Answer: Yes indeed it does.
Explanation: Newton's Three Laws of Motion explain how forces create motion in sport. These laws are usually referred to as the Laws of Inertia, Acceleration, and Reaction. maybe brainliest to level up.
What is the gravitational force between two students, Dylan and Sarah, if Dylan has a mass of 75 kg, Sarah has a mass of 54 kg, and their centers are separated by a distance of.45 m
Answer:
[tex]F=1.334*10^-^6[/tex]
Explanation:
From the question we are told that
Mass of Dylan [tex]M_d=75kg[/tex]
Mass of Sarah [tex]M_s=54kg[/tex]
Centers separated by[tex]C=0.45m[/tex]
Generally the equation for solving gravitational force is given by
[tex]F=\frac{G*m_1m_2}{r^2}[/tex]
[tex]F=\frac{6.67*10^-^1^1*75*54}{(0.45)^2}[/tex]
[tex]F=1.334*10^-^6[/tex]
Therefore Gravitational force is given as
[tex]F=1.334*10^-^6[/tex]
what can you conclude about the force of gravity?
Answer:
All objects, irrespective of their mass, experience the same acceleration g when falling freely under the influence of gravity at the same point on the Earth. If gravity is the only force acting on an object, the sum of kinetic energy and gravitational energy is constant. ...
Answer:
Gravity, also called gravitation, in mechanics, the universal force of attraction acting between all matter. ... On Earth all bodies have a weight, or downward force of gravity, proportional to their mass, which Earth's mass exerts on them. Gravity is measured by the acceleration that it gives to freely falling objects.The Earth's gravitational force accelerates objects when they fall. It constantly pulls, and the objects constantly speed up.Gravity is the force of attraction between two objects with mass and is dependent on the distance between these objects. Gravity is the force that repels two objects that have opposite charges. It is dependent upon the charges of the objects.
Explanation:
1. Sodium-24 has a half-life of 15 hours. How much sodium-24 will remain in an
26.0 g sample after 60 hours?
Answer:
1.625g
Explanation:
The half - life of a radioactive substance is the time taken for half of it to decay to half of its original composition.
For the given specie, the original composition is 26g and the half life is 15hrs:
At time 0hrs Mass = 26g
For the first half life at time 15hrs Mass = 13g
30hrs Mass = 6.5g
45hrs Mass = 3.25g
60hrs Mass = 1.625g
Which element of the integumentary system is responsible for acne in adolescents?
subcutaneous adipose tissue
sweat glands
squamous cells
sebaceous glands
Acne is a long-lasting condition that affects the sebaceous glands and hair follicles in the middle layer of the skin.Acne is characterized by clogged sebaceous glands, which produce pimples and cysts.
Which part of the integumentary system is responsible for acne?
A condition of the oil glands and hair follicles is acne (sebaceous glands).To keep the skin moist, the sebaceous glands release oils (sebum).Acne and cysts can develop as a result of clogged glands. Sebum makes the skin more resistant to moisture and lubricates it to defend against abrasion.Additionally, the sebaceous gland has a natural light protection action and transfers antioxidants into and onto the skin.It has pro- and anti-inflammatory properties as well as an intrinsic antimicrobial activity. Your skin, hair, and nails make up your integumentary system.It serves as the top layer of defense against bacteria and damage to your body. The face, upper chest, and back are among the skin regions where sebaceous follicle density is highest and are frequently affected by acne vulgaris. Sebum, an oily fluid that moisturizes your skin and hair, and dead skin cells clog hair follicles, resulting in acne.Bacteria can cause infection and inflammation, which worsen acne. Other sebaceous gland functions, such as sebaceous proinflammatory lipids, various locally produced cytokines, periglandular peptides and neuropeptides, such as corticotrophin-releasing hormone, which is produced by sebocytes, and substance P, which is expressed in sebocytes, are also linked to the development of acne.To learn more about acne in adolescents refer
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The large window air conditioner in Anita Breeze's room draws 11 amps of current. The unit runs for 8.0 hours during the course of a day. Determine the quantity of charge that passes through Anita's window AC during these 8.0 hours.
Answer:
quantity of charge that passes through Anita's window AC, Q = 316800 C
Explanation:
given data
current I = 11 amps
runs t = 8.0 hours
solution
we get here quantity of charge that passes through Anita's window AC
the quantity of charge passing Q = I × t .................1
Q = 11 A × ( 8 hr )× ( 3600 s/hr)
so
quantity of charge that passes through Anita's window AC, Q = 316800 C
The quantity of charge that passes through Anita's window AC, will be
Q = 316800 C
What is Current?
The current is defined as the quantity of charge flowing from the wire per unit time. Also defined as the rate of flow of the charge through any circuit. The formula will be given as:
[tex]\rm I=\dfrac{Q}{t}[/tex]
Now it is given that:
current I = 11 amps
runs t = 8.0 hours
The quantity of charge that passes through Anita's window AC during these 8.0 hours. Will be calculated as:
[tex]Q=I\times t[/tex]
[tex]Q=11\times 8\times 3600[/tex]
[tex]Q=316800 \ C[/tex]
Thus the quantity of charge that passes through Anita's window AC, Q = 316800 C
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A 0.015 kg marble moving to the right at 0.225 m/s makes an elastic head-on collision with a 0.03 kg shooter marble moving to the left at 0.18 m/s. After the collision, the smaller marble moves to the left at 0.315 m/s. What is the velocity of the 0.03 kg marble after the collision
Answer:
0.135m/s
Explanation:
Using the law of conservation of momentum
m1u1+m2u2 = m1v1 +m2v2
m1 and m2 are the masses
u1 and u2 are the initial velocities
v1 and v2 are thee final velocities
Substitute the given parameters and get v2
0.015(0.225)+0.03(0.18) = 0.015(0.315)+0.03v2
0.003375+0.0054 = 0.004725 + 0.03v2
0.008775 - 0.004725 = 0.03v2
0.00405 = 0.03v2
v2 = 0.00405/0.03
v2 = 0.135m/s
Hence the final velocity of the 0.03kg marble after collision is 0.135m/s
An investigator is using a laser to illuminate a distant target. He decides that he needs a smaller beam, so he puts a pinhole directly in front of the laser. He finds that this actually spreads the beam out, making matters worse.
Explain what is happening.
a. The pinhole has a size comparable with the laser wavelength, so the pinhole refracts the passing laser beam.
b. The pinhole has a size smaller than the laser wavelength, so the pinhole refracts the passing laser beam.
c. The pinhole has a size smaller than the laser wavelength, so the pinhole diffracts the passing laser beam.
d. The pinhole has a size comparable with the laser wavelength, so the pinhole diffracts the passing laser beam.
Answer:
correct answer is d
The pinhole has a size comparable with the laser wavelength, so the pinhole diffracts the passing laser beam.
Explanation:
The researcher does not have a refraction problem since the medium on both sides of the pinhole is the same with the same refractive index, the problem he is having is with the diffraction of the laser beam through the pinhole, let's analyze the diffraction process that is described by the expression
a sin θ = m λ
where a is the pinlole size, λ the wavelength of the laser and m an integer.
The laser extends from the maximum of diffraction to the first zero (m = 1) of diffraction
sin θ = λ/a
when analyzing this expression we have some interesting cases
* when λ « a. the sine approaches zero therefore we are in the case of optical geometry, in this case the laser passes through the hole without being diffracted
* when λ ≈ a. the sine function has values between 0 and 1, for which a diffraction of the laser beam occurs, which increases the diameter of the same
* when λ> a. The laser does not pass through the gap since the sine cannot have values greater than 1
After this analysis, we review the answers to the exercise to find that the correct answer is d
A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h; The maximum acceleration of the car is 5 ft/s2 and the maximum deceleration obtained by applying the brakes is 20 ft/s2. What is the shortest time in which the driver of the car can complete the passing operation? (assuming that the driver of the car does not pay any attention to the speed limit
Answer:
The shortest time in which the driver can complete the operation is approximately 6.32456 seconds
Explanation:
The given parameters are;
The speed of the truck and the car = 35 mi/h ≈ 51.33 ft./s
Let "t" represent the time it takes the car to pass the truck by 40 ft., we have;
The distance covered by the truck = 35 mi/h × t
The distance covered by the car = 35 mi/h × t + 80 ft = 51.33
Let the distance over which the car accelerates = d
We have;
d = 51.33×t₁ + 1/2×5×t₁²
51.33·t + 80 - d = (51.33 + 5·t₁)·(t - t₁) - 1/2·20·(t - t₁)²
51.33 = 51.33 + 5·t₁ - 20·(t - t₁)
∴ 5·t₁ = 20·(t - t₁)
5·t₁ = 20·t - 20·t₁
25·t₁ = 20·t
t₁ = 4·t/5
(51.33 + 5·t₁)·(t - t₁) - 1/2·20·(t - t₁)² + 51.33×t₁ + 1/2×5×t₁² = 51.33·t + 80
We get;
(51.33 + 5·(4·t/5))·(t - (4·t/5)) - 1/2·20·(t - (4·t/5))² + 51.33×(4·t/5) + 1/2×5×(4·t/5)² = 51.33·t + 80
(51.33 + 4·t)·(t/5) - 2·t²/5 + 51.33 × (4·t/5) + 8·t²/5 = 51.33·t + 80
51.33×t/5 + 51.33×4·t/5 + 4·t²/5 - 2·t²/5 + 8·t²/5 = 51.33·t + 80
51.33·t + 10·t²/5 = 51.33·t + 80
2·t² = 51.33·t + 80 - 51.33·t = 80
t² = 80/2 = 40
t = √40 = 2·√10 ≈ 6.32456
The shortest time in which the driver can complete the operation is "t" ≈ 6.32456 seconds
please help me this is timed .
Find x if a = 3.0 m/s^2
Answer:
x = 50 N
Explanation:
Given that we have a net force, a mass, and acceleration, we can use the fundamental formula for force found in newton's second law which is F = m × a.
Given a mass of 150 kg, and an acceleration 3.0m/s². We can substitute these two values in our formula to calculate the magnitude of these forces or it's net force to identify the unknown force acting on our known force for this situation to work.
_______
F (Net force) = F2 (Second force which we are given) - F1 (First force) = m × a
m (mass which we are given) = 150 kg
a (acceleration which we are given) = 3.0m/s
________
So F = m × a → F2 - F1 = m × a →
500 - F1 = 150 × 3.0 → 500 - F1 = 450 →
-F1 = -50 → F1 = 50
What do you think causes the changes in sunrise and sunset times over the course of the year?
Answer:
The combination of Earth's elliptical orbit and the tilt of its axis results in the Sun taking different paths across the sky at slightly different speeds each day. This gives us different sunrise and sunset times each day.
Explanation:
A low rider truck accelerates from 0 m/s to 30.0 m/s with a displacement of 50m. What is the vehicle's acceleration?
a. 2.00m/s
b. 5.00m/s
c. 9.0m/s
d. 10.0m/s
Answer:
a = 9 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}^{2}=v_{o}^{2} +2*a*x[/tex]
where:
Vf = final velocity = 30 [m/s]
Vo = initial velocity = 0
a = acceleration [m/s²]
x = displacement = 50 [m]
Now replacing:
[tex]30^{2}=0 +2*a*50\\100*a=900\\a=9[m/s^{2} ][/tex]
A constant force of 20. newtons applied to a box causes it to move at a constant speed of 4.0 meters per second. How much work is done in the boxin 6.0 seconds?
1) 240 joules
2) 120 joules
3) 480 joules
4) 80. joules
The work done is 480 Joules.
Option 3 is correct.
The work done is computed as multiplication of force and displacement.
Given that, Force = 20 N
Displacement = 4 meter per second
For 6 second,
Displacement [tex]=4*6=24m[/tex]
[tex]Workdone=Force*displacement\\\\Workdone=20*24=480J[/tex]
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A 10 kg weight is suspended in the air by a strong cable. How much work is done, per unit time, in suspending the weight
Answer:
There is no work done.
Explanation:
Given the following data;
Mass = 10 kg
To find the work done?
In Physics, work done can be defined as the amount of energy transfered when an object or body is moved over a distance due to the action of an external force.
Mathematically, work done is given by the formula;
Work done = force * distance
[tex] W = F * d[/tex]
Where,
W is the work doneF represents the force acting on a body. d represents the distance covered by the body.However, the weight suspended in the air by a strong cable does not move or experience any form of displacement. Therefore, there is no work done.
The resistance of a wire depends on its length i and on its cross sectional area A the resistance is
Answer:
The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area
Explanation:
A hammer strikes a nail with a 10 N force for 0.01 seconds. Calculate the impulse of the hammer.
Answer:
0.1Ns
Explanation:
Impulse is the product of Force and time
Impulse = Force * Time
Given
Force = 10N
Time = 0.01s
Substitute into the formula
Impulse = 10 * 0.01
Impulse = 10 * 1/100
Impulse = 10/100
Impulse = 0.1Ns
hence the impulse of the hammer is 0.1Ns
How much extra water does a 147-lb concrete canoe displace compared to an ultra lightweight 36-lb Kevlar canoe of the same size carrying the same load
Answer:
1.7791 ft^3
Explanation:
The computation of the extra water is as follows;
Here we use the following formula
Volume of the water = ÷ Weight of concrete canoe ÷ value of the weight of the specific water
= 147 lb ÷ 62.4 lb /ft^3
= 2.356 ft^3
Now the volume of water occupied in ultra lightweight
= 36 lb ÷ 62.4 lb /ft^3
= 0.5769 ft^3
Now the water volume displaced is
= 2.356 - 0.5769
= 1.7791 ft^3
A medicine ball has a mass of 8.0 kg and is thrown with a speed of 3.0 m/s. What is its kinetic energy?
O A 12 J
O
B. 24 J
C. 36 J
0 D. 192 J
Answer:
C. 36J
Explanation:
KE = ½ m V²
= ½ × 8 × 3²
= 4 × 9
= 36 J
Imagine you are a TV sports commentator for a major car race. What techniques could
you use to describe the action to your audience?
Answer:
friction , magnetism etc
All of the following are ways in which sports psychologists help athletes except __________.
A.
staying motivated
B.
managing fear of failure
C.
improving performance
D.
enhancing memory
Please select the best answer from the choices provided
A
B
C
D
Answer:
D-Enhancing memory
Explanation:
explain why work is not considered a form of energy.
What are the three longest wavelengths for standing waves on a270-cm-long string that is fixed at both ends
Answer:
The answer is below
Explanation:
a) What are the three longest wavelengths for standing waves on a 270-cm-long string that is fixed at both ends? b. If the frequency of the second-largest wavelength is 50.0 Hz, what is the frequency of the third-longest wave length?
Solution:
a) The wavelengths (λ) for standing waves is given by the formula:
[tex]\lambda_m=\frac{2*length\ of\ string}{m}\\\\Where\ m=1,2,3,.\ .\ .\\\\Given\ that\ length\ of\ string = 270\ cm=2.7\ m,\ m=1,2,3(three\ longest\ wavelengths)\\\\Hence:\\\\\lambda_1=\frac{2(2.7)}{1}=5.4\ m\\\\\lambda_2=\frac{2(2.7)}{2}=2.7\ m \\\\\lambda_3=\frac{2(2.7)}{3}=1.8\ m[/tex]
b) The frequency (f) and wavelength (λ) is given by:
fλ = constant
Hence:
[tex]f_2\lambda_2=f_3\lambda_3\\\\f_2=50\ Hz\\\\2.7*50=f_3(1.8)\\\\f_3=\frac{2.7*50}{1.8} \\\\f_3=75\ Hz[/tex]
The three longest wavelengths for the standing waves on a 270-cm long string that is fixed at both ends are:
1. 5.4 meters.
2. 2.7 meters.
3. 1.8 meters.
Given the following data:
Length of string = 270 cm to m = [tex]\frac{270}{100} =2.7\;m[/tex]To determine the three (3) longest wavelengths for these standing waves:
Mathematically, the wavelength for standing waves is given by the formula:
[tex]\lambda_n = \frac{2L}{n}[/tex]
Where:
[tex]\lambda_n[/tex] is the wavelength for standing waves.L is the length of string.Note: n = 1, 2, and 3.
When n = 1:
[tex]\lambda_1 = \frac{2\times 2.7}{1} \\\\\lambda_1 = 5.4 \;meters[/tex]
When n = 2:
[tex]\lambda_2 = \frac{2\times 2.7}{2} \\\\\lambda_2 = 2.7 \;meters[/tex]
When n = 3:
[tex]\lambda_3 = \frac{2\times 2.7}{3} \\\\\lambda_3 =\frac{5.4}{3} \\\\\lambda_3 = 1.8 \;meters[/tex]
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what tells you whether you are feeling hot, cold or in pain?
A. Nerves/Neurons
B. Spinal Cord
C. Brain
D. Skin
Answer:
i think its skin..think so this answre help you
Which three statements describe mechanical waves?
A. The waves can travel through empty space.
B. The waves transfer energy by causing particles of matter to
move.
C. The waves need matter to transfer energy.
D. The waves can transfer energy through solids, liquids, and gases.
The three statements that are correct about the mechanical waves are options B, C, and D.
Mechanical waves:Mechanical waves are produced due to disturbances in any matter. Which clearly indicates that the mechanical waves cannot travel through empty space.
The mechanical waves require matter to transfer energy through oscillation in the matter. So it is true that the waves transfer energy by causing particles of matter to move.
Also, the mechanical waves need matter to transfer energy, the waves can transfer energy through solids, liquids, and gases.
An example of a mechanical wave is the sound produced by the guitar strings.
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In a typical badminton swing the racket is in contact with the birdy for about 0.0010 seconds. If the 0.045kg birdy acquires a speed of 67 m/s, what is the force exerted by the racket on the ball
Answer:
3015 N
Explanation:
From Newton's second law, we know that;
F.t = mv
F = force on the ball= ?
m= mass of the ball
v= velocity
F= mv/t
F= 0.045 × 67/0.0010
F= 3.015/0.0010
F= 3015 N
The force exerted by the racket on the ball is 3015 N.
The force exerted by the racket on the ball is given in the formula below.
⇒ Formula:
F = mΔv/t............... Equation 1⇒ Where:
F = force exerted by the racketm = mass of the birdyΔv = change in speedt = time.From the question,
⇒ Given:
m = 0.045 kgΔv = 67 m/st = 0.0010 seconds.⇒ Substitute these values into equation 1
F = 0.045(67)/0.001F = 3015 NHence, The force exerted by the racket on the ball is 3015 N
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What is the velocity of a dropped object after it has fallen for 12 s?
Hellow!
For this use the next formula:
Vf = Vo + gt
Initial velocity is zero, so the formula simplificate:
Vf = gt
Data:
Vf = Final velocity = ?
g = Gravity = 9.8 m/s²
t = Time = 12 s
Replacing according our data:
Vf = 9.8 m/s² * 12 s
Vf = 117.8 m/s
The final velocity will be 117.8 meters per second.
If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
Complete Question
The speed of a transverse wave on a string of length L and mass m under T is given by the formula
[tex]v=\sqrt{\frac{T}{(m/l)}}[/tex]
If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
Answer:
[tex](m/l)=\frac{10}{V^2}[/tex]
Explanation:
From the question we are told that
Speed of a transverse wave given by
[tex]v=\sqrt{\frac{T}{(m/l)}}[/tex]
Maximum Tension is [tex]T=10.0N[/tex]
Generally making [tex](m/l)[/tex] subject from the equation mathematically we have
[tex]v=\sqrt{\frac{T}{(m/l)}}[/tex]
[tex]v^2=\frac{T}{(m/l)}[/tex]
[tex](m/l)=\frac{T}{V^2}[/tex]
[tex](m/l)=\frac{10}{V^2}[/tex]
Therefore the Linear mass in terms of Velocity is given by
[tex](m/l)=\frac{10}{V^2}[/tex]
15 points! Answer FAST!
When heat is transferred, it flows_____.
A. From the warmer object to the cooler object only if the cooler object is an insulator
B. From the cooler object to the warmer object regardless of the materials
C. From the warmer object to the cooler object regardless of the materials
D. From the warmer object to the cooler object only if the warmer object is a conductor.
Answer:heat brings it up then down
Explanation:
h. If the mass of an object is 390 g and its volume is 50 cm?, then its density is ....... g/cm' or ....... kg/m”.
- The relative density of the object in part "h" is
Edit: nvm I'm an idiot I remembered this but thanks anyway
Answer: See explanation
Explanation:
The density would like be calculated by dividing the mass by volume. This will be:
Density = Mass / Volume
Density = 390 / 50
Density = 7.8g/cm
A spring has an unstretched length of 16 cm . A 110 g mass hanging from the spring stretches it to an equilibrium length of 24 cm . Part APart complete Suppose the mass is pulled down to where the spring's length is 32 cm . When it is released, it begins to oscillate. What is the amplitude of the oscillation
Answer:
the amplitude of the oscillation is 8 cm
Explanation:
The computation of the amplitude of the oscillation is shown below:
= Length of the spring - length of the equilibrium
= 32 cm - 24 cm
= 8 cm
By deducting the length of the equilibrium from the length of the spring we can get the amplitude of the oscillation and the same is to be considered
hence, the amplitude of the oscillation is 8 cm
Which is greater, the energy of one photon of orange light or the energy of one quantum ofradiation having a wavelength of 3.36 * 10^-9
The question is incomplete, here is the complete question:
Which is greater, the energy of one photon of orange light or the energy of one quantum of radiation having a wavelength of [tex]3.36\times 10^{-9}m[/tex]
Answer: The energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.
Explanation:
To calculate the energy of one photon, we use the Planck's equation:
[tex]E=\frac{N_Ahc}{\lambda}[/tex]
where,
E = energy of radiation
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}mol^{-1}[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8 m/s[/tex]
[tex]\lambda}[/tex] = wavelength of radiation
For orange light:For 1 photon, the term [tex]N_A[/tex] does not appear
[tex]\lambda}[/tex] = 620 nm = [tex]620\times 10^{-9}m[/tex] (Conversion factor: [tex]1nm=10^{-9}m[/tex] )
Putting values in above equation, we get:
[tex]E=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{620\times 10^{-9}m}\\\\E=3.206\times 10^{-19}J[/tex]
For one quantum of radiation:[tex]\lambda}[/tex] = [tex]3.36\times 10^{-9}m[/tex]
Putting values in above equation, we get:
[tex]E=\frac{6.022\times 10^{23}mol^{-1}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{3.36\times 10^{-9}m}\\\\E=3.56\times 10^{7}J/mol[/tex]
Hence, the energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.