Answer:
There is no change in distance
Explanation:
Given that block slides a distance dd with initial velocity [tex]v_{ox}[/tex][tex]v_{ox}[/tex]
Work energy theorem states that Work done W is the difference between final kinetic energy [tex]KE_{f}[/tex] and initial kinetic energy [tex]KE_{i}[/tex]
W = [tex]KE_{f}[/tex] - [tex]KE_{i}[/tex] ...........(equation 1)
In the given problem, the work is done by frictional force
Hence the work done is given by
[tex]W_{Friction}[/tex] = -[tex]F_{Friction}[/tex] x dd
[tex]F_{Friction}[/tex] is negative since it is resistive force and dd is the distance
[tex]W_{Friction}[/tex] = - μ mg X dd
where μ is coefficient of friction.
Also we know that Kinetic energy KE = [tex]\frac{1}{2}[/tex] [tex]mv^{2}[/tex]
[tex]KE_{f}[/tex] = 0 since final velocity is zero
[tex]KE_{i}[/tex] = [tex]\frac{1}{2}[/tex] [tex]m(v_{ox} v_{ox}) ^{2}[/tex]
Substituting the corresponding values equation 1 becomes
-μ mg x dd = 0 - [tex]\frac{1}{2}[/tex] [tex]m(v_{ox} v_{ox}) ^{2}[/tex]
dd= [tex]\frac{(v_{ox}v_{ox} )^{2} }{2g}[/tex]x (1/μ) ........... (equation 2)
To find distance when mass of block is doubled and initial velocity is not changed:
Equation 2 shows that the distance dd is independent of mass, therefore there is no change in distance.
The block will slide only half the distance before stopping when its mass is doubled while the initial velocity remains the same.
When a block is pushed along the floor with an initial velocity [tex]\(v_{0x}\)[/tex] and slides a distance [tex]\(d\)[/tex] before stopping, its stopping distance is determined by the initial velocity and the frictional force opposing its motion.
The force of friction is given by [tex]\(F_{\text{friction}} = \mu \cdot N\)[/tex], where [tex]\(\mu\)[/tex] is the coefficient of friction and [tex]\(N\)[/tex] is the normal force. In this case, we are assuming a constant coefficient of friction.
The deceleration of the block can be calculated using Newton's second law: [tex]\(F = m \cdot a\)[/tex]. When the mass of the block is doubled, the force required to decelerate it to a stop is also doubled.
Let's assume the initial velocity [tex]\(v_{0x}\)[/tex] remains the same. The deceleration [tex]\(a\)[/tex] is proportional to the force, so [tex]\(a\)[/tex] is also doubled when the mass is doubled. This is because [tex]\(F\)[/tex] is directly proportional to [tex]\(m\)[/tex].
The stopping distance [tex]\(d'\)[/tex] can be calculated using the equation of motion:
[tex]\[v_f^2 = v_i^2 + 2as\][/tex]
Where:
[tex]\(v_f\)[/tex] is the final velocity (0 m/s, as the block stops),
[tex]\(v_i\)[/tex] is the initial velocity [tex](\(v_{0x}\)),[/tex]
[tex]\(a\)[/tex] is the deceleration (doubled),
[tex]\(s\)[/tex] is the stopping distance [tex](\(d'\)).[/tex]
Plugging in the values:
[tex]\[0 = v_{0x}^2 + 2(2a)d'\][/tex]
Now, we can solve for [tex]\(d'\):[/tex]
[tex]\[2ad' = -v_{0x}^2\][/tex]
[tex]\[d' = \frac{-v_{0x}^2}{2a}\][/tex]
Since [tex]\(a\)[/tex] is doubled, the stopping distance [tex]\(d'\)[/tex] will be halved compared to the initial distance [tex]\(d\).[/tex] So, the block will slide only half the distance before stopping when its mass is doubled while the initial velocity remains the same.
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An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required by the compressor. The gas constant of air is R = 0.287 kPa·m3/kg·K. The flow work required by the compressor is
The required flow work of the compressor is calculated using the flow work equation and the ideal gas law. First, we determine the final volume using the ideal gas law. Then we substitute these figures into the flow work equation, giving the result as 8.08 kJ/kg.
Explanation:The flow work required by the compressor is calculated using the equation
flow work = pressure * volume.
Given the initial pressure P1 = 120 kPa, volume V1 = 6 L and final pressure P2 = 1000 kPa,
we can substitute these values into the equation.
However, the volume at the end of compression is not given. To find this, we need to use the ideal gas law, P1V1/T1=P2V2/T2,
where T1 is the initial temperature and T2 is the final temperature.
Convert the temperatures to kelvins (T1 = 22 + 273 = 295 K, T2 = 400 + 273 = 673 K) and volume to m3 (V1 = 6 / 1000). Solving for V2 gives V2 = P1V1T2 / P2T1 = 0.00147 m3.
Now, substituting again into the flow work equation gives
flow work = (1000 kPa)(0.00147 m3) = 1.47 kJ.
This is the energy per unit volume, to find it per unit mass, divide it by the specific volume
v2 = V2 / m = R*T2 / P2 = 0.182 m3/kg,
Therefore, the required flow work = 1.47 kJ / 0.182 kg = 8.08 kJ/kg.
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The flow work required by the compressor is approximately 622.476 kJ/kg.
The flow work required by the compressor is given by the equation:
[tex]\[ w_{flow} = \int_{1}^{2} v \, dp \][/tex]
where [tex]\( v \)[/tex] is the specific volume of the air and [tex]\( dp \)[/tex] is the differential pressure change. For an ideal gas, the specific volume can be calculated using the ideal gas law:
[tex]\[ v = \frac{RT}{p} \][/tex]
where [tex]\( R \)[/tex] is the gas constant, [tex]\( T \)[/tex] is the absolute temperature in Kelvin, and [tex]\( p \)[/tex] is the pressure.
Given:
- Initial pressure [tex]\( p_1 = 120 \)[/tex] kPa
- Final pressure [tex]\( p_2 = 1000 \)[/tex] kPa
- Initial temperature [tex]\( T_1 = 22 + 273.15 = 295.15 \)[/tex] K (converting from Celsius to Kelvin)
- Final temperature [tex]\( T_2 = 400 + 273.15 = 673.15 \)[/tex] K
- Gas constant [tex]\( R = 0.287 \) kPa/m^3/kg/K[/tex]
Since the process is adiabatic and the air is being compressed, we can assume that the specific volume at the initial state [tex]\( v_1 \)[/tex] can be calculated using the initial conditions:
[tex]\[ v_1 = \frac{RT_1}{p_1} \][/tex]
Substituting the values:
[tex]\[ v_1 = \frac{0.287 \times 295.15}{120} \][/tex]
[tex]\[ v_1 = \frac{84.80}{120} \][/tex]
[tex]\[ v_1 = 0.7067 \text{ m}^3/\text{kg} \][/tex]
The flow work can be approximated if we assume the process to be isothermal at the inlet temperature (which is a common simplification for flow work calculation):
[tex]\[ w_{flow} = v_1(p_2 - p_1) \][/tex]
Substituting the values:
[tex]\[ w_{flow} = 0.7067(1000 - 120) \][/tex]
[tex]\[ w_{flow} = 0.7067 \times 880 \][/tex]
[tex]\[ w_{flow} = 622.476 \text{ kJ/kg} \][/tex]
A tilted bed is said to have a _____, describing the angle that the bed forms with the horizontal plane--and a _____,the compass direction that lies at right angles to the tilted bed.
Answer:
The answer to the question is
A tilted bed is said to have a _dip____, describing the angle that the bed forms with the horizontal plane--and a strike, the compass direction that lies at right angles to the tilted bed.
Explanation:
The dip of a tilted bed, describes the acute angle a tilted bed makes with the horizontal plane, by stating the numerical value of the angle from 0 to 90 degrees as well as pointing out the orientation of the downward dipping direction in the orientation towards N, S, E, W
The strike line represents the line formed to represent the intersection of a feature of a bed such as the bed rock surface with a horizontal plane.
The dip and the strike line of a tilted bed are always at right angles to each other on a geologic map.
A tilted geological layer's angle with the horizontal is called the 'dip' and the line it forms intersecting with the horizontal at 90 degrees is the 'strike'. These measurements help determine the orientation of rock layers suffering from deformation.
Explanation:A tilted bed is said to have a dip, describing the angle that the bed forms with the horizontal plane—and a strike, the compass direction that lies at right angles to the tilted bed. The dip is a measure of the steepest angle of descent relative to the horizontal plane and indicates the direction in which water would flow down the plane. The strike, on the other hand, is the direction of the line formed by the intersection of a rock layer's surface with the horizontal plane, which is always perpendicular to the dip direction. Survey instruments like a Brunton Compass are commonly used by geologists to measure strike and dip accurately to understand feature orientations within geological formations.
"____ formulated the law of falling bodies and used the telescope to confirm empirically the previously rhetorical model of a heliocentric universe?"
Answer:
Galileo
Explanation:
Galileo formulated the law of free falling bodies that distance traveled by falling bodies are proportional to the squares of the time elapsed.
Galileo confirmed the model of heliocentric universe given previously by Copernicus that the whole universe revolves around the sun . His observation of movement of Venus helped him prove this model.
SPEAR is a storage ring at the Stanford Linear Accelerator which has a circulating beam of electrons that are moving at nearly the speed of light (2.998 108 m/s). If a similar ring is about 92.0 m in diameter and has a 0.40 A beam, how many electrons are in the beam
Answer:
[tex]2.4\times 10^{12}[/tex]
Explanation:
We are given that
Speed of light,v=[tex]2.998\times 10^8 m/s[/tex]
Diameter of ring,d=92 m
Radius,r=[tex]\frac{d}{2}=\frac{92}{2}=46 m[/tex]
Current, I=0.40 A
We have to find the number of electrons in the beam.
We know that
Current,I=[tex]\frac{q}{t}[/tex]
Where q= ne
[tex]e=1.6\times 10^{-19} C[/tex]
Using the formula
[tex]0.40=\frac{1.6\times 10^{-19}n}{\frac{2\pi r}{v}}[/tex]
[tex]0.40=\frac{1.6\times 10^{-19}n\times v}{2\pi r}[/tex]
[tex]0.40=\frac{1.6\times 10^{-19}n\times 2.998\times 10^8}{2\pi\times 46}[/tex]
[tex]n=\frac{0.40\times 2\pi\times 46}{1.6\times 10^{-19}\times 2.998\times 10^8}=2.4\times 10^{12}[/tex]
A typical flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Take the mass of the insect to be 7.0g , and assume the wings move an average downward distance of 1.5cm during each stroke. Assuming 117 downward strokes per second, estimate the average power output of the insect.
Answer:
Average power output of insect is 2.42W
Explanation:
Workdone by constant force during displacement is given by:
W= F× d cos theta
Where theta is angle between F and d.
Power output due to the force over the interval time is given by:
P= Workdone/change in time
Ginen:
Mass of insect,m= 7.0g= 7/1000 = 0.07kg
Downward force applied by insect,F= 2mg
Distance moved by the wing each stroke=1.5cm=1.5/100= 0.015m
W= F× d cos theta
Where theta=0° since force is in the same direction as the displacement.
W= 2mg×d
W= 2× 0.07 × 9.8 × 0.015
W= 0.02058J
Power output = W/ change in time
Since wings make 117strokes each second time interval is 1/117 = 8.5×10^-3seconds
Power= 0.02058/(8.5×10^-3)
Power= 2.42W
One kind of baseball pitching machine works by rotating light and stiff rigid rod about a horizontal axis until the ball is moving toward the target. Suppose a 144 gg baseball is held 81 cm from the axis of rotation and released at the major league pitching speed of 81 mph.
a. What is the ball's centripetal acceleration just before it is released?
b. What is the magnitude of the net force that is acting on the ball just before it is released?
Answer:
(a). The ball's centripetal acceleration is [tex]16.17\times10^{2}\ m/s^2[/tex]
(b). The magnitude of the net force is 232.9 N.
Explanation:
Given that,
Mass of baseball = 144 g
Speed = 81 mph = 36.2 m/s
Distance = 81 cm
(a). We need top calculate the ball's centripetal acceleration just before it is released
Using formula of centripetal acceleration
[tex]a=\dfrac{v^2}{r}[/tex]
Where, v = speed
r = radius
Put the value into the formula
[tex]a=\dfrac{(36.2)^2}{81\times10^{-2}}[/tex]
[tex]a=1617.82\ m/s^2[/tex]
[tex]a=16.17\times10^{2}\ m/s^2[/tex]
(b). We need to calculate the magnitude of the net force that is acting on the ball just before it is released
Using formula of force
[tex]F=\dfrac{mv^2}{r}[/tex]
Put the value into the formula
[tex]F=\dfrac{144\times10^{-3}\times(36.2)^2}{81\times10^{-2}}[/tex]
[tex]F=232.9\ N[/tex]
Hence, (a). The ball's centripetal acceleration is [tex]16.17\times10^{2}\ m/s^2[/tex]
(b). The magnitude of the net force is 232.9 N.
Two boxes of masses m=35kg and m2=45kg, are hung vertically from opposite ends of a rope passing over a rigid horizontal metal rod. They system starts moving from rest. Assuming that friction between the rod and the rope is negligible, determine the magnitude of
(a) the acceleration of the boxes
(b) the tension in the rope
(c) the magnitude of each box's displacement after 0.5s
Answer:
a) 5.51m/s² b) 192.94N c) 1.38m each
Explanation:
Given two boxes of masses m1 = 35kg and m2 = 45kg hung vertically from opposite ends of a rope passing over a rigid horizontal metal rod, we will analyze the forces acting on each body.
According Newton's second law, Force = mass ×acceleration
The forces acting on body of mass m1 are the tension (T) and the frictional force (Ff) which opposes the tension.
Taking the sum of horizontal forces acting on mass m1, we will have;
T +(-Ff) = m1a
T - Ff = m1a... (1)
For the mass m2, the forces acting on the body are in the vertical direction and this forces are the weight (W) acting downwards and the tension(T) acting upwards. The sum of the forces in the body is given as ;
W + (-T) = m2a
W-T = m2a ...(2)
Since W = mg, equation 2 will become;
m2g - T = m2a...(2)
Solving equation 1 and 2 simultaneously to get the tension and the acceleration, we have;
T - Ff = m1a ... 1
m2g - T = m2a ... 2
Since friction is negligible, Ff = 0
Adding the two equation will give;
m2g-Ff = m2a+m1a
Since Ff =0
m2g = (m2+m1)a
a = m2g/m1+m2
a = 45(9.8)/45+35
a =441/80
a = 5.51m/s²
b) Substituting a = 5.51 into equation 1 to get the tension T in the rope will give;
T = m1a
T = 35×5.51
T = 192.94N
c) since velocity = displacement/time
Displacement = velocity × time
To get the velocity, since acceleration = velocity/time,
Velocity = acceleration ×time
Velocity = 5.51× 0.5
Velocity = 2.76m/s
Displacement of each box will be the same since they are moving with the same acceleration.
Displacement = 2.76m/s × 0.5s
Displacement of each boxes = 1.38m
The acceleration of the boxes is 0.98 m/s^2, the tension in the rope is 376.3 N, and the displacement of each box after 0.5 seconds is 0.12 m.
Explanation:This question deals with the physics of motion, particularly involving concepts of mass and friction. Given that the two boxes with different masses m1=35kg and m2=45kg are hung vertically from opposite ends of a rope over a rigid horizontal metal rod, and stating negligible friction, we can compute:
The acceleration (a) of the boxes using the equation: a = (m2 - m1)*g / (m1 + m2) where g is the acceleration due to gravity. Gravity is approximately 9.8 m/s^2. Substituting the given values, a = (45 - 35)*9.8 / (35 + 45) = 0.98 m/s^2. The tension (T) in the rope using the formula T = m1 * (g + a) or T = m2 * (g - a). Both formulas lead to the same result. Substituting the given values for the lighter 35kg box, T = 35*(9.8 + 0.98) = 376.3 N. The magnitude of each box's displacement after 0.5 seconds can be calculated using the equation s=0.5*at^2. Substituting the values, s = 0.5 * 0.98 * (0.5)^2 = 0.12 m.Learn more about Motion here:
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A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod ) What happens to end A of the rod when the ball approaches it closely this first time?
a.It is strongly repelled.
b.It is strongly attracted.
c.It is weakly attracted.
d.It is weakly repelled.
e.It is neither attracted nor repelled.
Answer:
Option B, it is strongly attracted
Explanation:
A Test Charge Determines Charge on Insulating and Conducting Balls, and the points made regarding conductors, it can be ascertained that in conductors, the electrons are free to move about. This means that when a charge is brought near to a conductor, the opposite charges all navigate to the sharpest point closest the charge and a strong attraction is created.
This shows that the rod will be strongly attracted. The density of the charges on the rod is mostly concentrated at the sharpest point.
End A of the rod will be strongly attracted to the negatively charged metal ball because of the process of charge induction, where opposite charges attract.
Explanation:When the negatively charged metal ball is brought near to end A of the rod, end A of the rod will be strongly attracted to the negatively charged ball. This is because of the principle of charge induction. When a charged body is brought near to another body, it will cause the charges in that body to redistribute. Opposite charges attract, so the near side of the rod (end A) will have a positive charge induced on it, and this positive charge will be attracted to the negative charge on the ball. So, the correct answer is option b. It is strongly attracted.
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(BRAINLIEST W/ WORK SHOWN!!!) Help with Physics question?
A sound wave traveling at 343 m/s takes 10.0 s to go from a speaker to a detector. How far apart are the two devices?
A.) 3.43 x 10^3
B.) 3.43 x 10^2
C.) 3.43 x 10^-1
D.) 3.43 x 10^4
Answer:
[tex]\boxed {3.43 x 10^{3}}[/tex]
Explanation:
We know that speed is defined as distance moved per unit time hence expressed as [tex]v=\frac {d}{t}[/tex] where v is speed in m/s, d is distance in m and t is time in seconds. Making d the subject of the above formula then
[tex]d=vt[/tex]
Substituting 343 m/s for d and 10 s for t then
[tex]d= 343\times10= 3430= 3.43 x 10^{3}[/tex]
Therefore, the distance between speaker and deter is [tex]\boxed {3.43 x 10^{3}}[/tex]
Which of the following characterizes a beta ray? Choose all that apply. is electromagnetic radiation is a product of natural radioactive decay is attracted to the positively charged plate in an electric field is attracted to the negatively charged plate in an electric field is composed of electrons
Explanation:
When a radioactive substance decays then the fast moving electrons emitted by it is known as beta ray. Basically, a number of beta particles are ejected by a beta ray.
Symbol of a beta particle is [tex]^{0}_{-1}e[/tex]. A beta ray is a natural decay of a radioactive element. As we know that opposite charges get attracted towards each other. So, a beta ray gets attracted towards a positively charged plate.
Therefore, we can conclude that following are the characterizes a beta ray:
a product of natural radioactive decay.is attracted to the positively charged plate in an electric field. is composed of electrons.Final answer:
Beta rays are negatively charged, attracted to the positively charged plate in an electric field, and are composed of electrons. They are a product of natural radioactive decay and are lighter and much less massive than alpha particles.
Explanation:
Beta rays are characterized by specific properties that distinguish them from other types of radiation produced during natural radioactive decay. According to Ernest Rutherford's research, which involved observing the behavior of radiation in magnetic and electric fields, beta particles are known to be negatively charged and relatively light. This means they are attracted to the positively charged plate in an electric field and are significantly deflected due to their lighter mass compared to alpha particles.
Beta rays are not electromagnetic radiation; this term is reserved for gamma rays, which are uncharged and therefore unaffected by electric fields. Beta rays are indeed a product of natural radioactive decay, specifically during a process known as beta decay, in which a nucleus emits an electron or a positron. Since they are composed of high-energy electrons, the identification that beta rays are composed of electrons is also correct.
A super ball is dropped from a height of 100 feet. Each time it bounces, it rebounds half the distance it falls. How many feet will the ball have traveled when it hits the ground for the fourth time
The total distance travelled by the ball after the fourth impact is 275 feet.
Explanation:
Given-
Height, h = 100 feet
Rebounds half the distance
Distance in feet for the fourth time, x = ?
For the first time, the distance travelled by the ball is, x = 100 feet
For the second time, it will bounce up to 50 feet and fall upto 50 feet( half of 100 feet)
So, the distance travelled after the second impact, x = 100 + 50 + 50 = 200 feet
For the third time, it will bounce up to 25 feet and fall upto 25 feet( half of 50 feet)
So, the distance travelled after the third impact, x = 200 + 25 + 25 = 250 feet
For the fourth time, it will bounce up to 12.5 feet and fall upto 12.5 feet( half of 25 feet)
So, the distance travelled after the fourth impact, x = 250 + 12.5 + 12.5 = 275 feet
Therefore, total distance travelled by the ball after the fourth impact is 275 feet.
The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 3.6 m/s in the positive x direction, and some time later has a velocity of 7.0 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time
Answer:
The work done on the canister by the 5.0 N force during this time is
54.06 Joules.
Explanation:
Let the initial kinetic energy of the canister be
KE₁ = [tex]\frac{1}{2} mv_1^{2}[/tex] = [tex]\frac{1}{2} *3*3.6^{2}[/tex] = 19.44 J in the x direction
Let the the final kinetic energy of the canister be
KE₂ = [tex]\frac{1}{2} mv_2^{2}[/tex] = [tex]\frac{1}{2} *3*7.0^{2}[/tex] = 73.5 J in the y direction
Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final
= 73.5 J - 19.44 J = 54.06 J
Explanation:
Below is an attachment containing the solution.
The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long is a year on mercury in units of earth years?
Answer:
[tex]T_1=0.24y[/tex]
Explanation:
Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:
[tex](\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3[/tex]
Where [tex]T_1[/tex] and [tex]T_2[/tex] are the orbital periods of Mercury and Earth respectively. We have [tex]D_1=0.39D_2[/tex] and [tex]T_2=1y[/tex]. Replacing this and solving for
[tex]T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y[/tex]
Final answer:
A year on Mercury is approximately 88 Earth days long, which means it is about 0.241 Earth years, due to Mercury's average distance from the Sun being 0.39 times that of Earth's.
Explanation:
The student's question relates to the orbital period of Mercury compared to Earth's, given its average distance from the Sun. Mercury’s orbit around the Sun takes approximately 88 Earth days, which constitutes a Mercury year. This is calculated using Kepler's third law of planetary motion, which relates the orbital period of a planet to its average distance from the Sun (orbital semi-major axis).
Because Mercury is 0.39 times as far from the Sun as Earth is, its orbital period is significantly shorter than Earth's. Earth's average distance from the Sun is approximately 1 astronomical unit (AU), making it the basis for measuring distances in our solar system. Therefore, a year on Mercury, in Earth years, is 88/365, or about 0.241 Earth years.
A high-speed dart is shot from ground level with a speed of 150 m/s at an angle 30° above the horizontal. What is the vertical component of its velocity after 4.0 s if air resistance is neglected?
Explanation:
The dart is project with 150 m/s from a point at an angle of 30⁰
The vertical component of velocity = 150 sin 30 = 75 m/s
Thus initial vertical velocity is = 75 m/s
The velocity after 4 s can be calculate by
v = u - g t
here u is the initial velocity and t is the time , g is the acceleration due to gravity .
Thus v = 75 - 10 x 4 = 35 m/s
The velocity after 4.0 seconds will be "35 m/s".
According to the question,
Speed,
150 m/sAngle,
30°Vertical component of velocity,
[tex]150 \ Sin 30^{\circ} = 75 \ m/s[/tex]After 4 seconds, the velocity will be:
→ [tex]v = u-gt[/tex]
By substituting the values, we get
[tex]= 75-10\times 4[/tex]
[tex]= 75-40[/tex]
[tex]= 35 \ m/s[/tex]
Thus the answer above is right.
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Enrico is having trouble telling the difference between the sound of a tuba and the sound of a piccolo. Even though a piccolo produces much briefer, faster sound waves than does a tuba, he has trouble picking out the differences in the _____ of these sounds. Please choose the correct answer from the following choices, and then select the submit answer button.
Answer: Pitch
Explanation:
Pitch of sound is defined as the factor that monitors the sound quality through produced vibrations rate.It helps in determination of sounds tone in terms highness or lowness.
According to the question,Enrico is finding difficulty in judging the difference between pitch sound of tuba and piccolo as per their tone in terms of high or low.
Enrico is having trouble telling the difference between the sound of a tuba and the sound of a piccolo. Even though a piccolo produces much briefer, faster sound waves than does a tuba, he has trouble picking out the differences in the pitch of these sounds.
Pitch: Pitch is the perceptual attribute of sound that allows us to distinguish between different frequencies. The sound of a tuba and a piccolo are different primarily because they produce sound waves at different frequencies. A tuba produces lower frequency sound waves (lower pitch), while a piccolo produces higher frequency sound waves (higher pitch). If Enrico is having trouble telling the difference between the sound of a tuba and a piccolo, it suggests he is having trouble distinguishing between their pitches.
Loudness: Loudness refers to the perceived volume or intensity of a sound. While the tuba and piccolo can be played at different volumes, the primary distinguishing factor between them is not loudness but pitch.
Hue: Hue is a term used in the context of color, not sound. It refers to the distinct characteristic of color that allows us to differentiate between colors such as red, blue, and green.
Amplitude: Amplitude refers to the height of the sound wave and is related to the loudness or volume of the sound. While amplitude can affect how loud a sound is, it does not directly differentiate between the characteristic sounds of a tuba and a piccolo.
Therefore, the appropriate term for distinguishing between the sounds of different instruments, such as a tuba and a piccolo, is pitch.
The complete question is:
Enrico is having trouble telling the difference between the sound of a tuba and the sound of a piccolo. Even though a piccolo produces much briefer, faster sound waves than does a tuba, he has trouble picking out the differences in the of these sounds.
O pitch
O loudness
O hue
O amplitude
How does sugar affect the attention of small children?
When you float in fresh water, the buoyant force that acts on you is equal to your weight. When you float higher in the denser water of the Dead Sea, the buoyant force that acts on you is:_______.a) greater than your weight.
b) less than your weight.
c) equal to your weight.
Final answer:
The buoyant force experienced by an object submerged in a fluid is determined by the Archimedes' principle. When floating in the denser water of the Dead Sea, the buoyant force is greater than one's weight.
Explanation:
The buoyant force experienced by an object submerged in a fluid is determined by the Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object. When a person floats in fresh water, the buoyant force acting on them is equal to their weight. However, when they float higher in the denser water of the Dead Sea, the buoyant force that acts on them is greater than their weight. Therefore, the correct answer is (a) greater than their weight.
Final answer:
The buoyant force acting on you when you float in the denser water of the Dead Sea is equal to your weight, which is consistent with Archimedes' principle. The correct option is c.
Explanation:
When you float higher in the denser water of the Dead Sea, the buoyant force that acts on you is: c) equal to your weight. Archimedes' principle tells us that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid that the object displaces. Given that the Dead Sea has a higher density due to its salt content, you displace less water to experience a buoyant force that equals your weight, compared to fresh water. As a result, you float higher in the Dead Sea, but the buoyant force itself remains equal to your weight. Hence, Option c is correct.
A 0.40-kg block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so that the spring stretches for 2.0 cm relative to its unstrained length. When the block is released, it moves with an acceleration of 8.0 m/s2. What is the spring constant of the spring
Answer:
160N/m
Explanation:
According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,
F = ke where
F is the applied force
k is the spring constant
e is the extension
From the formula k = F/e
Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.
The spring constant k = ma/e where
m is the mass of the block = 0.4kg
a is the acceleration = 8.0m/s²
e is the extension of the spring = 2.0cm = 0.02m
K = 0.4×8/0.02
K = 3.2/0.02
K = 160N/m
The spring constant of the spring is therefore 160N/m
Final answer:
The spring constant of the spring is calculated using Hooke's Law and Newton's second law of motion. By multiplying the mass of the block by its acceleration, we found the force, and then divided the force by the displacement to get the spring constant, which is 160 N/m.
Explanation:
To determine the spring constant of the spring, we need to apply Hooke's Law, which states that the force exerted by an ideal spring is directly proportional to its displacement from the equilibrium position (F = -kx), where 'F' is the force, 'k' is the spring constant, and 'x' is the displacement. Since the block is on a frictionless surface and we know the acceleration (a = 8.0 m/s2) and the mass (m = 0.40 kg), we can first find the force using Newton's second law (F = ma), and then use that force to calculate the spring constant 'k'.
The force exerted by the spring can be calculated as:
F = m * a
= 0.40 kg * 8.0 m/s²
= 3.2 N
The displacement (x) from the equilibrium position is given as 2.0 cm, which is 0.020 m in SI units. Using Hooke's Law, the spring constant can be calculated:
k = F / x
= 3.2 N / 0.020 m
= 160 N/m
An unstrained horizontal spring has a length of 0.43 m and a spring constant of 238 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.013 m relative to its unstrained length. Determine the possible algebraic signs and the magnitude of the charges. (a) the possible algebraic signs
Answer:
The charges on the spring are 1.23E-5 C and they have the same sign
Explanation:
Given
Coulomb laws states that the force exerted by a charge q on another charge Q at a distance r is given by
F = kqQ/r²
Where k = 8.99 * 10^9 Nm²/C²
r = 0.43 + 0.013
r = 0.443m
The force on the spring is calculated as;.
F = kx where x is the stretch length of the spring and k is the spring constant
The force acting on the spring = 238 * 0.013
F = 3.094N
By comparison;
F = kqQ/r² becomes
3.094 = F = kqQ/r²
kqQ/r² = 3.094 (Considering that a = Q)
kq²/r² = 3.094
8.99 * 10^9 * q²/0.443 = 3.094 -- make q² the subject of formula
q² = 3.094 * 0.443/8.99*10^9
q² = 1.524629588431E−10
q = √1.524629588431E−10
q = 0.000012347589191545C
q = 1.23E-5 C
The charges on the spring are 1.23E-5 C and they have the same sign
Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mass of 5.40 10-2 kg and a charge of 5.10 10-5 C. The particle has a speed of 2.00 m/s on surface A. A nonconservative outside force is applied to the particle, and it moves to surface B, arriving there with a speed of 3 m/s. How much work is done by the outside force in moving the particle from A to B
Answer:
0.247 J = 247 mJ
Explanation:
From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.
So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁) where m = mass of particle = 5.4 × 10⁻² kg, q = charge of particle = 5.10 × 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.
So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁)
= 1/2 × 5.4 × 10⁻²kg × ((3m/s)² - (2 m/s)²) + 5.10 × 10⁻⁵ C(7850 - 5650)
= 0.135 J + 0.11220 J
= 0.2472 J
≅ 0.247 J = 247 mJ
A car is rounding a circular curve of radius r on a banked turn. As the drawing indicates, there are two forces acting on the car, its weight mg and the normal force FN exerted on it by the road. Which force, or force component, provides the centripetal force that keeps the car moving on the circular path?
1. The vertical component, FNcosθ of the normal force.
2. The horizontal component, FNsinθ of the normal force.
3. Both the normal force, FN, and the weight, mg, of the car.
4. The normal force, FN.
5. The weight, mg, of the car.
6. The horizontal component, FNsinθ of the norma.l
Final answer:
The horizontal component of the normal force, represented as FNsinθ, provides the centripetal force necessary for a car to round a circular curve on a banked turn without friction.
Explanation:
When a car is rounding a circular curve on a banked turn, the force that provides the centripetal force necessary to keep the car moving on the circular path is the horizontal component of the normal force exerted on it by the road. This can be represented as FNsinθ, where θ is the banking angle. The weight of the car, mg, acts vertically downwards and does not contribute to the centripetal force in this ideal frictionless scenario. For ideal banking, where the angle is perfect for the speed and radius of the turn, the net external force equals the horizontal centripetal force required for circular motion. Therefore, the horizontal component of the normal force is the only force component that acts towards the center of the curvature, providing the centripetal acceleration.
Consider the following possible stages in the evolution of a star like our sun: black dwarf, giant, main-sequence, planetary nebula, supernova, white dwarf. Rank the stages in the order they occur. Leave out any stages that will not occur in the evolution of a star of similar mass to the sun
Answer:
Main sequence
Giant
Planetary nebula
White dwarf
Black dwarf
Explanation:
Main Sequence:
Stars are called main sequence stars when their core temperature reaches up to 10 million kelvin and their start the nuclear fusion reactions of hydrogen into helium in the core of the star. For example sun is known as to be in the stage of main sequence as the nuclear fusion reactions are happening in its core.
Giant:
Next step is the Giant phase. When the stars run out of their fuel that is hydrogen for the nuclear fusion reactions then they convert into Giant stars. Giant stars have the larger radius and luminosity then the main sequence stars.
Planetary nebula:
Planetary nebula consists of glowing gases and plasma, it ejects from the red giant stars that run out of their fuel.
White dwarf:
When the stars run out of their fuel then they shed the outer layer planetary nebula, the remaining core part that left behind is called as white dwarf. It's the most dense part as the most of the mass is concentrated in this part.
Black dwarf:
When the white dwarf cool down completely that it no longer emitt heat and light then it is called as black dwarf.
These were the possible stages that includes in the evolution of stars
Nana manages to stop their decent. While stopped, with the axe in the ice, her coefficient of static friction is LaTeX: \mu_s=2.8μ s = 2.8. What is the maximum acceleration ice Popo can climb up the rope, without causing them to slide again?
Answer:
Acceleration, a= 1.65m/s^2
Acceleration is downward.
Explanation:
y: N- m1gcosalpha=0
x: Ff- FT- m1gsinalpha= m1× alpha
Ff= uN= um1gcosalpha
FT=m2g
Acceleration ,a = g(ucosalpha- (m2/m1)-sin alpha)
a= (m2/m1) + sinalpha= 1.976
Cos alpha= 0.766
U2= 2.8
a= g(2.8×0.766)- 1.976)
a= g× 0.1688
a= 9.8× 0.1688
a=1.65m/s^2
Which of the following statements about diffusion is true?
"1. It requires integral proteins in the cell membrane. 2. It is very rapid over long distances. 3. It is a passive process in which molecules move from a region of higher concentration to a region of lower concentration. 4. It is an active process in which molecules move from a region of lower concentration to one of higher concentration. 5. It requires an expenditure of energy by the cell."
Answer:
3. It is a passive process in which molecules move from a region of higher concentration to a region of lower concentration.
Explanation:
Diffusion can be defined as the movement of solute or gaseous molecules from the region in which they have higher concentration to the regions in which they have lower concentration until they become evenly distributed across the two regions.
Diffusion is a passive process, that is, it requires no energy.
Option 3 is the correct option.
Answer:
It is a passive process in which molecules move from a region of higher concentration to a region of lower concentration
Explanation:
Diffusion is the movement of a substance from an area of high concentration to an area of low concentration. It is an important process for living things.
A single substance tends to move from an area of high concentration to an area of low concentration until the concentration is equal across a space
A Ferris wheel with a radius of 5 m is rotating at a rate of one revolution every 2 minutes. How fast is a rider rising when the rider is 9 m above ground level
Explanation:
Below is an attachment containing the solution.
The rider is rising at a speed of approximately 15.71 m/min when the rider is 9 m above ground level on a Ferris wheel with a radius of 5 m rotating at a rate of one revolution every 2 minutes.
Explanation:To find the speed at which the rider is rising, we can use the concept of angular velocity. The angular velocity of the Ferris wheel can be calculated by taking the circumference of the circle formed by the rider's position and dividing it by the time it takes to complete one revolution. In this case, the circumference is equal to 2π multiplied by the radius of the Ferris wheel. The time it takes to complete one revolution is given as 2 minutes.
The formula for angular velocity is ω = θ/t, where ω is the angular velocity, θ is the angle swept, and t is the time taken. Since one revolution is equal to 360 degrees or 2π radians, the angular velocity can be calculated as:
ω = (2π rad)/(2 min) = π rad/min
Now, to find how fast the rider is rising, we can use the relationship between linear velocity (v) and angular velocity (ω) given by v = rω, where r is the radius of the Ferris wheel. In this case, the radius is given as 5 m. Plugging in the values, the linear velocity is:
v = (5 m)(π rad/min) = 5π m/min ≈ 15.71 m/min
Therefore, when the rider is 9 m above ground level, the rider is rising at a speed of approximately 15.71 m/min.
Learn more about Angular velocity here:https://brainly.com/question/29557272
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Astronomers estimate that new stars form in our galaxy at the rate of about
Explanation:
The quantity at which the speed of formation of the star depends is called the "star formation rate". Astronomers estimate that in our Galaxy the star formation rate is about 3 solar masses per year According to "solar mass per year" the mass of interstellar gas and dust related to about 3 times the mass of the Sun goes into stars each year).However, the whole of the mass doesn't necessarily go in the formation of 1 star.Stars form in the Milky Way at a rate of about 1 solar mass per year, which means it would take a few hundred billion years for all gas to be turned into stars, far exceeding the universe's age of 14 billion years. Pulsars and supernova events influence this star-forming process and the interstellar medium.
Explanation:Astronomers have estimated that new stars form in our galaxy, the Milky Way, at a rate of about 1 solar mass per year. If we consider the amount of interstellar gas available to form new stars, and no new gas was added, we can calculate how long it would take for all the gas to be converted into stars. Given that the Milky Way contains about a few hundred billion solar masses of gas, it would take a few hundred billion years for all the interstellar gas to be used up at the current rate of star formation, which is significantly longer than the age of the universe (14 billion years).
Stars, including pulsars, form and die within the galaxy at varying rates. For example, one new pulsar is born approximately every 25 to 100 years, aligned with the rate of supernovae occurrences. Supernova explosions contribute to the recycling of interstellar material, affecting star formation and the interstellar medium.
A 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 K. It expands from initial volume 19 L to final volume V2. A total of 1.7 kJ of heat is added during the expansion process. What is V2? Let the ideal-gas constant R = 8.314 J/(mol • K).
32 L
41 L
26 L
35 L
Answer:
26 L
Explanation:
According to the first law of thermodynamics, for an ideal gas:
[tex]\Delta U=Q-W[/tex]
where
[tex]\Delta U[/tex] is the change in internal energy of the gas
Q is the heat absorbed by the gas
W is the work done by the gas
The internal energy of a gas depends only on its temperature. Here the temperature of the gas is kept constant (330 K), so the internal energy does not change, therefore
[tex]\Delta U=0[/tex]
So we have
[tex]Q=W[/tex]
The heat added to the gas is
[tex]Q=1.7 kJ = 1700 J[/tex]
So this is also equal to the work done by the gas:
[tex]W=1700 J[/tex]
For a process at constant temperature, the work done by the gas is given by
[tex]W=nRT ln\frac{V_2}{V_1}[/tex]
where:
n is the number of moles
R is the gas constant
T is the temperature of the gas
[tex]V_1[/tex] is the initial volume
[tex]V_2[/tex] is the final volume
In this problem, we have:
W = 1700 J is the work done by the gas
n = 2.00 mol
T = 300 K is the gas temperature
[tex]V_1=19 L[/tex] is the initial volume of the gas
And solving the equation for V2, we find the final volume of the gas:
[tex]V_2=V_1 e^{\frac{W}{nRT}}=(19)e^{\frac{1700}{(2.0)(8.314)(330)}}=26 L[/tex]
The greater the masses of two objects, the __ the resultant gravitational force; the greater the distance between the two objects, the ___the resultant gravitational force.
Answer:
The greater the masses of two objects, the greater the resultant gravitational force; the greater the distance between the two objects, the smaller the resultant gravitational force.
Explanation:
The gravitational force between two objects is an attractive force, whose magnitude is:
[tex]F=G\frac{m_1 m_2}{r^2}[/tex]
where
G is the gravitational constant
m1, m2 are the masses of the two objects
r is the distance between the objects
From the equation above, we observe that:
- The magnitude of the force is directly proportional to the product of the masses
- The magnitude of the force is inversely proportional to the square of the distance between the masses
Therefore, we can say that:
The greater the masses of two objects, the greater the resultant gravitational force; the greater the distance between the two objects, the smaller the resultant gravitational force.
Final answer:
The gravitational force between two objects increases with the masses and decreases with the distance between them. The correct terms for the blanks are 'stronger' for the effect of mass and 'weaker' for the effect of distance.
Explanation:
The question is related to the concept of gravitational force in physics. We know from Newton's law of universal gravitation that the gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This can be represented by the formula:
[tex]F_{gravity} = G \frac{ (M_1 M_2)}{R^2}[/tex],
where [tex]F_{gravity}[/tex] is the gravitational force, G is the gravitational constant, [tex]M_1[/tex] and [tex]M_2[/tex] are the masses of the two objects, and R is the distance between the centers of the two objects.
Thus, the sentence correctly filled out would be: The greater the masses of two objects, the stronger the resultant gravitational force; the greater the distance between the two objects, the weaker the resultant gravitational force.
The ball was kicked in the air and it iss about to hit the . if horizonta componenet of its final velocity is 10m/s and vertical component of its final velocity is -10m/s .what is the magnotide of th final vellocity of the ball?
Answer:
The magnitude of the final velocity is 14.14 m/s
Explanation:
The horizontal component of the final velocity and vertical component of the final velocity, forms a perpendicular vector. To determine the magnitude of the final velocity, we sum the two vectors.
To add two perpendicular vector, Pythagoras principle is used.
[tex]V^2 = V_x^2 +V_y^2\\\\V = \sqrt{V_x^2 +V_y^2}\\\\V = \sqrt{(10)^2 +(-10)^2} =14.14 m/s[/tex]
The magnitude of the final velocity is 14.14 m/s
Why do the pvc plastic wells used in this weeks elisa need to be sticky on the inner walls and how may your results change if a different non-adherent plastic was used instead?
Answer:
The stickiness in the inner walls allows them to be easily coated with the desired antigens, this translates in the use of a smaller amount of antigen. If the walls weren't sticky there's a possibility the antigen won't stick to them and therefore the result of the ELISA can be a false negative.
I hope you find this information useful and interesting! Good luck!