Answer:
30 kg
Explanation:
Newton's second law:
F = ma
60 N = m (2 m/s²)
m = 30 kg
Answer:
[tex]30\:g[/tex]
Explanation:
Isaac Newton's second law of motion
[tex]\displaystyle ma = F_{net}[/tex]
[tex]\displaystyle \frac{60}{2} = m \\ \\ 30 = m[/tex]
* Now, that bar above 18 is what is known as bar notation, indicating repetition in the digits.
I am joyous to assist you anytime.
How does universal gravitation affect you and objects around you?
Answer:
Newton proposed the "law of Universal Gravitation". He said that gravity is universal in nature and it affects all the objects. Gravity is present between you and every object having mass around you. The "Strength of the gravity" between two objects with mass is depends on the product of masses and distance between the objects.
The "Gravitational force" increases with the increase in mass of the objects and decreases with the increase in distance between the objects. The gravity exerts a force on every object. The direction of force of gravity is towards the centre of the Earth.
23. Challenge: A gas is heated so that it expands from a volume of 1.0L to a volume of 1.5 l. If the
initial temperature of the gas was 5.0°C, then what is the final temperature of the gas?
417 K is the “final temperature” of the gas.
Explanation:
According to “Charles law” the change in “volume” of a given mass of gas expanded is “directly proportional” to the “temperature” of the given gas expanded. When we keep “pressure” of the gas as constant. Mathematically,
[tex]v \alpha T[/tex]
[tex]\frac{v}{T}=\text { constant }[/tex]
If gas is expanded from initial volume to final volume and initial temperature to final temperature then,
[tex]\frac{v_{1}}{T_{1}}=\frac{v_{2}}{T_{2}}[/tex]
[tex]\begin{array}{l}{\text { Where, } v_{1} \text { and } v_{2} \text { are the initial and final volumes of the gas expanded recpectively,}} \\ {T_{1} \text { and } T_{2} \text { are the initial and final temperatures of the gas expanded recpectively }}\end{array}[/tex]
Given that,
Initial volume is 1.0L
Final volume is 1.5L
[tex]\text { Initial temperatureis } 5.0^{\circ} \mathrm{C}=5+273=278 \mathrm{K}[/tex]
To find final temperature of the gas
Substitute the given values,
[tex]\frac{1}{278}=\frac{1.5}{T_{2}}[/tex]
[tex]T_{2}=1.5 \times 278[/tex]
[tex]T_{2}=417 K[/tex]
Therefore, final temperature of the gas after expanding is 417 K.
To calculate the final temperature of an expanding gas, you can use Charles's Law, which relates volume and temperature. By substituting the given values into the Charles's Law equation and solving for the final temperature in Kelvin, and then converting it to Celsius, you can find the final temperature of the gas.
Explanation:To find the final temperature of a gas that has expanded, we can use Charles's Law. This law states that at constant pressure, the volume of a given mass of an ideal gas increases or decreases by the same factor as its temperature (in Kelvin) increases or decreases. The formula is: V1/T1 = V2/T2, where V is volume and T is temperature (in Kelvin).
The initial volume (V1) is 1.0L, and the final volume (V2) is 1.5L. The initial temperature (T1) is 5.0°C, which is 278.15K. To find the final temperature in Kelvin (T2), rearrange the formula to get T2 = V2*T1/V1. Substituting in the values gives T2 = (1.5L * 278.15K) / 1.0L. Calculating this gives us a value for T2. To convert T2 back to Celsius, subtract 273.15 from the Kelvin temperature.
What happens to a low-frequency photon when it hits a metal plate if its energy is less than the work function Wo?
A
An electron is emitted that has higher energy than the incident photon.
B. An electron is emitted from the metal surface.
C
An electron is emitted from the metal surface but is immediately reabsorbed.
D.
An electron is not emitted from the metal surface.
Answer:
D
Explanation:
plato
A 42 kg object is 2.5 m from a 55kg object. How much is the force of attraction between them?
Answer:
Force of attraction is 2,46*10^-8N[tex]F= G*(m1*m2/r^2)\\where \\\\G=6,67*10^-11 (N*m^2/kg^2)\\m1=42kg\\m2=55kg\\r=2,5m\\therefore\\F=6,67*10^-11*(42*55/2.5^2)\\
F=2,46*10^-8 N[/tex]
Explanation:
Using the Law of Universal Gravitation, proposed by Newton.
Which part provides the resistance force in the
photo?
A (spring scale)
-В
B (wooden cube and clay)
C (ruler)
OD (clay block)
Answer:
D- clay block
Explanation:
Answer:
The correct answer is D.
Explanation:
The clay block provides the force of resistance in the photo, since the weight it exerts on the right end of the ruler on which it is supported makes the system stay in balance. The weight of the spring balance (object A) is transferred to its point of support on the ruler, while the wood and clay cube only exerts one point of support, but does not exert any force on the ruler. The weight exerted by the clay block downward counteracts the weight of the spring balance.
Have a nice day!
You want to make a ride so you do not want to exceed 1.1g’s, if the radius of the turns are 10m, then what is the maximum speed the ride can go at?
The maximum speed is 10.4 m/s
Explanation:
For a body in uniform circular motion, the centripetal acceleration is given by:
[tex]a=\frac{v^2}{r}[/tex]
where
v is the linear speed
r is the radius of the circular path
In this problem, we have the following data:
- The maximum centripetal acceleration must be
[tex]a=1.1 g[/tex]
where [tex]g=9.8 m/s^2[/tex] is the acceleration of gravity. Substituting,
[tex]a=(1.1)(9.8)=10.8 m/s^2[/tex]
- The radius of the turn is
r = 10 m
Therefore, we can re-arrange the equation to solve for v, to find the maximum speed the ride can go at:
[tex]v=\sqrt{ar}=\sqrt{(10.8)(10)}=10.4 m/s[/tex]
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Final answer:
To find the maximum speed that a ride can go without exceeding 1.1g's with a turn radius of 10m, the centripetal acceleration formula is used, and the calculation results in a maximum speed of approximately 10.44 m/s.
Explanation:
To calculate the maximum speed at which a ride can go without exceeding 1.1g's of acceleration, we use the formula for centripetal acceleration a = v^2 / r, where a is the centripetal acceleration, v is the speed, and r is the radius of the turn, which is given as 10m.
We want to keep the acceleration a at no more than 1.1 times the gravitational acceleration g (9.8 m/s^2). Thus, we have: 1.1g = v^2 / r. Plugging in the numbers, we get 1.1 × 9.8 m/s^2 = v^2 / 10 m. To find the maximum speed v, we solve for v, which gives us v = √(1.1 × 9.8 m/s^2 × 10 m). After calculating, we find the maximum speed to be approximately 10.44 m/s.
A skidder dragging of 520 kg walk-through forest at a constant speed of 3.5 m/s. If the skidder is applying a force of 1850 N what is the coefficient of friction between the log and the ground
Answer:
The coefficient of friction between the log and the ground is, μₓ = 0.3630
Explanation:
Given data,
The skidder dragging the mass, m = 520 kg
The acceleration of the mass, a = 3.5 m/s
The force applied by the skidder, Fₓ = 1850 N
The kinetic frictional force is given by,
Fₓ = μₓ η
μₓ = Fₓ / η
Where μₓ - the coefficient of the kinetic friction
η is the normal force acting on the mass due to gravitational force.
η = mg
= 520 x 9.8
= 5096 N
Substituting the values in the above equation,
μₓ = 1850 N / 5096 N
= 0.3630
Hence, the coefficient of friction between the log and the ground is, μₓ = 0.3630
A race car accelerates from 0 m/s to 30.0 m/s with a displacement of .25 km. what is the vehicle's acceleration?
Answer:
The acceleration of the car, a = 60 m/s²
Explanation:
Given data,
The initial velocity of the car, u = 0 m/s
The final velocity of the car, v = 30 m/s
The displacement of the car, s = 0.25 km
Using the III equations of motion
v² = u² + 2as
a = (v² - u²) / 2s
= 30 / 2 x 0.25
= 60 m/s²
Hence, the acceleration of the car, a = 60 m/s²
Venus has a radius of 6.05 x 106 m,
and a surface gravity, g = 8.87 m/s2.
Based on this information, what is the
mass of Venus?
The mass of Venus is [tex]4.87\cdot 10^{24}kg[/tex]
Explanation:
The gravitational acceleration at the surface of a planet is given by:
[tex]g=\frac{GM}{R^2}[/tex]
where:
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}[/tex] is the gravitational constant
M is the mass of the planet
R is the radius of the planet
For Venus, we know that:
[tex]R=6.05\cdot 10^6 m[/tex] is the radius of the planet
[tex]g=8.87 m/s^2[/tex] is the gravity at the surface
Solving the equation for M, we find Venus mass:
[tex]M=\frac{gR^2}{G}=\frac{(8.87)(6.05\cdot 10^6)^2}{6.67\cdot 10^{-11}}=4.87\cdot 10^{24}kg[/tex]
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The mass of Venus is calculated using the gravitational acceleration and radius of the planet, applying Newton's law of gravitation. With the given values for the surface gravity and radius of Venus, the mass is approximately 4.87 x 10¹´ kg.
Explanation:To calculate the mass of Venus, we use the equation for gravitational acceleration on the surface of a planet, which is derived from Newton's law of universal gravitation: g = G * M / r², where g is the gravitational acceleration, G is the gravitational constant (6.674×10⁻¹¹ N·m²/kg²), M is the mass of the planet, and r is the radius of the planet.
Reorganizing the equation for M gives us: M = g * r² / G. Substituting in the values given for Venus (g = 8.87 m/s², r = 6.05 x 10⁶ m, G = 6.674×10⁻¹¹ N·m²/kg²), we can calculate the mass of Venus.
M = (8.87 m/s² * (6.05 x 10⁶ m)²) / (6.674×10⁻¹¹ N·m²/kg²)
After performing the calculation, we find that the mass of Venus is approximately 4.87 x 10¹´ kg.
A force of 6.0 N acts on a 3.0 kg object at rest for 10 seconds.
c) What was the beginning momentum of the object?
The beginning momentum of the object is zero
Explanation:
The momentum of an object is given by:
[tex]p=mv[/tex]
where
m is the mass of the object
v is its velocity
For the object in this problem, we have:
m = 3.0 kg is its mass
v = 0, because it is initially at rest, so its initial velocity is zero
Therefore, the initial momentum of the object is
[tex]p=(3.0)(0)=0[/tex]
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Two carts, one of mass 2m and one of mass m, approach each other with the same speed, v. When the carts collide, they
hook together. Assume that positive momentum is to the right.
When two carts collide and hook together, their momenta combine according to the law of conservation of momentum.
Explanation:The subject of this question is Physics and the grade level is High School.
In this scenario, we have two carts of different masses approaching each other with the same speed. When the carts collide, they hook together and move as a single system. The total momentum before the collision is equal to the total momentum after the collision, according to the law of conservation of momentum.
The equation for conservation of momentum in this case would be: (2m * v) + (m * (-v)) = (3m * v'), where v' is the final velocity of the hooked carts.
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v' = 0. The combined cart comes to a complete rest.
Let's analyze the scenario of two carts colliding and hooking together:
Initial Conditions:
Cart 1 (mass 2m) has velocity v to the right (positive momentum)
Cart 2 (mass m) has velocity v to the left (negative momentum)
Collision and Resulting Motion:
Upon collision, the carts hook together, forming a single system with a combined mass of 3m. The carts will then move together with a common velocity, v'.
Since the collision is inelastic, kinetic energy is not conserved. Some of the kinetic energy is converted into other forms, such as heat or sound.
Conservation of Momentum:
Despite the loss of kinetic energy, momentum is conserved in an isolated system. The total momentum before the collision must equal the total momentum after the collision.
Initial momentum = (2m * v) + (-m * v) = 0
Final momentum = (3m * v')
Therefore, v' = 0. The combined cart comes to a complete rest.
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Two vectors, X and Y, form a right angle. Vector X is 48 inches long and vector Y is 14 inches long. The length of the
resultant vector is
inches.
Mark this and retum
Save and Exit
Submit
Answer:
The length of the resultant vector is 50 inches
Explanation:
Use the Pythagorean theorem to find the answer, since the addition of these two perpendicular vectors will have a magnitude (length) equal to the hypotenuse of the right angle triangle formed by the two:
[tex]|Resultant|\,=\,\sqrt{x^2+y^2} \\|Resultant|\,=\,\sqrt{48^2+14^2}\\|Resultant|\,=\,\sqrt{2304+196}\\|Resultant|\,=\,\sqrt{2500}\\|Resultant|\,=\,50\,in[/tex]
The length of the resultant vector is 50 inches
Answer:
50 inches Edge 2021
Explanation:
A speed boat travels with a speed of 6 m/s travels directly north to cross a river flowing west with a current traveling at a speed of 3 m/s. What is the resultant velocity of the boat?
Answer:
6.7 m/s @ 116.6°
Explanation:
Use Pythagorean theorem to find the magnitude:
v² = vx² + vy²
v² = (-3 m/s)² + (6 m/s)²
v = √45 m/s
v ≈ 6.7 m/s
Use trigonometry to find the direction:
θ = atan(vy / vx)
θ = atan(6 m/s / -3 m/s)
θ = atan(-2)
θ ≈ 116.6° from the +x axis, or 63.4° north of west.
Describe what would happen to the volume of a gas if
the pressure on it were decreased and then the gas's temperature
were increased.
The volume of the gas has increased
Explanation:
To solve this problem, we can use the ideal gas equation, which can be written as:
[tex]\frac{pV}{T}=const.[/tex]
where:
p is the pressure of the gas
V is its volume
T is its absolute temperature
This equation can also be rewritten as follows:
[tex]\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}[/tex]
where:
[tex]p_1[/tex] is the initial pressure of the gas
[tex]V_1[/tex] is the initial volume
[tex]T_1[/tex] is the initial temperature
[tex]p_2[/tex] is the final pressure
[tex]V_2[/tex] is the final volume
[tex]T_2[/tex] is the final temperature
In this problem:
- The pressure of the gas is decreased, so [tex]p_2 < p_1[/tex]
- The temperature of the gas is increased, so [tex]T_2 > T_1[/tex]
We can rewrite the equation making V2 the subject, to see what happens to the volume:
[tex]V_2 = \frac{p_1 V_1 T_2}{p_2 T_1}=\frac{p_1}{p_2}\frac{T_2}{T_1} V_1[/tex]
Where we have:
[tex]\frac{p_1}{p_2}>1[/tex]
[tex]\frac{T_2}{T_1}>1[/tex]
Therefore, this implies that
[tex]V_2 > V_1[/tex]
So the volume of the gas has increased.
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Answer:
Gases respond to changes in pressure, temperature, and volume in predictable ways, and according to Boyle's law, when the volume of gas decreases, the pressure increases. Therefore, when the pressure is decreased and temperature increased, the volume would also increase.
Consider the circuit. Switches are added at points A, B, C, and D. All the switches are closed EXCEPT the switch at position D, which is left open. What is the result of this?
Answer:
A I just had it on usa test prep
Explanation:
A car goes 152 miles in 2 hours and 5 minutes. Calculate the car’s speed in units of m/s. 1 mi = 1600 m.
Answer:
The speed of the car is 32.4m/s
Explanation:
1mi = 1600m
152mi = 243 200m
2hr 5min = 7200s + 300s
= 7500s
speed = distance/time
= 243200/7500
= 32.4m/s (3s.f)
Supposing you were in space..in a weightless environment, would it require a force to set an object in motion? Explain..
it wouldn't, right? because it's in space, no gravity to pull it, or act upon it..so it would just float?
Answer:
Yes, a force is require to set an object in motion.
Explanation:
In space, even if you feel weightless, you are subject to motion. If you are orbiting the Earth, you are under the constant influence of Earth having a free-fall acceleration equal to the centripetal acceleration. To disturb this orbital motion, an external force is required. According to Newton's laws of motion, a force is required to change the state of the rest of a body or to change the velocity or direction if it is moving with uniform velocity along a straight line. Whenever there is a change in velocity or direction of a body there is a force acting on it.Which are examples of convection currents ? Select three options.
marshmallows toasting over a campfire
a pot being heated by an electric burner
rice moving in a pot of water that is being heated
a radiator that emits warm air and draws in cool air
a hot air balloon rising and falling in the atmosphere
Answer:
(B) A pot being heated by an electric burner
(D) A radiator that emits warm air and draws in cool air
(E) A hot air balloon rising and falling in the atmosphere
These are some of the examples of the convection currents.
Explanation:
Earlier, electrons were believed to have positive charges and then electric current were discovered. But later after the invention of electric current and current which is termed to be the flow of electrons and is usually flows from negative to positive terminal. But its convention is not discarded in which current moves from positive terminal to negative and it is called convention current. The direction of current shown in the circuit is said to be the convention current.
Hence, the following are the examples of convention current.
1. Boiling water - The energy travels into the pot from the burner, boiling down the water. Then this warm water is accumulating on the top and colder one is heading down to absorb it, triggering a circular motion.
2. Radiator - Place hot air at the peak and pull cool air at the bottom.
3. Hot air balloon - The air is warmed up by a heating element within the balloon, so the air jumps upwards. This induces the balloon to increase in size due to the inside trapping of the warm air. He removes a few of the warm air when the pilot commences to dive, and cold air takes place, enabling the parachute to drop.
Your answer is.
(B) a pot being heated by an electric burner
(D) a radiator that emits warm air and draws in cool air
(E) a hot air balloon rising and falling in the atmosphere
Hope this helps you and you pass your test good luck my friend!
A 38,500 kg sphere is located 2.55 m from a 15,400 kg sphere. What is the gravitational force, rounded to the nearest thousandth of a Newton, between the two spheres?
The gravitational force between the two spheres is 0.006 N
Explanation:
The magnitude of the gravitational force between two objects is given by
[tex]F=G\frac{m_1 m_2}{r^2}[/tex]
where
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}[/tex] is the gravitational constant
[tex]m_1, m_2[/tex] are the masses of the two objects
r is the separation between them
For the two spheres in this problem, we have
[tex]m_1 = 38,500 kg[/tex]
[tex]m_2 = 15,400 kg[/tex]
r = 2.55 m
Substittuting into the equation, we find
[tex]F=(6.67\cdot 10^{-11})\frac{(38,500)(15,400)}{2.55^2}=0.006 N[/tex]
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Which of the following is a scalar quantity?
a. 35m
b. 20 m to the right
c. 30 m to the North
d. All of the above
The only scalar quantity is a. 35 m
Explanation:
In physics, there are two types of quantities:
- Scalar: a scalar quantiy is a quantity having only a magnitude, so it is just a number followed by a unit. Examples of scalar quantities in physics are:
Speed
Energy
Distance
Time
- Vector: a vector quantity is quantity having both a magnitude and a direction. Examples of vector quantities in physics are:
Velocity
Force
Acceleration
Displacement
Let's now analyze each given option, to evaluate if it is a scalar or a vector:
a. 35m --> it is only a unit (no direction), so it is a scalar
b. 20 m to the right --> it has a direction, so it is a vector
c. 30 m to the North --> it has a direction, so it is a vector
So, the only correct option is a).
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Fysical suo
A young girl is on a swing that completes 20.0 cycles in 25 seconds. What are its
frequency and period?
Answer:
0.8 - 1.25
Explanation:
[tex]f = n \div t[/tex]
[tex]period = 1 \div f[/tex]
f = 20/25 = 0.8 Hz
period = 1/0.8 = 1.25 s
The frequency of the girls motion is 0.8 Hz and her period of oscillation is 1.25 s
The given parameters;
number of cycles made by the girl during the swing, n = 20 cyclestime of her swing, t = 25 sThe frequency of the girls motion is calculated as follows;
[tex]f = \frac{number \ of \ complete \ cycle}{time \ of \ motion} \\\\f = \frac{n}{t} \\\\f = \frac{20}{25} \\\\f = 0.8 \ Hz[/tex]
The period of the girls oscillation is calculated as follows;
[tex]T = \frac{t}{n} \\\\T = \frac{25 \ s}{20} \\\\T = 1.25 \ s[/tex]
Thus, the frequency of the girls motion is 0.8 Hz and her period of oscillation is 1.25 s
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If only 10 pounds is required to lift a 500-lb block, how much chain must be played out to lift the engine 3.0 inches?
0.060 in
150 in
290 in
6.0 in
need help on more asap
Answer:
150 inches (12.5 ft)
Explanation:
The work done to lift the 500 pound block 3 inches should be the same as that to lift the 10 pond object a given distance.
The following is the equation one needs to solve:
[tex]10 \,lb\,* \,x\,=\,500\,lb\,*\,3\,in\\10 \,lb\,* \,x\,=\,1500\,lb\,in\\[/tex]
therefore solving for the distance "x" gives as the answer (in inches):
[tex]10 \,lb\,* \,x\,=\,1500\,lb\,in\\x\,=\,\frac{1500\,lb\,in}{10\,lb} \\\\x\,=150\,in[/tex]
which can also be given in feet as: 12.5 ft
Answer:
150 in"
Explanation:
What does sodium form an ion a+1 charge
Answer:
It has only 1 electron in its valence shell and if that electron is removed it will acquire noble gas configuration
Explanation:
Sodium forms an ion of charge +1 as in its ionic state it has inert gas configuration
It forms an ion of charge +1 when the valence shell electron is removed to gain the noble gas configuration and the electrons can be removed mostly from its valence shell because inner shell electrons will be tightly bounded to the nucleus of the atom and it requires high energy to remove the electron present in the inner shell when compared to the valence shell (outermost shell)
the sun and stars appear to move due to the earths rotation true or false
Answer:
True. The sun and stars look like they are moving because the earth rotates.
ANSWER FAST!! WILL GIVE CORRECT ANSWER BRAINLIEST! Which of the following could not be responsible for the acceleration of an object? wind friction gravity density
Answer:
The correct answer to this question is D) Density
EXPLANATION:
From Newton's second laws of motion, we know that the external force acting on a body is equal to the product of mass with the acceleration of the body.
Mathematically Force F = m × a
Here, m is the mass of the body.
a is the acceleration of the body.
Hence, it is the force which is responsible for creating acceleration in an object.
In the given question, we have four factors i.e wind, friction, gravity and density.
Out of the four options, wind, friction and gravity can exert force on an object and can create acceleration in it.
The density of a body is defined as the mass of that body per unit volume.
Mathematically it is written as -
Density d = \frac{m}{v} where v is the volume occupied by the body.
Hence, density will not be responsible for the acceleration of the body.
Answer: Your answer is option D: density
the wind can move objects depending on how strong it is and how small the object is.
Gravity pulls things towards the earth which will move it.
you needs movement to cause friction and friction causes heat which causes movement
but density has to do with the object itself not with movement.
Explanation:
A driver in a 2000 kg Porsche wishes to pass to pass a slow-moving school bus on a four-lane road. What is the average power in watts required to accelerate the sports car from 30 m/s to 60 m/s in 9 seconds
Answer:
The power require to accelerate the car is, P = 299700 watts
Explanation:
Give data,
The mass of the car, m = 2000 kg
The initial velocity of the sports car, u = 30 m/s
The final velocity of the sports car, v = 60 m/s
The time period of acceleration, t = 9 s
The acceleration of the car, a = (v-u) / t
= (60 - 30) / 9
= 3.33 m/s²
The displacement of the car,
S = ut + ½ at²
= 30 x 9 + ½ x 3.33 x 9²
= 405 m
The force acting on the car, F = m x a
= 2000 x 3.33
= 6660 N
The work done by the car, W = F S
= 6660 x 405
= 2697300 J
The power of the car, P = W / t
= 2697300 / 9
= 299700 watts
Hence, the power require to accelerate the car is, P = 299700 watts
When a mass of
m = 304g
is attached to a spring and the mass-spring system is set into oscillatory motion, the period of the motion is T = 0.467
Determine the amplitude (in meters) of the oscillation, if the total energy of the oscillating system is 0.163 J
The amplitude is 0.077 m
Explanation:
The period of a spring-mass system is given by
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
where
m is the mass
k is the spring constant
For the system in this problem, we have
m = 304 g = 0.304 kg is the mass
T = 0.467 s is the period
Solving for k, we find the spring constant:
[tex]k=(\frac{2\pi}{T})^2 m=(\frac{2\pi}{0.467})^2(0.304)=55.0 N/m[/tex]
The total energy of an oscillating system is given by
[tex]E=\frac{1}{2}kA^2[/tex]
where A is the amplitude of the oscillation. In this problem, we know that
k = 55.0 N/m
E = 0.163 J is the total energy
Solving for A, we find the amplitude:
[tex]A=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(0.163)}{55.0}}=0.077 m[/tex]
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Two mechanical waves that have positive displacements from the equilibrium position meet and coincide. What kind of
interference occurs?
a constructive
C. complete destructiv
b. destructive
none
When two mechanical waves that have positive displacements from the equilibrium position meet and coincide, a constructive interference occurs.
Option A
Explanation:Considering the principle of superposition of waves; the resultant amplitude of an output wave due to interference of two or more waves at any point is given by individual addition of their amplitudes at that point. Two waves with positive displacements refer to the fact that crest of the both the waves are on the same side of displacement axis, either both are positive or both are negative, similarly with their troughs.
If such two waves with their crest on crest meet at any point, by superposition principle. their individual amplitude gets added up and hence the resultant wave after interference is greater in amplitude that both the individual waves. This is termed as a constructive interference. Destructive interference on the other hand is a condition when one of the two waves has a positive displacement and other has a negative displacement (a condition of one’s crest on other’s trough); resulting in amplitude subtraction.
Object A has mass m and is traveling forward with speed v. Which set of changes to object A will not affect its momentum?
mass reduced to 0.5m, speed increased to 2v
mass reduced to 0.25m, speed increased to 1.25v
mass increased to 2m, speed increased to 2v
the direction of velocity reversed, speed kept at v
The situation that will not affect momentum is:
mass reduced to 0.5m, speed increased to 2v
Explanation:
The momentum of an object is defined as
[tex]p=mv[/tex]
where
m is the mass of the object
v is its velocity
Let's call m the initial mass of object A and v its initial velocity, so its initial momentum is
[tex]p=mv[/tex]
Let's now analyze each situation to see how the momentum changes:
- mass reduced to 0.5m, speed increased to 2v
New momentum is:
[tex]p'=(0.5m)(2v)=mv=p[/tex] --> so momentum is unchanged
- mass reduced to 0.25m, speed increased to 1.25v
New momentum is:
[tex]p'=(0.25m)(1.25v)=0.31mv = 0.31 p[/tex] --> momentum has changed
- mass increased to 2m, speed increased to 2v
New momentum is:
[tex]p'=(2m)(2v)=4mv=4p[/tex] --> momentum has changed
- the direction of velocity reversed, speed kept at v
Since momentum is a vector, the direction also matters: therefore, if the direction of the velocity changes, the direction of the momentum changes as well, so the momentum has changed.
So the only situation that will not affect the momentum is
mass reduced to 0.5m, speed increased to 2v
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solve this for 100 points work must be shown and must be all the problems PLZ I NEED IT ASAP
1) Angular velocity: D) 73.3 rev/s
2) Centripetal acceleration: D) [tex]5053 m/s^2[/tex]
3) Net force on the car: E) [tex]2.19\cdot 10^4 N[/tex]
4) Orbital speed: D) [tex](\frac{GM}{r})^{\frac{1}{2}}[/tex]
5) Total mechanical energy: E) [tex]-\frac{GMm}{2r}[/tex]
6) Gravitational force: C) [tex]1.32\cdot 10^{-10}N[/tex]
7) Upward force: C) 188 N
8) Total upward force at B: C) 122 N
Explanation:
1)
The relationship between centripetal acceleration and angular velocity is given by
[tex]a_c = \omega^2 r[/tex]
where:
[tex]a_c[/tex] is the centripetal acceleration
[tex]\omega[/tex] is the angular velocity
r is the radius of the circular path
In this problem,
[tex]a_c = 1950 g = 1950(9.8)=19,110 m/s^2[/tex]
r = 9 cm = 0.09 m
Solving for [tex]\omega[/tex],
[tex]\omega=\sqrt{\frac{a_c}{r}}=\sqrt{\frac{19,110}{0.09}}=460.8 rad/s[/tex]
And keeping in mind that [tex]1 rev = 2 \pi rad[/tex], we can convert into rev/s:
[tex]\omega=460.8 rad/s \cdot \frac{1}{2\pi rad/rev}=73.3 rev/s[/tex]
2)
Again, the centripetal acceleration is given by
[tex]a_c = \omega^2 r[/tex]
where in this problem we have:
[tex]\omega=16 rev/s \cdot 2\pi (rad/rev) = 100.5 rad/s[/tex] is the angular velocity of the piece of dust
r = 0.50 m is the distance of the piece of dust from the axis of rotation
And substituting into the equation, we find the centripetal acceleration of the piece of dust:
[tex]a_c = (100.53)^2(0.5)=5053 m/s^2[/tex]
3)
Since the car is moving in circular motion, the net force is equal to the centripetal force, so it is given by:
[tex]F=m\frac{v^2}{r}[/tex]
where
m = 732 kg is the mass of the car
v = 20.0 m/s is the linear speed of the car
r = 13.0 m is the radius of curvature of the loop-the-loop
Here we have
m = 732 kg
v = 20.0 m/s
r = 13.0 m
Substituting into the equation, we find:
[tex]F=(732)\frac{20.0^2}{13.0}=2.25\cdot 10^4 N[/tex]
The closest option is E) [tex]2.19\cdot 10^4 N[/tex]
4)
The gravitational force between the satellite and the Earth provides the centripetal force that keeps the satellite in circular motion, therefore we can write:
[tex]m\frac{v^2}{r}=G\frac{Mm}{r^2}[/tex]
where the term on the left is the centripetal force while the term on the right is the gravitational force, and where
m is the mass of the satellite
v is its speed
r is the orbital radius
G is the gravitational constant
M is the Earth's mass
Solving for v,
[tex]v=\sqrt{\frac{GM}{r}} = (\frac{GM}{r})^{\frac{1}{2}}[/tex]
5)
The kinetic energy of the satellite is given by
[tex]K=\frac{1}{2}mv^2[/tex]
And substituting its speed, [tex]v=\sqrt{\frac{GM}{r}}[/tex], we get
[tex]K=\frac{GMm}{2r}[/tex]
The potential energy of the satellite is equal to the work that must be done against gravity to bring the satellite from r to infinite, and it is
[tex]U=-\frac{GMm}{r}[/tex]
The total mechanical energy is equal to the sum of kinetic and potential energy:
[tex]E=K+U=\frac{GMm}{2r}+(-\frac{GMm}{r})=-\frac{GMm}{2r}[/tex]
6)
The gravitational force between two objects is given by
[tex]F=G\frac{m_1 m_2}{r^2}[/tex]
where
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}[/tex] is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
Here we have:
[tex]m_1 = 12 kg\\m_2 = 14 kg[/tex]
and
r = 9.22 m
Substituting,
[tex]F=(6.67\cdot 10^{-11})\frac{(12)(14)}{(9.22)^2}=1.32\cdot 10^{-10}N[/tex]
7)
To solve the problem, we can apply the equation of the equilibrium of momenta about point B:
[tex]Wx_W - F_A x_A=0[/tex]
where:
W = 300 N is the weight of the plank
[tex]x_W = 5 m[/tex] is the distance of the point of application of the weight from B (half the length of the plank)
[tex]F_A[/tex] is the upward force applied at A
[tex]x_A = 10 - 2 = 8 m[/tex] is the distance of the point of application of [tex]F_A[/tex] from B
Solving for [tex]F_A[/tex],
[tex]F_A = \frac{W x_W}{x_A}=\frac{(300)(5)}{8}=188 N[/tex]
8)
We can solve this part instead by applying the equation of equilibrium of moments around point A:
[tex]-Wx_W - W_B x_B + F_B x_{F_B} = 0[/tex]
where
W = 300 N is the weight of the plank
[tex]x_W = 5 - 2 = 3 m[/tex] is the distance of the point of application of the weight from A
[tex]W_B = 9.0 N[/tex] is the weight of the can of paint at B
[tex]x_B = 10 - 2 = 8 m[/tex] is the distance of the can of paint from A
[tex]F_B[/tex] is the upward force exerted by block B
[tex]x_{F_B}= 10 -2 = 8 m[/tex] is the distance of the point of application of [tex]F_B[/tex] from A
Solving for [tex]F_B[/tex],
[tex]F_B = \frac{W_B x_B + W x_A}{x_{F_B}}=\frac{(9)(8)+(300)(3)}{8}=122 N[/tex]
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