The new gravitation force at the new location is 40 N
Explanation:
The weight of the astronaut is given by the equation
[tex]F=mg[/tex] (1)
where
m is the mass of the astronaut
g is the acceleration of gravity
The acceleration of gravity at a certain distance [tex]r[/tex] from the centre of the Earth is given by
[tex]g=\frac{GM}{r^2}[/tex]
where G is the gravitational constant and M is the Earth's mass. So we can rewrite eq.(1) as
[tex]F=\frac{GMm}{r^2}[/tex]
When the astronaut is on the Earth's surface, [tex]r=R[/tex] (where R is the Earth's radius), so his weight is
[tex]F=\frac{GMm}{R^2}=640 N[/tex]
Later, he moves to another location where his distance from the Earth's surface is 3 times the previous distance, so the new distance from the Earth's centre is
[tex]r'=3R+R=4R[/tex]
Therefore, the new weight is
[tex]F'=\frac{GMm}{(4R)^2}=\frac{1}{16}\frac{GMm}{R^2}=\frac{F}{16}[/tex]
Which means that his weight has decreased by a factor 16: therefore, the new weight is
[tex]F'=\frac{640}{16}=40 N[/tex]
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Final answer:
The gravitational force on the astronaut 3 times farther from the Earth's surface than the radius would be 40 N, as at this distance the gravitational force is reduced to 1/16th of its original value.
Explanation:
The question asks: If an astronaut weighs 640 N on the earth's surface, what is the gravitation force between him and the earth if he is 3 times the distance from the Earth's surface? This requires understanding of Newton's law of universal gravitation, which can be expressed as F = G(m1m2)/r², where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers of mass.
As the astronaut moves to a distance 3 times farther from the surface of the Earth, the distance (r) from the center of the Earth becomes 4r, because the initial distance from the surface to the center of the Earth (the radius of the Earth, or 1r) is included. According to the inverse-square law, if the distance increases by a factor of n, the force decreases by a factor of n2. Thus, at 3 times the distance from the surface, or 4 times the radius of the Earth, the gravitational force becomes 1/16th of what it was at the surface.
Therefore, the new gravitational force is 640 N / 16 = 40 N.
A 88.00 kg man is swinging on a rope, and the rope has a length of 12.0 meters. The mans speed at the bottom of his swing is 10.20m/s. If the swing can be treated like an object moving in a vertical circle, what is the maximum amount of tension in the rope during the swing?
The maximum tension in the rope during the swing is the sum of the gravitational force and the centripetal force, which can be calculated with the given mass, rope length, and speed, resulting in a tension of 1602.48 N.
Explanation:To calculate the maximum tension in the rope during the swing, we need to consider the forces acting on the man at the lowest point of the swing. These forces are the gravitational force (weight) and the centripetal force required to maintain the circular motion. The tension in the rope is the sum of these two forces.
The gravitational force (Fg) on the man can be calculated using the equation Fg = m imes g, where m is the man's mass (88.00 kg) and g is the acceleration due to gravity (9.81 m/s2). The centripetal force (Fc) required for circular motion can be calculated using the equation Fc = m imes v2 / r, where v is the man's speed (10.20 m/s) at the bottom of the swing and r is the radius of the swing, which is equal to the length of the rope (12.0 meters).
Therefore, the maximum tension (T) in the rope would be:
T = Fg + Fc
T = (88.00 kg imes 9.81 m/s2) + (88.00 kg imes (10.20 m/s)2 / 12.0 m)
Calculating each part:
Fg = 863.28 N (2 decimal places)
Fc = 739.20 N (2 decimal places)
Thus, the maximum tension in the rope would be:
T = 863.28 N + 739.20 N
T = 1602.48 N (rounded to two decimal places)
Tension, centripetal force, and gravitational force are crucial components in determining the tension in the rope during circular motion.
Spaceship 1 and spaceship 2 have equal masses of 300 kg. spaceship 1 has a speed of 0 m/s, and spaceship 2 has a speed of 6 m/s. they collide and stick together. what is their speed?
In a collision between two 300 kg spaceships, one at rest and the other moving at 6 m/s, the final velocity after they stick together is 2 m/s.
When two objects with masses of 300 kg collide and stick together, their momentum is conserved. Utilizing the principle of conservation of momentum, we can calculate their final velocity after the collision.
In this case, since both spaceships have equal masses and one is at rest while the other moves with a known speed, the final velocity after the collision would be 2 m/s.
A level curve on a country road has a radius of 150 m. What is the maximum speed at which this curve can be safely negotiated on a rainy day when the coefficient of friction between the tires on a car and the road is 0.40?
The maximum speed of the car is 24.3 m/s
Explanation:
For a car moving along an unbanked turn, the frictional force provides the centripetal force required to keep the car in circular motion. Therefore, we can write:
[tex]\mu mg = m\frac{v^2}{r}[/tex]
where the term on the left is the frictional force while the term on the right is the centripetal force, and where
[tex]\mu=0.40[/tex] is the coefficient of friction between the tires and the road
m is the mass of the car
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
v is the speed of the car
r = 150 m is the radius of the curve
Solving for v, we find the (maximum) speed at which the car can move along the turn:
[tex]v=\sqrt{\mu gr}=\sqrt{(0.40)(9.8)(150)}=24.3 m/s[/tex]
For speed larger than this value, the frictional force is no longer enough to keep the car along the turn.
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Final answer:
On a rainy day, the maximum speed at which a curve with a radius of 150 m can be safely negotiated is 12.43 m/s. This calculation is based on the centripetal force provided by the friction between the tires and the road surface.
Explanation:
To determine the maximum speed at which the curve can be safely negotiated, we need to consider the centripetal force acting on the car. The centripetal force is provided by the friction between the tires and the road surface. The formula for centripetal force is Fc = mv²/r, where m is the mass of the car, v is the velocity, and r is the radius of the curve.
On a rainy day, the coefficient of friction between the tires and the road is given as 0.40. The maximum frictional force is given by the equation Ff = μN, where μ is the coefficient of friction and N is the normal force.
Since the normal force is equal to the gravitational force acting on the car, we can calculate it using N = mg, where m is the mass of the car and g is the acceleration due to gravity.
To find the maximum speed, we need to equate the centripetal force and the frictional force: mv²/r = μN. We can rearrange this equation to solve for the maximum velocity v: v = sqrt(μrg).
Substituting the given values, v = sqrt(0.40 * 9.8 * 150) = 12.43 m/s.
Which of the following is a characteristic of sexually active teens?
Group of answer choices
single and step-parent families
poor school performance
same-sex parents
A and B are correct
A, B, and C are correct
Out of all given options, poor school performance is a characteristic of sexually active teens.
Answer: Option B
Explanation:
Socio-demographic characteristics include age, gender and the highest level of education of the respondent. Most respondents had mothers with completed high school levels or diploma. Just over a quarter of respondents did not know their mother's highest education. A low level of education and some family situations made the teen to be sexually active.
Teen sexual activity is "a fact in most African countries, as in most Western societies." There is a relationship between the sexual activity of young girls and their school resignation. Therefore, additional consequences of early sexual intercourse are women (and sometimes men) who drop out of school when they become pregnant.
Sexually active teens are more likely to come from single and step-parent families and have poor school performance.
Explanation:The characteristic of sexually active teens is A and B are correct, which means they are more likely to come from single and step-parent families and have poor school performance.
While there may be studies suggesting correlations between certain factors, making broad statements about the behavior of sexually active teens based solely on family structure and school performance can be misleading.
Teenage sexual behavior is influenced by a variety of factors, including individual, family, cultural, and socioeconomic aspects. Family structure alone does not determine a teen's likelihood of engaging in sexual activity. Additionally, attributing poor school performance solely to family structure oversimplifies the complex interplay of factors influencing academic achievement.
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2. (06.09A) Which expression is equivalent to one over five m − 20? (5 points) one over five (m − 4) one over five (m − 100) 5(m − 4) 5(m − 100)
Answer:
A
Explanation:
Answer:
a
Explanation:
a
Which would be the easiest way to move a heavy box up 4 feet? To lift it up. To slide it up a 8 foot ramp. To slide it up a 4 foot ramp. To slide it up a 2 foot ramp.w
A motorbike is travelling with a velocity of 3m/s. It accelerates at a rate of 9.3m/s for 1.8s. Calculate the distance it travels in this time
Explanation:
Consider the kinematic equation,
[tex]x = vt + \frac{1}{2} a {t}^{2} [/tex]
where x is the distance traveled, v is the initial velocity, a is the acceleration and t is time. By plugging in known values and solving for x,
[tex]x = 3(1.8) + \frac{1}{2} (9.3) {1.8}^{2} [/tex]
through simple algebra we get
[tex]x = 20.466[/tex]
where this is the distance traveled in meters.
A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.
a) Find the acceleration of gravity (g) at 2.60 x 10^6 m above the Earth.
b) Find the critical velocity to maintain a circular orbit at 2.60 x 10^6 m above the surface of Earth. (in both m/sec & km/hr)
c) Find the period of the orbit.
d) How many orbits will the satellite make in one day?
e) Find the escape velocity for a satellite to leave this orbit and escape the gravitational pull of the Earth.
f) Find the escape velocity for a rocket leaving the surface of the Earth.
a) The acceleration of gravity is [tex]4.96 m/s^2[/tex]
b) The critical velocity is 6668 m/s (24,006 km/h)
c) The period of the orbit is 8452 s
d) The satellite completes 10.2 orbits per day
e) The escape velocity of the satellite is 9430 m/s
f) The escape velocity of the rocket is 11,191 m/s
Explanation:
a)
The acceleration of gravity for an object near a planet is given by
[tex]g=\frac{GM}{(R+h)^2}[/tex]
where
G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
h is the height above the surface
In this problem,
[tex]M=5.98\cdot 10^{24} kg[/tex] (mass of the Earth)
[tex]R=6.37\cdot 10^6 m[/tex] (Earth's radius)
[tex]h=2.60\cdot 10^6 m[/tex] (altitude of the satellite)
Substituting,
[tex]g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2[/tex]
b)
The critical velocity for a satellite orbiting around a planet is given by
[tex]v=\sqrt{\frac{GM}{R+h}}[/tex]
where we have again:
[tex]M=5.98\cdot 10^{24} kg[/tex] (mass of the Earth)
[tex]R=6.37\cdot 10^6 m[/tex] (Earth's radius)
[tex]h=2.60\cdot 10^6 m[/tex] (altitude of the satellite)
Substituting,
[tex]v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s[/tex]
Converting into km/h,
[tex]v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h[/tex]
c)
The period of the orbit is given by the circumference of the orbit divided by the velocity:
[tex]T=\frac{2\pi (R+h)}{v}[/tex]
where
[tex]R=6.37\cdot 10^6 m[/tex]
[tex]h=2.60\cdot 10^6 m[/tex]
v = 6668 m/s
Substituting,
[tex]T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s[/tex]
d)
One day consists of:
[tex]t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s[/tex]
While the period of the orbit is
T = 8452 s
So, the number of orbits completed by the satellite in one day is
[tex]n=\frac{t}{T}=\frac{86400}{8452}=10.2[/tex]
e)
The escape velocity for an object in the gravitational field of a planet is given by
[tex]v=\sqrt{\frac{2GM}{R+h}}[/tex]
where here we have:
[tex]M=5.98\cdot 10^{24} kg[/tex]
[tex]R=6.37\cdot 10^6 m[/tex]
[tex]h=2.60\cdot 10^6 m[/tex]
Substituting, we find
[tex]v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s[/tex]
f)
We can apply again the formula to find the escape velocity for the rocket:
[tex]v=\sqrt{\frac{2GM}{R+h}}[/tex]
Where this time we have:
[tex]M=5.98\cdot 10^{24} kg[/tex]
[tex]R=6.37\cdot 10^6 m[/tex]
[tex]h=0[/tex], because the rocket is located at the Earth's surface, so its altitude is zero.
And substituting,
[tex]v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s[/tex]
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The angle of incidence and the angle of reflection are measured from the ____, the perpendicular to the surface.
A. planar
B. normal
The angle of incidence and the angle of reflection are measured from the B) normal to the surface.
Explanation:
Reflection is a phenomenon that occurs when a wave hits a boundary between two mediums and bounces back into the original medium.
The direction of the wave after reflection changes according to the law of reflection, which states that:
The incident ray, the reflected ray and the normal to the surface all lie in the same planeThe angle of reflection is equal to the angle of incidence ([tex]\theta_r = \theta_i[/tex]In this framework, the two angles are defined as follows:
the angle of incidence is the angle between the incident ray and the normal to the surfacethe angle of reflection is the angle between the reflected ray and the normal to the surfaceTherefore, the complete sentence is
The angle of incidence and the angle of reflection are measured from the B) normal to the surface.
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Answer: The answer is B.
Explanation: i got it correct on Usatestprep.
10. Which of the following is an acceleration?
a. 12 m/s2 down
b. 5 m/s up
c. 8N West
Answer:
a. 12 m/s² down
Explanation:
Acceleration has units of length per time squared. Acceleration is a vector, so it also has a direction.
What are possible units for impulse? Check all that apply.
Okg•m
Okg.
ONós
ON•m
O Nom
The units for impulse are
[tex]N\cdot s[/tex]
Explanation:
We have two definition for impulse:
1) Impulse is defined as the product between the force exerted on an object and the time interval during which the force is applied:
[tex]I=F \Delta t[/tex]
where
F is the force, measured in Newtons (N)
[tex]\Delta t[/tex] is the time interval, measured in seconds (s)
From this definition it follows that Impulse can be measured in [tex]N\cdot s[/tex], which corresponds to option 3)
2)
Impulse is also defined as the change in momentum of an object:
[tex]I=\Delta p[/tex]
where
[tex]\Delta p[/tex] is the change in momentum of the object
The change in momentum has the same units of the change in momentum, so [tex]kg \cdot m \cdot s^{-1}[/tex], which corresponds to none of the options given
So the only correct option is
[tex]N\cdot s[/tex]
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Which one of the above forms of em radiation does our body detect as heat?
Answer: Infrared radiation
Explanation:
Infrared radiation is not visible to the human eye, since its wavelengths are outside the visible spectrum (between 700 nm and 1 mm), however this type of electromagnetic radiation is detected by the human body as heat.
These waves can be divided into:
- Near infrared or long wave infrared: it is the least sensitive to color and is easily absorbed by water.
- Medium or medium wave infrared: it is also insensitive to color and easily absorbed by water and many types of plastics and paints.
- Far infrared or short wave infrared: it is more penetrating than the long wave and is good for heating metals, these waves also can pass through clear materials. This light has many uses, including heating lamps in physiotherapy and medical treatments, heat sensing devices, among others.
What assumption do error theorist claims at the heart of morality
Answer:
a) The assumption that error theorists claim is at the heart of morality is that sincere moral judgments are always in error, therefore there is no moral knowledge possible.
Explanation:
To sterilize a 53.0 g glass baby bottle, we must raise its temperature from 25.0°C to 92.0°C. How much heat transfer (in J) is required?
The heat energy required is 2983 J
Explanation:
When an amount of energy Q is supplied to a certain sample of substance of mass m, the temperature of the substance increases by an amount of [tex]\Delta T[/tex] according to the equation :
[tex]Q=mC_s \Delta T[/tex]
where
Q is the heat supplied
m is the mass of the object
[tex]C_s[/tex] is the specific heat capacity of the substance
[tex]\Delta T[/tex] is the change in temperature
For the glass bottle in this problem:
[tex]m=53.0 g[/tex]
[tex]\Delta T = 92.0 - 25.0 =67.0^{\circ}C[/tex] is the change in temperature
[tex]C_s = 0.84 J/g^{\circ}C[/tex] is the specific heat of glass
Substituting into the equation, we find
[tex]Q=(53.0)(0.84)(67)=2983 J[/tex]
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What factors limit the accuracy of a potentiometer?
Answer:
Whats a potentiometer.
Explanation:
Some of the factors that can limit the accuracy of a potentiometer are internal temperature and Resistance change in the resistor
A potentiometer is a variable resistor that can be adjusted manually and has three terminals. A potentiometer is also called a pot
Further ExplanationPotentiometers have different shapes and they are used in different in different application. A potentiometer can be used to control the volume of a radio. There are several types of a potentiometer.
some of the common ones include
Single-turn pot: it is a single rotation of about 270 degreesMulti-turn pot: this type of pot is used where high precision and resolution are needed. Dual gang pot: this type of pot is used in audio volume control.Concentric pot: this type of pot is mostly used in older car radios, particularly in radios that have its volume and the tone control combinedServo pot: this type of pot is used where manual and automatic adjustment is needed.However, some of the materials that can be used to construct potentiometers include
Metal firmCarbon compositionWirewoundCermetAs stated earlier, the two factors that can limit the accuracy of a potentiometer are:
Internal temperatureResistance change in the resistorLEARN MORE:
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potentiometeraccuracyfactorslimittypes of potentiometerWhen reporting the speed, we need to include the value, units and direction the object was traveling.
True or False??
Answer:
When reporting the speed, we need to include the value, units and direction the object was traveling.
It is a FALSE
Explanation:
Justification :
Speed is a scalar quantity.Scalar quantity required only magnitude not the direction.The definition of speed is total distance travelled divided by total time taken to cover that distance.Therefore, speed required value and units not the direction.By velocity which is similar to speed required value, units and direction. And velocity is a vector quantity hence it require direction.For example without speed of a car is 50 m/s. Here 50 is the value and m/s is the unit of a speed.You pull a suitcase along the floor by exerting 43N at an angle. The force of friction is 27 N and the suitcase moves at a constant speed. What angle does the handle make with the horizontal?
The angle of the force is [tex]51.1^{\circ}[/tex]
Explanation:
To solve this problem, we can apply Newton's second law along the horizontal direction of motion of the suitcase:
[tex]\sum F_x = ma_x[/tex]
where
[tex]\sum F_x[/tex] is the net force along the x-axis
m is the mass of the suitcase
[tex]a_x[/tex] is the acceleration along the x-axis
The suitcase is moving at constant speed, so the acceleration is zero:
[tex]a_x=0[/tex]
Therefore the net force must also be zero:
[tex]\sum F_x = 0[/tex] (1)
We have two forces acting along the horizontal direction:
- The component of the push (forward) in the horizontal direction, [tex]F cos \theta[/tex], with
F = 43 N
[tex]\theta[/tex] = angle of the force with the horizontal
- The force of friction, [tex]F_f = 27 N[/tex], backward
So the net force can be written as
[tex]\sum F_x = F cos \theta - F_f[/tex] (2)
Combining (1) and (2),
[tex]F cos \theta - F_f = 0[/tex]
And so we can find the angle:
[tex]\theta = cos^{-1}(\frac{F_f}{F})=cos^{-1}(\frac{27}{43})=51.1^{\circ}[/tex]
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13. If both the directions of velocity and acceleration are negative, ...........
a velocity of the object increases
6 velocity of the object decreases
velocity of object is constant
the object stops motion .. Explain please
Answer:
B)velocity of object decreases
Explanation:
Consider the positive x axis as positive direction
Assume a body moving in negative x-axis direction
It's acceleration also alone negative x-axis direction
So according to our consideration
velocity and acceleration values are negative
That is both are towards negative x direction
But as both velocity and acceleration are in same direction, MAGNITUDE of velocity increases
But as magnitude increases in negative direction, velocity value decreases
But speed value increases(As speed is scalar and velocity is a vector)
3. A roller coaster car at the top of a hill, right before it plunges downward
has
a. potential energy
b. kinetic energy.
c. momentum
d. electrical energy.
The correct answer is a. potential energy. At the top of a hill, right before it plunges downward, a roller coaster car has potential energy due to its position relative to the ground. This potential energy is converted into kinetic energy as the car moves downward and gains speed.
a .101 kg particle undergoes simple harmonic motion along the horizontal x axis between the points x1 = - .219 m and x2 = .495 m. the period of oscillation is .513 s find the frequency, f, the equilibrium position, xep, the amplitude, a, the maximum speed, v max, the maximum magnitude of acceleration, a max, the force constant, k, and the total mechanicalenergy, e tot.
1) Frequency: 1.95 Hz
2) Equilibrium position: 0.138 m
3) Amplitude: 0.357 m
4) Maximum speed: 4.36 m/s
5) Maximum acceleration: [tex]53.1 m/s^2[/tex]
6) Spring constant: 15.0 N/m
7) Total mechanical energy: 0.96 J
Explanation:
1)
The frequency of a simple harmonic motion is equal to the reciprocal of the period:
[tex]f=\frac{1}{T}[/tex]
where
f is the frequency
T is the period
For the particle in this problem, we have
T = 0.513 s (period)
So the frequency of motion is
[tex]f=\frac{1}{0.513}=1.95 Hz[/tex]
2)
In a simple harmonic motion, the object oscillates between two maximum positions [tex]+x_m[/tex] and [tex]-x_m[/tex] which are equidistance from the equilibrium position. So, the equilibrium position is the midpoint between these two positions.
For the particle in this problem, the two extreme positions are:
[tex]x_1 = -0.219 m[/tex]
[tex]x_2 = 0.495 m[/tex]
So the mid-point (the equilibrium position) is
[tex]x_{ep} = \frac{x_1 + x_2}{2}=\frac{-0.219+0.495}{2}=0.138 m[/tex]
3)
The amplitude of a simple harmonic motion is the maximum displacement of the object, measured from the equilibrium position.
This means that we can calculate the amplitude simply as the difference between one of the extreme positions and the equilibrium position.
Taking
[tex]x_2 = 0.495 m[/tex]
and
[tex]x_{ep} = 0.138 m[/tex]
We find the amplitude:
[tex]A=x_2 - x_{ep} = 0.495-0.138 =0.357 m[/tex]
4)
In a simple harmonic motion, the maximum speed is given by
[tex]v_{max}=\omega A[/tex]
where
[tex]\omega[/tex] is the angular frequency
A is the amplitude
The angular speed can be calculated from the frequency as follows:
[tex]\omega=2\pi f=2 \pi (1.95 Hz)=12.2 rad/s[/tex]
The amplitude is
A = 0.357 m
So, the maximum speed is
[tex]v_{max} = (12.2)(0.357)=4.36 m/s[/tex]
5)
The maximum acceleration in a simple harmonic motion is given by
[tex]a_{max}= \omega^2 A[/tex]
Where we already know that:
[tex]\omega=12.2 rad/s[/tex] is the angular frequency
A = 0.357 m is the amplitude of motion
Substituting, we find the maximum acceleration:
[tex]a_{max}=(12.2)^2(0.357)=53.1 m/s^2[/tex]
6)
The angular speed in a simple harmonic motion can be written as
[tex]\omega=\sqrt{\frac{k}{m}}[/tex]
where
k is the spring constant
m is the mass
Here we know that:
[tex]\omega=12.2 rad/s[/tex] is the angular speed
m = 0.101 kg is the mass of the particle
So we can solve the formula for k, the spring constant:
[tex]k=\omega^2 m =(12.2)^2(0.101)=15.0 N/m[/tex]
7)
Since the energy is conserved, the total mechanical mechanical energy is just equal to the maximum potential energy of the system, which occurs when the particle is at maximum displacement (x=A) and the speed is zero (so the kinetic energy is zero), therefore it is given by:
[tex]E=\frac{1}{2}kA^2[/tex]
where
k = 15.0 N/m is the spring constant
A = 0.357 m is the amplitude
And substituting,
[tex]E=\frac{1}{2}(15.0)(0.357)^2=0.96 J[/tex]
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Which type of committee has members from both the House and the Senate?
Answer:the answe would be joint comitee
Explanation:
A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long as 2.20 m have been recorded. If a frog jumps 2.20 m and the launch angle is 54.0°, find the frog's launch speed and the time the frog spends in the air. Ignore air resistance.
(a) the frog's launch speed (in m/s)
(b)the time the frog spends in the air (in s)
Answer:
[tex]v_0[/tex]=4.761 m/s
t=0.786 sec
Explanation:
In a projectile motion (or 2D motion), the object is launched with an initial angle [tex]\theta[/tex] and an initial velocity [tex]v_0[/tex]
The components of the velocity are
[tex]v_{ox}=v_ocos\theta[/tex]
[tex]v_{oy}=v_osin\theta[/tex]
Similarly the displacement has the components
[tex]x=v_{ox}.t=v_ocos\theta.t[/tex]
[tex]y=v_osin\theta.t-\frac{gt^2}{2}[/tex]
The last formula is valid only if the object is launched at ground level, as our frog does.
There are two times where the value of y is zero, when t=0 (at launching time) and when it lands back from the air. We need to find that time t by making y=0
[tex]0=v_osin\theta.t-\frac{gt^2}{2}[/tex]
Dividing by t (assuming t different from zero)
[tex]0=v_osin\theta-\frac{gt}{2}[/tex]
Then we find the total flight as
[tex]t=\frac{2v_osin\theta}{g}[/tex]
Replacing this time in the formula of x
[tex]x=v_ocos\theta\frac{2v_osin\theta}{g}[/tex]
We can solve for [tex]v_o[/tex]
[tex]\displaystyle v_0=\sqrt{\frac{xg}{sin2\theta}}[/tex]
Knowing that x=2.20 m and [tex]\theta=54[/tex]°
[tex]\displaystyle v_0=\sqrt{\frac{xg}{sin2\theta}}=4.761m/s[/tex]
We now compute t
[tex]t=\frac{2v_osin\theta}{g}=0.786\ sec[/tex]
Final answer:
The launch speed of the frog is 4.81 m/s, and the time the frog spends in the air is 0.737 seconds.
Explanation:
A) Using the kinematic equation for projectile motion, we can find the launch speed. The launch speed, in this case, would be 4.81 m/s.
B) To determine the time the frog spends in the air, we can use the time of flight formula for projectile motion. The time in the air for the frog is 0.737 seconds.
if f(x)=4/x+2 and g is the inverse of f,then g'(10)=
Answer:
g'(10) = [tex]\frac{-1}{16}[/tex]
Explanation:
Since g is the inverse of f ,
We can write
g(f(x)) = x (Identity)
Differentiating both sides of the equation we get,
g'(f(x)).f'(x) = 1
g'(10) = [tex]\frac{1}{f'(x)}[/tex] --equation[1] Where f(x) = 10
Now, we have to find x when f(x) = 10
Thus 10 = [tex]\frac{4}{x}[/tex] + 2
[tex]\frac{4}{x}[/tex] = 8
x = [tex]\frac{1}{2}[/tex]
Since f(x) = [tex]\frac{4}{x}[/tex] + 2
f'(x) = -[tex]\frac{4}{x^{2} }[/tex]
f'([tex]\frac{1}{2}[/tex]) = -4 × 4 = -16
Putting it in equation 1, we get:
We get g'(10) = -[tex]\frac{1}{16}[/tex]
The speed in case a is twice that of case b. The circles radius in case a is two times that of case b. How does the acceleration in case a compare to the acceleration in case b?
The acceleration of an object in circular motion is calculated by the formula ac = v²/r. Given that in case a, both the speed and radius are twice that of case b, the acceleration in case a is twice that of case b. This shows the squared relationship between speed and acceleration in circular motion.
Explanation:In physics, the acceleration of an object moving in a circular path, also known as centripetal acceleration, is calculated by the formula ac = v²/r where v is the velocity or speed of the object and r is the radius of the circle.
In case a, the speed is twice that of case b and the circle's radius is also two times of case b. If we plug these values into our formula, we can see that the acceleration in case a (let's call it acA) could be represented as acA = (2v)² / 2r = 4v²/2r = 2v²/r. In case b, the acceleration (acB) would be represented as acB = v² / r.
Hence, comparing both, you can see the acceleration in case a is twice that in case b. Therefore, even if the speed and the radius are increased by the same factor, the acceleration increases by a factor of that increase due to the squared relationship between velocity and acceleration in circular motion.
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The acceleration in case A is twice that of case B.Given the conditions that the speed and radius of circle in case A are twice those in case B.
Let's analyze the relationship between acceleration, speed, and radius in circular motion. We know the acceleration of an object in uniform circular motion can be given by the equation:
|a| = |v|²/r
From the problem, we have two cases:
[tex]Case A: Speed (|v_A|) is \twice \ that \ of \ case \ B (|v_B|).\\\\Case A: Radius (|r_A|) \ is \ also \ two\ times \ that\ of \ case\ B (|r_B|).[/tex]
Let's denote the acceleration in case A as [tex]|a_A|[/tex] and in case B as [tex]|a_B|[/tex]
For case A:
[tex]|a_A| = |v_A|^2 / |r_A|[/tex]
For case B:
[tex]|a_B| = |v_B|^2 / |r_B|[/tex]
Given that [tex]|v_A| = 2|v_B|[/tex] and [tex]|r_A| = 2|r_B|[/tex], substituting these into the equations gives:
[tex]|a_A| = (2|v_B|)^2 / (2|r_B|)[/tex]
Simplifying this, we get:
[tex]|a_A| = 4|v_B|^2 / 2|r_B| = 2(|v_B|^2 / |r_B|) = 2|a_B|[/tex]
Therefore, the acceleration in case A is twice that of case B.
a 3,000 kg car is traveling with a constant velocity of 10 m/s. how much work must be applied to the car to change its speed to 15 m/s?
187500 Joules of work must be done to the 3000 kg car to change its speed from 10 m/s to 15 m/s.
Explanation:
Work is the measurement of the force on an object that overcomes a resisting force (such as friction or gravity) times the distance the object is moved. If there is no distance, there is no work, no matter what the effort.
Work done = Force * displacement
When you apply enough force on an object to overcome a resistive force, such that you move that object, you are doing work on that object. There is a relationship between that work and mechanical energy.
When an object is accelerated, you are doing work against inertia, such that the work equals the change in kinetic energy of the object.
Work done = Change in K.E
In the given case:
mass = m = 3000 kg
initial velocity = v₁ = 10 m/s
final velocity = v₂ = 10 m/s
Work = W = (1/2)m(v₂)² - (1/2)m(v₁)²
[tex]= \frac{1}{2}m(v2^{2} - v1^2)\\= \frac{3000}{2}(15^{2} - 10^{2})\\= 1500(225 - 100)\\= 1500*125\\= 187500 kg m^{2}s^{-2} \\= 187500 Nm\\= 187500 J[/tex]
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A force F~ = Fx ˆı + Fy ˆ acts on a particle that
undergoes a displacement of ~s = sx ˆı + sy ˆ
where Fx = 8 N, Fy = −1 N, sx = 6 m, and
sy = 2 m.
Find the work done by the force on the
particle.
Answer in units of J.
Answer:
W = 46 J
Explanation:
We need to find the angle between the two vectors Force vector and displacement vector.
First we will find the angle α of the force vector
[tex]tan\alpha =\frac{1}{8} \\\\\\alpha =7.125 deg\\[/tex]
Then we find the angle β of the displacement vector
[tex]tan\beta=\frac{2}{6} \\\\beta = 18.43 deg\\[/tex]
With these two angles we can find the angle between the two vectors
∅ = α + β = 25.56 deg
The definition of work is given by the expression
[tex]W=F*d*cos (theta)[/tex]
The absolute value of F will be:
[tex]F=\sqrt{8^{2}+1^{2} } \\F= 8.06 N[/tex]
The absolute value of d will be:
[tex]d=\sqrt{(6 )^{2}+(2)^{2} } \\d= 6.32m\\[/tex]
Now we have:
[tex]W=8.06*6.32*cos(25.56)\\W=46 J[/tex]
Pierce conducts an experiment in which waves collide in a way that the energy increases, which means that has occurred.
It means Constructive interference has occurred.
Explanation:
Waves travelling from opposite sides can collide to results into constructive interference or destructive interference.
In constructive interference, the crust of the waves merge.This results into increase in energy.In destructive interference the two waves cancel each other to cause loss of energy.
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Answer:
Constructive Interference
Explanation:
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An ocean wave has a wavelength of 10 m and a frequency of 4.0 Hz. What is the velocity of the
wave? Show the appropriate equation from your book and show your work with units.
The velocity of the wave is 40 m/s
Explanation:
The relationship between the velocity of a wave and its frequency and wavelength is given by
[tex]v=f \lambda[/tex]
where
v is the velocity
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the ocean wave in this problem, we have:
[tex]\lambda = 10 m[/tex] (wavelength)
f = 4.0 Hz (frequency)
Therefore, its velocity is
[tex]v=(4.0)(10)=40 m/s[/tex]
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The velocity of an ocean wave with a wavelength of 10 m and a frequency of 4.0 Hz is 40 m/s.
The subject of this question is Physics, and more specifically, it deals with the topic of wave motion and wave velocity. The velocity of a wave can be calculated using the equation v = f × lambda, where v is the velocity of the wave, f is its frequency, and lambda is its wavelength.
Given that the wavelength lambda is 10 m and the frequency f is 4.0 Hz, we can substitute these values into the equation to find the wave velocity:
v = f × lambda = 4.0 Hz × 10 m = 40 m/s.
Therefore, the velocity of the ocean wave is 40 meters per second (m/s).
An unknown source plays a pitch of middle C (262 Hz). How fast would the sound wave from this source have to travel to raise
the pitch to C sharp (271 Hz)? Use 343 m/s as the speed of sound.
a. 2.8 m/s
c. 11.4 m/s
b. 15.6 m/s
d. 8.2 m/s
Answer:
The sound travels at [tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]
Option: c
Explanation:
Unknown source plays of middle C (fs) = 262 Hz
The sound wave from this source have to travel to raise the pitch to C sharp is (fd) = 272 Hz
[tex]\begin{array}{l}{velocity of sound in air(v)=343 \mathrm{m} / \mathrm{s}, f_{\mathrm{s}}=262 \mathrm{Hz}, f_{\mathrm{d}}=271 \mathrm{Hz}} \\ {velocity of receiver(v_{\mathrm{d}})=0 \mathrm{m} / \mathrm{s},velocity of source( v_{\mathrm{s}}) \text { is unknown }}\end{array}[/tex]
[tex]\text { Speed of sound } \mathrm{V}_{\mathrm{S}}=343 \mathrm{m} / \mathrm{s}[/tex]
[tex]f_{\mathrm{d}}=f_{\mathrm{s}}\left(\frac{v-v_{\mathrm{d}}}{v-v_{\mathrm{s}}}\right)[/tex]
[tex]\frac{f_{d}}{f_{s}}=\left(\frac{v-v_{d}}{v-v_{s}}\right)[/tex]
[tex]\left(v-v_{s}\right)=\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)[/tex]
[tex]v_{s}=v-\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)[/tex]
Substitute the given values in the formula,
[tex]v_{s}=343+\frac{262}{271}(343-0)[/tex]
[tex]v_{s}=343+0.966(343)[/tex]
[tex]v_{s}=343-331.33[/tex]
[tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]
Therefore, The sound travels at [tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]
Assume that a hockey player skates 60 feet in 3 seconds. What is the player’s speed? If the hockey player travels this distance from East to West, what is his velocity?
Answer:
20ft/secs
Explanation:
60 feet/ 3 secs