus and makes a quite obvious mess in front of the face of the driver. This is a clear case of Newton's third law of motion. The firefly hit the bus and the bus hits the firefly. Which of the two forces is greater: the force on the firefly or the force on the bus?
Answer:
Both forces have the same magnitude
Explanation:
1) Notation
[tex]F_{bus}[/tex]= force exterted by the bus to the firefly
[tex]F_{firefly}[/tex]= force exterded by the firefly to the bus
2) Analysis for the situation
The reason why is from the Third law of Newton that states: "For every action, there is an equal reaction force on the opposite direction to the original action force".
For this special case on math terms we have:
[tex]F_{bus}=F_{firefly}[/tex]
The fact that the firefly splatter on this case is because since have a smaller mass, it is less able to stand up to the larger acceleration from the interaction.
3) Conclusion
Based on this analysis, each force would have the same magnitude, so none force is greater than the other.
The force on the firefly and the force on the bus are equal in magnitude but opposite in direction, demonstrating Newton's third law of motion.
Explanation:When a firefly collides with the windshield of a bus, we have a real-life illustration of Newton's third law of motion. This law states that for every action there's an equal and opposite reaction.
In the case of the firefly and the bus, both objects exert forces on each other that are equal in magnitude but opposite in direction.
The mess that is made when the firefly hits the bus might suggest to us that the force on the firefly is greater, but according to Newton's third law, the force on the firefly and the force on the bus are actually the same.
This symmetry in nature is fundamental to our understanding of physics and can be observed in many situations, like a professor walking across a room or a car accelerating down a road.
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The concept that large masses warp the shape of spacetime and thus change the path of an object moving through that spacetime applies to all objects. As such, Einstein’s theory of general relativity predicts that light will also curve around massive objects as it travels through space. Does Newton’s law of gravity agree with that prediction? Note that the mass of a photon is zero.
Answer: Newton's law of gravity theory disagrees with Einstein's theory. The last one were named Theory of General Relativity and was discovered and proposed by the physicist Einstein in the year of 1915.
Explanation: Until the beginning of 20 century, the physics were ruled by Isaac Newton's ideas. He believed that the gravity was a force caused by the objects mass on the space, made them to be draw towards each other. Newton thought that the greater the mass of the object, the more intense was its attraction, which would justify the planet's moovements around the sun and how the gravity between them maintain the planets on solar orbit. Concluding, he believed gravity was a immediate force of action, regardless of the distance of the bodies.
Contrary to Newton's law of gravity, in 1915, the physicist Einstein created the Theory of General Relativity, wich discovered that gravity was, in fact, the deformation caused by the attraction of massive celestial bodies. This deformation, related to the Sun, for example, creates a curvature on the space-time and this curvature are followed by the other planets.
So, we can conclude that Newton's law of gravity disagree's with the Theory of General Relativity, once the first believes that light force of attraction are transmitted instantly and, as Einstein already prooved, the gravity influency propagates in the speed of light.
What formal regions are located in the western hemisphere
Answer: Part of Europe, part of Africa, part of Antarctic and the entire America.
Explanation: Western hemisphere or west hemisphere encompasses all regions located west of the longitude 0 °, or meridian of Greenwich.
The formal regions located in the western hemisphere is Europe, Africa, Antarctic and America.
Final answer:
Western Europe can be divided into several smaller regions, including northern Europe, southern Europe, Central Europe, and the British Isles. The western hemisphere also includes various formal regions such as Central America, the Caribbean islands, and parts of South America.
Explanation:
Western Europe can be divided into several smaller regions, including northern Europe, southern Europe, Central Europe, and the British Isles. These regions are based on geographical location or cultural similarities. Additionally, the western hemisphere includes various formal regions such as Central America, the Caribbean islands, and parts of South America.
A 45 kg boy runs at a rate of 2.5 m/s and jumps on top of a stationary skateboard that has a mass of 4 kg. After jumping onto the board, the boy and the board travel off together. Assuming that momentum is conserved, what is the final speed of the boy and the board?
Answer:
v = 2.29 m/s
Explanation:
As we know that the external force on the system of mass of boy + board is ZERO
So here we can use momentum conservation
now we have
[tex]m_1v_1 + m_2v_2 = (m_1 + m_2) v[/tex]
now we have
[tex]45 (2.5) + 4(0) = (45 + 4) v[/tex]
now we have
[tex]v = \frac{45}{49} (2.5)[/tex]
[tex]v = 2.29 m/s[/tex]
Answer:
V = 2.29 m/s
Explanation:
Given that,
Mass of the boy, [tex]m_1=45\ kg[/tex]
Mass of the skateboard, [tex]m_2=4\ kg[/tex]
Initial speed of the boy, v = 2.5 m/s
Let V is the final velocity of the boy and the board. The net momentum of the system remains constant. Using the conservation of linear momentum to find it as :
[tex]45\times 2.5=(45+4)V[/tex]
[tex]V=\dfrac{45\times 2.5}{(45+4)}[/tex]
V = 2.29 m/s
So, the velocity of the boat after Batman lands in it 2.29 m/s. Hence, this is the required solution.
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.80 s apart.How far away did the impact occur? The speed of sound in air is 343 m/s, and in concrete is 3000 m/s.
Answer:
309 m
Explanation:
time apart = 0.8 s
speed of sound in air = 343 m/s
speed of sound in concrete = 3000 m/s
lets assume it time it takes to travel trough concrete = T
the time it takes to travel through air = T + 0.8 since they are 0.8 s apart
the distance traveled by both waves is the same, so we can equate the distance for both waves
distance = speed x time
for concrete, distance = 3000 x Tfor air, distance = 343 x (T + 0.8)now equating the two distance together we have
3000 x T = 343 x (T + 0.8)
3000T = 343T + 274.4
3000T - 343T = 274.4
T = 0.103 s
recall that distance = 3000 x T = 3000 x 0.103 = 309 m
Keller is driving at a speed of 60 MPH. At noon he sees the following sign: Belmont 3 miles Arville 17 miles Blison 35 miles Longwood 52 miles Swan's Lake 120 miles If he continues to drive the same speed, how far away will Longwood be at 12:20pm?
Answer:
32 miles
Explanation:
Assuming that the speed of 60 mph is constant
Distance = Speed × Time
Time is 20 minutes
Converting to hours
[tex]\frac{20}{60}=\frac{1}{3}\ h[/tex]
[tex]Distance=60\times \frac{1}{3}\\\Rightarrow Distance=20\ mi[/tex]
The car covered 20 mi in 20 minutes.
Distance to Longwood would be [tex]52-20=32\ mi[/tex]
Distance to Longwood is 32 miles
A fluid flows through a pipe whose cross-sectional area changes from 2.00 m2 to 0.50 m2 . If the fluid’s speed in the wide part of the pipe is 3.5 m/s, what is its speed when it moves through the narrow part of the pipe?
Answer:
v₂ = 7/ (0.5)= 14 m/s
Explanation:
Flow rate of the fluid
Flow rate is the amount of fluid that circulates through a section of the pipeline (pipe, pipeline, river, canal, ...) per unit of time.
The formula for calculated the flow rate is:
Q= v*A Formula (1)
Where :
Q is the Flow rate (m³/s)
A is the cross sectional area of a section of the pipe (m²)
v is the speed of the fluid in that section (m/s)
Equation of continuity
The volume flow rate Q for an incompressible fluid at any point along a pipe is the same as the volume flow rate at any other point along a pipe:
Q₁= Q₂
Data
A₁ = 2m² : cross sectional area 1
v₁ = 3.5 m/s : fluid speed through A₁
A₂ = 0.5 m² : cross sectional area 2
Calculation of the fluid speed through A₂
We aply the equation of continuity:
Q₁= Q₂
We aply the equation of Formula (1):
v₁*A₁= v₂*A₂
We replace data
(3.5)*(2)= v₂*(0.5)
7 = v₂*(0.5)
v₂ = 7/ (0.5)
v₂ = 14 m/s
Using the Continuity Equation that states the mass flow rate must be constant in fluid flow, the speed of the fluid when it moves through the narrow part of the pipe is 14 m/s.
Explanation:The question pertains to the physics principle of fluid dynamics, specifically the concept of continuity in fluid flow. According to the Continuity Equation, the mass flow rate must remain constant across different sections of a pipe. In this particular scenario, the fluid is changing speed due to a reduction in pipe cross-sectional area. This principle is represented by the equation A₁V₁ = A₂V₂, where A₁ and A₂ represent the cross-sectional areas of the wide and narrow sections of the pipe respectively, and V₁ and V₂ represent the fluid's velocity in these sections.
Applying the equation to this scenario, we have A₁ (2.00 m²) multiplied by V₁ (3.5 m/s) equal to A₂ (0.50 m²) multiplied by V₂, the value we want to find. Solving for V₂, we get V₂ = A₁V₁ / A₂ = (2.00 m²)(3.5 m/s) / 0.50 m² = 14 m/s.
Hence, the speed of the fluid when it moves through the narrow part of the pipe is 14 m/s.
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Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 349 m/s, piano A produces a wavelength of 0.766 m, while piano B produces a wavelength of 0.776 m. How much time separates successive beats?
Answer:
Time period between the successive beats will be 0.1703 sec
Explanation:
We have given speed of the sound v = 349 m/sec
Wavelength of piano [tex]A\lambda _A=0.766m[/tex]
Wavelength of piano [tex]B\lambda _B=0.776m[/tex]
So frequency of piano A [tex]f_1=\frac{v}{\lambda _1}=\frac{349}{0.766}=455.61Hz[/tex]
Frequency of piano B [tex]f_2=\frac{v}{\lambda _1}=\frac{349}{0.776}=449.74Hz[/tex]
So beat frequency f = 455.61 - 449.74 = 5.87 Hz
So time period [tex]T=\frac{1}{f}=\frac{1}{5.87}=0.1703sec[/tex]
So time period between the successive beats will be 0.1703 sec