Answer:
The time is [tex]t_t = 3.7583 \ s [/tex]
Explanation:
From the question we are told that
The angle is [tex]\theta = 30^o[/tex]
The horizontal distance is [tex]d = 120 \ ft[/tex]
Generally when the ball is at maximum height before descending the velocity is zero and this velocity can be mathematically represented as
[tex]v = u + at[/tex]
here a = -g the negative sign is because the direction of motion is against gravity
So
[tex]v = v_i + at[/tex]
Here[tex] v_i [/tex] is the vertical component of the initial velocity of the ball which is mathematically
represented as
[tex]v_i = usin(\theta )[/tex]
So
=> [tex]0 = usin(\theta ) -9.8t[/tex]
Generally the total time taken to travel the 120 ft is mathematically represented as
[tex]t_t = \frac{120}{v_h}[/tex]
Here [tex]v_h[/tex] is the horizontal component of the initial velocity which is mathematically represented as
[tex]v_h = u cos(\theta )[/tex]
So
[tex]t_t = \frac{120}{ u cos(\theta )}[/tex]
Generally the time taken to reach the maximum height is
[tex]t = \frac{t_t}{2}[/tex]
=> [tex]t = \frac{120}{ u cos(\theta )} * \frac{1}{2} [/tex]
=> [tex]t = \frac{60}{ u cos(\theta )} [/tex]
So
[tex]0 = usin(\theta ) -9.8* [\frac{60}{ u cos(\theta )}][/tex]
[tex] usin(\theta ) = 9.8* [\frac{60}{ u cos(\theta )}][/tex]
[tex] usin(\theta ) * u cos(\theta) =60* 9.8 [/tex]
[tex] u^2 sin(\theta ) cos(\theta) =60* 9.8 [/tex]
[tex] u^2 sin(30 ) cos(30) =60* 9.8 [/tex]
[tex] u^2 * \frac{1}{2}* \frac{\sqrt{3}}{2} =588.6 [/tex]
[tex] u^2 *\sqrt{3} =2354.4 [/tex]
[tex] u^2 = 1359.31 [/tex]
[tex] u = 36.87 \ ft/s [/tex]
Substituting this value into the equation for total time
[tex]t_t = \frac{120}{36.87 cos(30 )}[/tex]
[tex]t_t = 3.7583 \ s [/tex]
How much work is required to move it at constant speed 5.0 m along the floor against a friction force of 290 N?
Answer:
The answer is 1450 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distanceFrom the question
force = 290 N
distance = 5 m
We have
workdone = 290 × 5
We have the final answer as
1450 JHope this helps you
Find the work done by a 75.0 kg person in climbing a 2.50 m flight of stairs at a constant speed.
Answer:
1,839.375 Joules
Explanation:
Work is said to be done is the force applied to an object cause the object to move through a distance.
Workdone = Force * Distance
Workdone = mass * acceleration due to gravity * distance
Given
Mass = 75.0kg
acceleration due to gravity = 9.81m/s²
distance = 2.50m
Substitute the given parameters into the formula:
Workdone = 75.0*9.81*2.50
Workdone = 1,839.375Joules
Hence the workdone is 1,839.375 Joules
A 126 N force is applied at an angle of 25.00 to a 8.50 kg block pressed against a rough vertical wall and the block slides down the wall at constant velocity. Calculate the coefficient of kinetic friction between the block and the wall.
Answer:
μk = 0.58
Explanation:
If the block is sliding down at constant speed, this means that no net force is acting upon it in the vertical direction.As the block is pressed on the wall, this means that it doesnt accelerate in the horizontal direction either, so no net force acts upon it in this direction also.In this direction, we have only two forces acting, equal and opposite each other, one is the normal force (exerted by the wall) and the other is the horizontal component of the applied force.If the applied force forms an angle of 25º with the wall (which is vertical), this means that we can get its projection along the horizontal direction, using simple trigonometry , as follows:[tex]F_{apph} = F_{app} * sin\theta = 126 N * sin 25 = 53.3 N[/tex]
⇒ [tex]F_{n} = - F_{apph} = -53.3 N[/tex]
In the vertical direction, we have three forces acting on the block: the weight pointing downward, the kinetic friction force (as we know that the block is sliding), and the vertical component of the applied force, in the same direction as the friction one.As we have already said, the sum of these forces must be 0.[tex]F_{g} + F_{appv} + F_{ff} = 0 (1)[/tex] where Fg is the weight of the block, Fappv is the vertical component of the applied force, and Fff is the kinetic friction force.Replacing these forces by their mathematical expressions, we have:[tex]F_{g} = m_{b} * g = 8.5 Kg * (-9.8 m/s2) = -83.3 N[/tex]
[tex]F_{appv} = F_{app}* cos\theta = 126 N * cos 25 = 114.2 N[/tex]
[tex]F_{ff} = \mu_{k}* F_{n} =\mu_{k} * (-53.3 N)[/tex]
Replacing in (1), and solving for μk, we finally get:μk = 0.58
my heart strike him to dead.what figure of speech is that?
Answer:
Hyperbole
Explanation:
this is an extreme exaggeration or overstatement/ magnification
Given that water at standard pressure freezes at 0∘C, which corresponds to 32∘F, and that it boils at 100∘C, which corresponds to 212∘F, calculate the temperature difference ΔT in degrees Fahrenheit that corresponds to a temperature difference of 1 K on the Kelvin scale. Give your answer to two significant figures.
Answer:
In two significant figure 360K
Explanation:
The temperature difference (ΔT) can be calculated as the boiling temperature minus the freezing temperature in Fahrenheit.
Hence,
ΔT = 212 - 32
ΔT = 180°F
To convert to °F to kelvin, we use the formula below
= (°F - 32) × 5/9 + 273.15
= (180°F - 32) × 5/9 + 273.15
= 355.37K ⇔ 360K
Time it takes stone to fall from the height of 80 m is approximately equal to *
A. 1 s
B. 2 s
C. 4 s
D. 8 s
Answer:
D
Explanation:
Answer:
c.4s
Explanation:
If two identical objects are dropped one after 1 s delay with respect to another then in the absence of the air resistance *
A. distance between two falling objects will keep increasing
B. distance between two falling objects will keep decreasing
C. will stay the same
D. All of the above
Answer:
A.No. Assuming no other factors (such as air resistance) the first object will have a velocity of 32 feet/second when the other is dropped. Since they will both have the same acceleration, the first distance between them will increase by 32 feet per second.
Explanation:
Metals that have shine and luster?
Answer:
luster
Explanation:
A cart with an unknown mass is at rest on one side of a track. A student must find the mass of the cart by using Newton’s second law. The student attaches a force probe to the cart and pulls it while keeping the force constant. A motion detector rests on the opposite end of the track to record the acceleration of the cart as it is pulled. The student uses the measured force and acceleration values and determines that the cart’s mass is 0.4kg . When placed on a balance, the cart’s mass is found to be 0.5kg . Which of the following could explain the difference in mass?
Answer choices:
A) The track was not level and was tilted slightly downward.
B) The student did not pull the cart with a force parallel to the track.
C) The wheels contain bearings that were rough and caused a significant amount of friction.
D) The motion sensor setting was incorrect. The student set it up so that motion away from the sensor would be the negative direction.
Answer: The correct answer is A) The track was not level and was tilted slightly downward.
Explanation: This is because of the two values: 0.4 kg and 0.5 kg. I won't go into much detail but due to this difference of mass, we know that the track was not level.
"The track was not level and was tilted slightly downward" could explain the difference in the mass.
Mostly because the university student or learners calculates a mass of just over the spring quantity, the vehicle speed seems to have been higher than there would have had to be.Option B, as well as Option C, are wrong because the acceleration would've been smaller in each of these 2 circumstances, so that computed mass would've been larger.
Thus Option A is appropriate.
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6) The magnitude of the force the Sun exerts on Uranus is 1.41 x 1021 newtons. Explain how it is possible for the Sun to exert agreater force on Uranus than Neptune exerts on Uranus.
Answer and Explanation:
TL: DR The Sun is much more massive than Neptune — more than enough to make up for the somewhat smaller distance between the two planets at the closest approach.
[The surprise in this answer (to me, a non-astronomer), is that the gap between the orbits of Neptune and Uranus is large — half the distance from Uranus to the Sun.]
The ratio of gravitational attraction of the Sun on Uranus versus Neptune on Uranus is directly proportional to the ratio of the Sun’s mass to Neptune’s and inversely proportional to the ratio of the square of the distances (let’s use the closest approach of the two planets to one another to calculate a maximum attraction).
Numbers:
Sun’s mass: 2 x 10^30 kg
Neptune’s mass: 1 x 10^26 kg
Distance of Sun to Uranus: 3 x 10^9 km
Closest approach of Uranus and Neptune: 1.5 x 10^9 km
Without doing any arithmetic, we see that even at their closest approach, Uranus and Neptune are separated by about one-half of the Uranus to Sun distance. Squaring that ratio, we see that if the Sun and Neptune had the same mass, the attraction between the Sun and Uranus would only be about 1/4 of that between the Sun and Neptune; however, the Sun has 20000 times the mass of Neptune, so the attraction between Uranus and the Sun is about 5000 times stronger than the maximum attraction between Uranus and Neptune.
The explanation of the possibility of why sun exerts a greater force on Uranus than Neptune exerts on Uranus is; because the force was calculated to be greater.
The formula for calculating the Force of Gravity between two masses is:
F = G*m₁*m₂/r²
Where;
F = force of gravity
G = gravitational constant = 6.674 × 10⁻¹¹ N•m²/kg²
m₁ = mass of the larger object
m₂ = mass of the smaller object
r = the distance between the centers of the two masses
Now, from online values, we have the following;
mass of Neptune; m₁ = 102.413 × 10²⁴ kg
mass of Uranus; m₂ = 86.813 × 10²⁴ kg
average distance between the centers of Neptune and Uranus; r = 1.62745 × 10¹² m
Thus, force exerted by Neptune on Uranus is;
F = (6.674 × 10⁻¹¹ × 102.413 × 10²⁴ × 86.813 × 10²⁴)/(1.62745 × 10¹²)²
F = 2.240 × 10¹⁷ N
We are told that the force the Sun exerts on Uranus is 1.41 the force the Sun exerts on Uranus is 1.41 × 10²¹ N.
That is greater than the force Neptune exerts on Uranus.
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What observations did the solar system this geocentric models of the solar system help to explain
Answer:
Geocentric model of the solar system helped to explain retrograde motions of planets. Explanation: Geocentric model of planets was proposed by Ptolemy. It stated that all sun, planets and stars revolve round the earth in circular orbits.
Answer:
retrograde motion
Explanation:
i reverse searched the image
If I am driving down the highway going north at 50 miles per hour, and another car is driving south at 75 miles per hour. How fast is the car coming toward me?
Its an exam >.
I WILL GIVE BRAINLIEST
A 12 kg bowling ball would require what force to accelerate it down an alley at a rate of 2.5 m/s ^ 2
Answer:
hi
Explanation:
hiijjjjjjjjjjjjjjj
A motorcyclist goes around an un-banked (i.e., flat) circular turn of radius 31m, at a constant speed of 110km/hr (convert this to m/s). What is the minimum coefficient of static friction needed to keep the tires from slipping? Explain why the answer is (or is not) plausible.
Answer:
[tex]\mu_{s} = 3.071[/tex]
This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.
Explanation:
From Second Newton's Law we understand that centripetal acceleration experimented by motocyclist is due to force derived from static friction. And normal force of the ground on motocyclist equals weight of motocyclist due to the flatness of circular turn. The equations of equilibrium of the motocyclist is:
[tex]\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (Eq. 1)
[tex]\Sigma F_{y} = N-m\cdot g = 0[/tex] (Eq. 2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
[tex]m[/tex] - Mass of the motocyclist, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v[/tex] - Speed of the motorcyclist, measured in meters per second.
[tex]R[/tex] - Radius of the circular turn, measured in meters.
The static coefficient of friction is cleared in (Eq. 1):
[tex]\mu_{s} = \frac{m\cdot v^{2}}{N\cdot R}[/tex]
From (Eq. 2) we get that normai force is:
[tex]N = m\cdot g[/tex]
And we expand the resulting expression in (Eq. 1):
[tex]\mu_{s} = \frac{m\cdot v^{2}}{m\cdot g\cdot R}[/tex]
[tex]\mu_{s} = \frac{v^{2}}{g\cdot R}[/tex] (Eq. 3)
If we know that [tex]v = 30.556\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 31\,m[/tex], the expected static coefficient of friction is:
[tex]\mu_{s} = \frac{\left(30.556\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (31\,m)}[/tex]
[tex]\mu_{s} = 3.071[/tex]
This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.
How many strings of length 10 over the alphabet (a, b, c, d] have at least one b somewhere in the string?
a) 310
b) 410 - 310
c) 10.4
d) 10.39
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is B
Explanation:
The number of alphabet is n= 4 (a , b , c , d )
Generally the total number of string of length 10 over the 4 alphabets is
[tex]N = 4^{10}[/tex]
Gnerally the number of string of length 10 that does not include b is
[tex]T = 3^{10}[/tex]
Generally the number of strings of length 10 over the 4 alphabets that have at least one alphabet b somewhere in the string is
[tex]G = N - T[/tex]
=> [tex]G = 4^{10} - 3^{10}[/tex]
A small child weighs 60 N. If mommy left him sitting on top of the stairs, which are 12 m high, how much energy does the child have!
Please help ASAP
Answer:
6000 joules
Explanation:
I jus learned dis
Answer:6000j
Explanation:
Hope that helps
How do compounds differ from mixtures such as lemonade
Answer:
A mixture is a combination of two or more substances in any proportion. This is different from a compound, which consists of substances in fixed proportions. ... The lemonade pictured above is a mixture because it doesn't have fixed proportions of ingredients.
Explanation:
A 50 kg bicyclist starts his ride down the road with an acceleration of 1m/s2 in air with a density of 1.2 kg/m3. If his velocity at a given moment is 2m/s, how much force is he exerting? Assume the area of his body is 0.5m^2.
a. The bicyclist is exerting 1.1 N of force.
b. The bicyclist is exerting 49 N of force.
c. The bicyclist is exerting 50 N of force.
d. The bicyclist is exerting 51 N of force.
Answer:
b. The bicyclist is exerting 49 N of force
Explanation:
Given;
mass of bicyclist start, m = 50 kg
acceleration, a = 1 m/s²
density of air, ρ = 1.2 kg/m³
velocity, v = 2 m/s
Area of the bicyclist body, A = 0.5 m²
The drag force on the bicyclist is given by ;
Fd = 0.5CρAv²
where;
C is drag coefficient = 0.9 for bicycle
Fd = 0.5 x 0.9 x 1.2 x 0.5 x 2²
Fd = 1.1 N
The force of the bicyclist is given by;
F = ma
F = 50 x 1
F = 50 N
The effective force exerted by the bicyclist is given by;
Fe = F - Fd = 50 N - 1.1 N
Fe = 49 N
Therefore, the force exerted by the bicyclist is 49 N
The word acid comes from the Latin word
Compare the amount of thermal energy required to MELT a solid with the amount of thermal energy released when the same liquid becomes a solid.
A region around the nucleus of an atom where electrons are likely to be found
Answer:
The region where an electron is most likely to be is called an orbital. Each orbital can have at most two electrons. Some orbitals, called S orbitals, are shaped like spheres, with the nucleus in the center.
Explanation:Hope this helps :)
By definition, a region around the nucleus of an atom where electrons are likely to be found is called an orbital.
First of all, an atom is the smallest constituent unit of ordinary matter that has the properties of a chemical element.
All atoms are made up of subatomic particles: protons and neutrons, which are part of their nucleus, and electrons, which revolve around them. Protons are positively charged, neutrons are neutrally charged, and electrons are negatively charged.
In other words, every atom consists of a nucleus in which neutrons and protons meet and energy levels where electrons are located.
This is, the atomic nucleus is the central part of the atom that is made up of protons and neutrons, while the orbitals or peripheral region is an area where electrons are found.
In summary, a region around the nucleus of an atom where electrons are likely to be found is called an orbital.
Learn more:
https://brainly.com/question/10866484?referrer=searchResultshttps://brainly.com/question/1275002?referrer=searchResultsbrainly.com/question/1814899?referrer=searchResults brainly.com/question/2449569?referrer=searchResultsAn electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is 7.9 times the magnitude of the tangential acceleration. What is the angle
Answer:
The angle is 3.95 rad.
Explanation:
The angle can be calculated as follows:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta [/tex]
Where:
[tex]\omega_{f}[/tex]: is the final angular speed
ω₀: is the initial angular speed = 0 (it starts from rest)
α: is the angular acceleration
θ: is the angle=?
The centripetal acceleration is:
[tex]a_{c} = \omega_{f}^{2}*r[/tex]
And the tangential acceleration is:
[tex] a_{T} = \alpha*r [/tex]
Since the magnitude of the centripetal acceleration is 7.9 times the magnitude of the tangential acceleration:
[tex]a_{c} = 7.9a_{T}[/tex]
[tex]\omega_{f}^{2}*r = 7.9*\alpha*r \rightarrow \alpha = \frac{\omega_{f}^{2}}{7.9}[/tex]
Now, the angle is:
[tex]\omega_{f}^{2} = 2(\frac{\omega_{f}^{2}}{7.9})\theta[/tex]
[tex] \theta = \frac{7.9}{2} = 3.95 rad [/tex]
Therefore, the angle is 3.95 rad.
I hope it helps you!
The angular distance traveled by the electric drill is 3.95 radians.
The given parameters;
initial angular speed, [tex]\omega_i[/tex] = 0centripetal acceleration, [tex]a_c[/tex] = 7.9aThe angular distance traveled by the electric drill is calculated as follows;
[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta[/tex]
The relationship between centripetal acceleration, tangential acceleration and angular speed is given as;
[tex]a_c = \omega ^2 r\\\\a = \alpha r\\\\a_c = 7.9a= 7.9\alpha r\\\\7.9\alpha r = \omega^2 r\\\\\alpha = \frac{\omega ^2}{7.9}[/tex]
Substitute the value of angular acceleration into the first equation;
[tex]\omega _f^2 = 0 + 2(\a (\frac{\omega _f^2}{7.9})\theta\\\\2\theta \omega_f^2 = 7.9\omega_f ^2\\\\\theta = \frac{7.9}{2} \\\\\theta = 3.95 \ rad[/tex]
Thus, the angular distance traveled by the electric drill is 3.95 radians.
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Hollywood and video games often depict the bad guys being "blown away" when they’re shot by a bullet (i.e. once hit, their feet leave the ground and they fly backwards). Assuming that even if a handgun cartridge did generate enough momentum for the bullet to do this, why is it still nonsense on-screen?
Answer:
Taking a look at Newton's third law of motion which states "for every force exerted, their is an opposite force equal in magnitude and opposite in direction on the first force".
Similarly if a bullet had enough forces behind it to hurl someone through the air when they were hit, a similar force would act on the person holding the gun that fired the bullet.
What we load into the gun is called a 'cartridge' Each piece is composed of four basic substance the casing, the bullet, the primer, and the powder.
The primer explodes lighting the powder which causes a buildup of pressure behind the bullet. This powder can be used in rifle cartages because the bullet chamber is designed to withstand greater pressures.
It is difficult in practice to measure the forces within a gun bagel, but the one easily measured parameter is the velocity with which the bullet exits muzzle velocity, therefore assuming that even if a handgun cartridge which generate enough momentum for the bullet to do this, it is still nonsense on screen in Hollywood and video.
Plz help me fast WITH EXTRA POINTS AFTER SUBMITTING
Answer:
4 bobux
Explanation:
one bobux
two bobux
three bobux
four bobux
Is it true or false that the displacement always equals the product of the average velocity and the time interval?
Answer:
True.
Explanation:
Applying the definition of average velocity, we know that we can always write the following expression:[tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex] (1)
By definition, Δx is just the displacement, and Δt is the time interval.So, just rearranging terms in (1), we get:[tex]\Delta x} = v_{avg}* {\Delta t}[/tex]
An ideal gas increases in temperature from 22°C to 42°C by two different processes. In one process, the temperature increases at constant volume, and in the other process the temperature increases at constant pressure. Which process requires more heat or are the required amount of heat same in both?
Answer:
a- More heat is required for the constant-pressure process than for the constant-volume
Explanation:
we have to solve using the thermodynamic first law. this is the heat applied to the system
dQ = dU + dW
definition of terms:
dU = change in internal energy
dW = work done
we have it that
change in internal energy dU is directly proportional to work done dW
but when we are in constant volume process, work done of the gas is zero
therefore
dQ of constant pressure is > than that of constant volume
so constant pressure process requires more heat
The process that requires more heat is the constant-pressure process than the constant-volume process.
According to the first law of thermodynamics, the heat that's applied to the system will be the addition of the change in internal energy and the work done.
In a constant-volume process, the work done on the gas is equal to zero. More heat will be required for the constant-pressure process than for the constant-volume process.
Also, it should be noted that the change in the thermal energy of the gas will be the same for the constant-pressure process and the constant-volume process.
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The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the center of the earth. (a) What magnitude and sign of charge would a 60-kg human have to acquire to overcome his or her weight by the force exerted by the earth’s electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 m? Is use of the earth’s electric field a feasible means of flight? Why or why not?
Answer:
a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.
b) The force of repulsion between two people is [tex]13.851\times 10^{6}[/tex] newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.
Explanation:
a) From Second Newton's Law, we form this equation of equilibrium:
[tex]\Sigma F = F_{E}-W = 0[/tex] (Eq. 1)
Where:
[tex]F_{E}[/tex] - Electrostatic force exerted on human, measured in Newton.
[tex]W[/tex] - Weight of the human, measured in Newton.
If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:
[tex]q\cdot E-m\cdot g = 0[/tex]
[tex]q = \frac{m\cdot g}{E}[/tex] (Eq. 2)
[tex]E[/tex] - Electric field, measured in Newtons per Coloumb.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravity acceleration, measured in meters per square second.
[tex]q[/tex] - Electric charge, measured in Coulomb.
As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that [tex]m = 60\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]E = -150\,\frac{N}{C}[/tex], the charge that a 60-kg human must have to overcome weight is:
[tex]q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }[/tex]
[tex]q = -3.923\,C[/tex]
The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.
b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:
[tex]F = \kappa \cdot \frac{q^{2}}{r^{2}}[/tex] (Eq. 3)
Where:
[tex]\kappa[/tex] - Electrostatic constant, measured in Newton-square meter per square Coulomb.
[tex]q[/tex] - Electric charge, measured in Coulomb.
[tex]r[/tex] - Distance between two people, measured in meters.
If we know that [tex]\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}[/tex], [tex]q = -3.923\,C[/tex] and [tex]r = 100\,m[/tex], then the force of repulsion between two people is:
[tex]F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right][/tex]
[tex]F = 13.851\times 10^{6}\,N[/tex]
The force of repulsion between two people is [tex]13.851\times 10^{6}[/tex] newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.
I need help with this answer
use the hubble's law to determine the distance to a quasar receding at 75% the speed of light. The speed of light is 300,000 km/sec. assume Hubble's constant is
Complete question:
use the hubble's law to determine the distance to a quasar receding at 75% the speed of light. The speed of light is 300,000 km/sec. assume Hubble's constant is 2.2 x 10⁻⁵ km/s/Lyr
Answer:
The distance to the quasar is 1.02 x 10¹⁰ Lyr
Explanation:
Given;
speed of light, v = 300, 000 km/sec
Hubble's constant, H₀ = 2.2 x 10⁻⁵ km/s/Lyr
percentage of the quasar recession = 75% of speed of light
Hubble's Law is given by;
[tex]v =H_od\\\\d = \frac{v}{H_o}\\\\d= \frac{(0.75*300,000)}{2.2*10^{-5}}.Lyr\\\\d = 1.02*10^{10} \ Lyr[/tex]
Therefore, the distance to the quasar is 1.02 x 10¹⁰ Lyr
See Conceptual Example 6 to review the concepts involved in this problem. A 12.0-kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume a radial distance of 86.4 cm is between the branch and the point where the monkey's mass is located. As the monkey swings through the lowest point on the circle, it has a speed of 1.33 m/s. Find (a) the magnitude of the centripetal force acting on the monkey and (b) the magnitude of the tension in the monkey's arm.
Answer:
(a) 24.56 N
(b) 142.28 N
Explanation:
(a)
The designation assigned to something like the net force pointed toward the middle including its circular route seems to be the centripetal force. The net stress only at lowest point constitutes of the strain throughout the arm projecting upward towards the middle as well as the weight pointed downwards either backwards from the center.
The centripetal function is generated from either scenario by Equation:
⇒ [tex]Fc = \frac{mv^2}{r}[/tex]
On putting the values, we get
⇒ [tex]=\frac{12\times 1.33^2}{0.864}[/tex]
⇒ [tex]=24.56 \ N[/tex]
(b)
Use T to denote whatever arm stress we can get at the bottom including its circle:
⇒ [tex]Fc = T - mg =\frac{ mv^2}{r}[/tex]
⇒ [tex]T = mg + Fc[/tex]
⇒ [tex]=12\times 9.81+24.56[/tex]
⇒ [tex]=142.28 \ N[/tex]