Answer:
Explanation:
A fire alarm notification appliance is an active fire protection component of a fire alarm system. The primary function of the notification appliance is to alert persons at risk.
If want the audible public mode signal to be hear clearly then, we need to have a sound level that is at least 15dB above the average ambient sound level or 5dB above the maximum sound level of at least 1minute
In this case the,
The average ambient sound level is 62dB,
And the maximum sound level is 68dB
Then, the public mode signal should be at least
1. 62dB+ 15dB=77dB
Or
2. 68dB +5dB =73dB.
Then the public mode signal hearing must be at least 77dB.
How can astrophysicists tell whether a star is receding from or approaching earth?
Answer:
Doppler shift of the starlight
Explanation:
To predict the movement of a star, we compare the spectra of elements found in star (H, He Na etc.), first spectra which are obtained from star and second spectra from laboratory. If spectral lines of the spectra obtained from star, are shifting towards red end (called red shift) then star is going away from earth and if shifting is towards blue (called blue shift), then star is approaching the earth. This is Doppler's shift.
In a vertical dive, a peregrine falcon can accelerate at 0.6 times the free-fall acceleration g (that is, at 0.6 g ) in reaching a speed of about 108 m / s. If a falcon pulls out of a dive into a circular arc at this speed and can sustain a radial acceleration of 0.6 g , what is the radius R of the turn?
The radius R of the turn is 1.984 km.
Explanation:
As the falcon is experiencing a centripetal motion, the acceleration exhibited by the falcon will be centripetal acceleration. The formula for centripetal acceleration is
[tex]a=\frac{v^{2} }{R}[/tex]
Here a is the acceleration for centripetal motion, v is the velocity and R is the radius of the circular path.
As the centripetal acceleration is given as 0.6 g, the velocity is given as 108 m/s, then the radius of the path can be determined as
[tex]0.6 \times 9.8=\frac{(108)^{2}}{R}[/tex]
[tex]R=\frac{(108)^{2}}{0.6 \times 9.8}=\frac{11664}{5.88}=1983.67\ \mathrm{m}[/tex]
So, the radius of the turn is 1.984 km.
A 73 kg person is riding in a car moving at 20 m/s when the car runs into a bridge abutment. Calculate the magnitude of the average force (in Newtons) on the person if he is stopped by a padded dashboard that compresses an average distance of 0.16 m
Answer:
The magnitude of average force [tex]F_{avg} =[/tex] [tex]9.125 \times 10^{4} N[/tex]
Explanation:
Given :
Mass of person [tex]m = 73[/tex] Kg
Initial velocity of car [tex]v_{o} = 20 \frac{m}{s}[/tex]
Average distance [tex]x = 0.16[/tex] m
According to the second law of newton.
[tex]F_{avg} = m a[/tex]
From the kinematic equations,
[tex]v^{2}- v_{o} ^{2} = 2ax[/tex]
Where [tex]v_{o} =[/tex] initial velocity, in our case final velocity is zero ([tex]v = 0[/tex])
[tex]a =- \frac{400}{2 \times 0.16} = -1250 \frac{m}{s^{2} }[/tex]
So average force is given by,
[tex]F _{avg} = 73 \times -1250 = -91250[/tex] N
But magnitude of average force is,
[tex]F_{avg} = 9.125 \times 10^{4}[/tex] N
From the following statements about mechanical waves, identify those that are true for transverse mechanical waves only, those that are true for longitudinal mechanical waves only, and those that are true for both types of waves.A. In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy.B. Many wave motions in nature are a combination of longitudinal and transverse motion.C. In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy.D. All of the above
Answer: D all of the above
Explanation:
All of the statements in the options are the character's exhibited by both longitudinal amd transverse waves.
Examples of transverse waves are water waves, light waves, radio waves, and also waves produced in strings and ropes.
Examples of longitudinal waves includes : vibrating turn fork,, drum head etc
Statement A is true for longitudinal waves only, statement C is true for transverse waves only, and statement B is true for both types of waves. These are based on the nature of particle movement in relation to the direction of wave energy flow.
Explanation:In the given statements, Statement A, 'In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy', is true solely for longitudinal mechanical waves. This is because, in longitudinal waves, the disturbances (or oscillations) in the medium occur in the same direction as the wave propagation, with notable examples being sound waves in air and water.
Statement C, 'In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy', on the other hand, pertains exclusively to transverse mechanical waves. Transverse waves are characterized by disturbances in the medium that are perpendicular to the direction of the wave propagation, common examples of which are waves seen on stringed instruments and electromagnetic waves, such as visible light.
Statement B, 'Many wave motions in nature are a combination of longitudinal and transverse motion', applies to both longitudinal and transverse mechanical waves. Certain seismic waves generated during earthquakes, for instance, possess both longitudinal and transverse components, demonstrating that waves can indeed display composite behaviors. Thus, in essence, this statement is true for both types of mechanical waves.
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Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller flywheel of radius 30 cm has a cord that has a pulling force of 50 N on it. What pulling force needs to be applied to the cord connecting the larger flywheel of radius 50 cm such that the combination does not rotate?
Answer:
Explanation:
Torque on smaller wheel
= F x r
50 x .30
= 15 Nm
Torque on larger wheel
= F x .5
For equilibrium
F x .5 = 15
F = 15 / .5
= 30 N
In a pickup game of dorm shuffleboard,students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 3.5 kg book is pushed from rest through a distance of 0.90 m by the horizontal 25 N force from the broom and then has a speed of 1.60 m/s,what is the coefficient of kinetic friction between the book and floor?
Answer:
Coefficient of friction between the book and floor is 0.582.
Explanation:
Using the velocity formula;
v^2 = 2as
a = v^2/(2s)
a = 1.6^2/(2*0.9)
a = 2.56/1.8
a = 1.42 m/s^2
the force necessary to give the book the acceleration is
F = ma = 3.5*1.42 (m is mass of the book i.e. 3.5 kg)
F = 4.98 N
The difference in the force is the friction force, which is
Ff = 25 - 4.98 = 20 N
Ff = mgμ
where μ is coefficient of friction and g is acceleration due to gravity that is 9.8 m/s^2
μ = Ff/mg
μ = 20/(3.5*9.81)
μ = 0.582
Coefficient of friction between the book and floor is 0.582.
This question involves the concepts of the equation of motion and Newton's second law of motion.
The coefficient of kinetic friction between the book and the floor is "".
First, we will find the acceleration of the block by using the third equation of motion:
[tex]2as = v_f^2-v_i^2[/tex]
where,
a = acceleration = ?
s = distance covered = 0.9 m
vf = final speed = 1.6 m/s
vi = initial speed = 0 m/s
Therefore
[tex]a=\frac{(1.6\ m/s)^2-(0\ m/s)^2}{2(0.9\ m)}\\\\a=1.42\ m/s^2[/tex]
Hence, from Newton's second law of motion:
[tex]Net\ Force = Frictional\ Force + F\\Net\ Force = \mu mg+ma[/tex]
where,
Net Force = 25 N
μ = coefficient of kinetic friction = ?
m = mass of the book = 3.5 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]25\ N = \mu(3.5\ kg)(9.81\ m/s^2)+(3.5\ kg)(1.42\ m/s^2)\\\\\mu=\frac{25\ N - 4.98\ N}{34.33\ N}\\\\\mu = 0.58[/tex]
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Our Sun undergoes slight orbital motion mostly due to the gravitational force exerted by Jupiter. If our solar system only contained Saturn, how would the Sun's orbital period differ?
Answer:
it would be longer
As you watch the video, notice that the size of the tidal bulges varies with the Moon's phase, which depends on its orbital position relative to the Sun. Which of the following statements accurately describes this variation? A. Low tides are lowest at both full moon and new moon.
B. High tides are highest at both full moon and new moon.
Answer:
Both
A. Low tides are lowest at both full moon and new moon.
B. High tides are highest at both full moon and new moon.
Explanation:
Tides are formed as a consequence of the differentiation of gravity due to the moon across to the Earth sphere.
Since gravity variate with the distance:
[tex]F = G\frac{m1\cdot m2}{r^{2}}[/tex] (1)
Where m1 and m2 are the masses of the two objects that are interacting and r is the distance Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.
For example, see the image below, point A is closer to the moon than point b and at the same time the center of mass of the Earth will feel more attracted to the moon than point B. Therefore, that creates a tidal bulge in point A and point B.
On the other hand, a full moon it gets when Sun, the Earth and the moon are in a line and the moon is reflecting the sunlight.
When the Moon is between the Earth and the Sun it will be illuminated in its back, so it is not possible to see it from the Earth (that is called new moon).
In those two cases mentioned above, the Sun tidal force contributes to the tidal force of the moon over the earth making high tides higher and low tides lower.
3. A powerful motorcycle can produce an acceleration of while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What force does the motorcycle exert backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg
Answer:
Total force = 1257.5N
Explanation:
Acceleration = 3.50 m/s2
Total Mass = 245kg
Force = Mass x acceleration
Force = 245 x 3.50 = 857.5N
Opposing Force = 400N
Total Force = Force of Motorcycle + Opposing force
Total force = 857.5N + 400N
Total force = 1257.5N
A student of mass 57.4 kg, starting at rest, slides down a slide 17.2 m long, tilted at an angle of 28.1° with respect to the horizontal. If the coefficient of kinetic friction between the student and the slide is 0.108, find the force of kinetic friction, the acceleration, and the speed she is traveling when she reaches the bottom of the slide. (Enter the magnitudes.)
Explanation:
(a) Formula to calculate the force of kinetic friction is as follows.
f = [tex]\mu N[/tex]
= [tex]\mu mg Cos (\theta)[/tex]
Putting the given values into the above formula as follows.
f = [tex]\mu mg Cos (\theta)[/tex]
= [tex]0.118 \times 57.4 kg \times 9.8 \times Cos (28.1^{o})[/tex]
= [tex]0.118 \times 57.4 kg \times 9.8 \times 0.882[/tex]
= 58.54 N
Hence, the force of kinetic friction is 58.54 N.
(b) Net force experienced by the block will be as follows.
F = [tex]mg Sin (\theta) - f[/tex]
ma = [tex]mg Sin (\theta) - \mu mg Cos (\theta)[/tex]
or, a = [tex]g[Sin (\theta) - \mu Cos (\theta)][/tex]
= [tex]9.8[Sin(28.1) - Cos(28.1)][/tex]
= [tex]9.8 \times (0.471 - 0.882)[/tex]
= 4.03 [tex]m/s^{2}[/tex]
Therefore, the acceleration is 4.03 [tex]m/s^{2}[/tex].
(c) According to the third equation of motion,
[tex]v^{2} = u^{2} + 2as[/tex]
= [tex]0 + 2 \times 4.03 \times 17.2[/tex]
= 138.63 m/s
Hence, the speed she is traveling when she reaches the bottom of the slide is 138.63 m/s.
Answer:
Explanation:
mass, m = 57.4 kg
distance, d = 17.2 m
angle of inclination, θ = 28.1°
initial velocity, u = 0 m/s
coefficient of kinetic friction, μk = 0.108
(a) N is the normal reaction acting on the student.
N = mg Cosθ
N = 57.4 x 9.8 x Cos 28.1
N = 496.2 N
Friction force = μk x N
Friction force = 0.108 x 496.2 = 53.6 N
Let a is the acceleration
ma = mg Sinθ - friction force
ma = 57.4 x 9.8 x Sin 28.1 - 53.6
a = 3.7 m/s²
Let the speed is v.
v² = u² + 2ad
v² = 0 + 2 x 3.7 x 17.2
v = 11.3 m/s
The acceleration of gravity on the moon is one-sixth what it is on Earth. The radius of the moon is one-fourth that of the Earth. What is the moon's mass
Answer: moon mass = earth mass/96
Explanation:
The moon mass will be 1/96th of the earth mass. There is an attached detailed solution to this.
Laminar flow, where water moves in approximately straight-line paths, characterizes ________.
a. fast-moving streams
b. slow-moving streams
c. the edges of channels
d. the bottoms of channels
e. All of these
Answer:
b. slow-moving streams.
Explanation:
In Fluid Mechanics, the Reynolds numbers indicates the existence of turbulence in fluid streams. Low Reynolds numbers are related with laminar flow. The Reynolds formula is:
[tex]Re = \frac{\rho_{water} \cdot L_{c}}{\mu_{water}} \cdot v[/tex]
The Reynolds number is directly proportional to fluid speed. Hence, slow-moving streams are a sound example of laminar flow. The correct answer is B.
A ball with mass M, moving horizontally at 4.00 m>s, collides elastically with a block with mass 3M that is initially hanging at rest from the ceiling on the end of a 50.0-cm wire. Find the maximum angle through which the block swings after it is hit.
Answer:
θ = 53.7°
Explanation:
Given:
- The mass of ball = M
- The mass of object = 3M
- The wire length L = 0.5 m
- The velocity of ball vi = 4.0 m/s
- The velocity of ball vf
- The velocity of object Vf
Find:
Find the maximum angle through which the block swings after it is hit.
Solution:
- When two objects collide with no external force acting on the system the linear momentum of the system is conserved. The initial (Pi) and final (Pf) linear momentum are equal:
Pi = Pf
M*vi = M*vf + 3M*Vf
vi = vf + 3*Vf
4 = vf + 3*Vf
- For elastic collision between two particles the relative velocities before and after collision have the same magnitude but opposite sign; so,
vi - 0 = Vf - vf
4 = Vf - vf
- Solve the above two equation simultaneously.
8 = 4*Vf
Vf = 2 m/s
vf = -2 m/s
- When the ball hits the object it swing under the influence of gravity only. Hence, no external force acts on the object so we can apply the conservation of energy as the object attains a height h.
ΔK.E = ΔP.E
0.5*(3M)*Vf^2 = (3M)*(g)*(h)
h = Vf^2 / 2*g
- Plug in the values:
h = 2^2 / 2*9.81
h = 0.2039 m
- We can see that the maximum angle can be given as θ according trigonometric relation as follows:
θ = arccos [ ( L - h ) / L ]
θ = arccos [ ( 0.5 - 0.2039 ) / 0.5 ]
θ = 53.7°
The maximum angle through which the block swings after it is hit θ is = 53.7°
Calculation of Mass
Given:
The mass of ball is = M
The mass of object is = 3M
The wire length L is = 0.5 m
The velocity of ball vi is = 4.0 m/s
Then The velocity of ball vf
After that The velocity of object Vf
Now we Find:
Find the maximum angle through which the block swings after it is hit that is :
When two objects collide with no external force acting on the system the linear momentum of the system is conserved. Then The initial (Pi) and also final (Pf) linear momentum are equal:
Pi is = Pf
M*vi is = M*vf + 3M*Vf
vi is = vf + 3*Vf
4 is = vf + 3*Vf
Now For elastic collision between two particles the relative velocities before and also after collision have the same magnitude but opposite signs; so,
vi - 0 is = Vf - vf
4 is = Vf - vf
Solve the above two equations simultaneously.
8 is = 4*Vf
Vf is = 2 m/s
vf is = -2 m/s
When the ball hits the object it swings under the influence of gravity only. Hence proof, no external force acts on the object so we can apply the conservation of energy as the object attains a height h.
ΔK.E is = ΔP.E
0.5*(3M)*Vf^2 is = (3M)*(g)*(h)
h is = Vf^2 / 2*g
Then we Plug in the values is:
h is = 2^2 / 2*9.81
h is = 0.2039 m
Now We can see that the maximum angle can be given as θ according trigonometric relation as follows:
θ is = arccos [ ( L - h ) / L ]
θ is = arccos [ ( 0.5 - 0.2039 ) / 0.5 ]
θ is = 53.7°
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1. Meiosis produces daughter cells that are
A) Haploid
B) Diploid
C) The same as the parent cell
D) Have more chromosomes than the parent cell
2. What is the purpose of mitosis?
A) Somatic Cell production
B) Gametic cell production
C) To make gametes
D) To make sperm and eggs
3. What is the purpose of meiosis?
A) Somatic Cell production
B) Gametic cell production
C) Liver cell production
D) Brain cell production
4. Which of the following about cell generation is true?
A) Mitosis is a form of sexual reproduction.
B) Meiosis is a form of asexual reproduction.
C) Mitosis produces identical cells to parent cells.
D) Meiosis produces identical cells to parent cells.
5. S=short fur and s=long fur, SS short fur and Ss long fur mate. What is the phenotypic ratio of short fur to long fur?
A) 3:1
B) 4:0
C) 1:3
D) 2:2
6. L=long wings and l=short wings. How can two long winged bees produce short winged bees?
A) 2 long winged bees cannot produce short winged bees
B) Both parent bees are homozygous dominant LL
c) Both parent bees are heterozygous Ll
D) Both parent bees are homozygous recessive ll
7. Which of the following is NOT true of DNA?
A) DNA is made up of 4 base chemicals
B) DNA is genetic material passed from parent to offspring
C) Long strands of DNA make up chromosomes
D) DNA is made up of chromosomes
8. DNA is now accepted as the molecule for heredity instead of proteins. How did this happen?
A) Scientists thought DNA seemed cooler
B) Each hypothesis was tested and DNA supported the data better
C) Highly respected scientists introduced it, everyone accepted
D) Proteins are known for producing muscles and that's it
9. Which of the following is most likely a concern about genetically modified crops?
A) There may be an increase in global food prices.
B) There may be a decrease in the range of environmental conditions in which a crop can survive.
C) There may be a decrease in the amount of time that fruits or vegetables stay fresh in the supermarket
D) There may be an increase in pest populations that are resistant to control measures.
10. What is NOT true of GMOs
A) They are proven to cause cancer
B) GM plants are pest resistant lowering the need for pesticide
C) GM plants produce higher yields for farmers
D) The long term effects are unknown
11) What are alleles?
A) characteristics
B) a form of a gene
C) a gamete
D) the study of heredity
I am willing to give brainliest and 100 points whoever EXPLAINS and gives the ANSWER to each question. You are not rushed, because I have 11:59, by the end of today to submit the assignment. If you write the answers and no explanation, you will get reported. So please answer and explain the questions clearly so I understand them. I am putting the subject as physics, but it is really science. good luck.
Answer:
1. A Haploid
2. A. Somatic Cell production
3. B. Gametic cell production
4. C. Mitosis produces identical cells to parent cells.
5. B 4:0
6. C Both parent bees are heterozygous Ll
7. D DNA is made up of chromosomes
8. B Each hypothesis was tested and DNA supported the data better
9. D There may be an increase in pest populations that are resistant to control measures.
10. A They are proven to cause cancer.
11. B a form of a gene
Explanation:
Answer was too long, so is attached in a word document
Atmospheric pressure at sea level has a value of 100kpa. The density of sea water is 1020kg/m-3. At which depth in the sea would the total pressure be 110kpa?
Answer:
1 m
Explanation:
P = Patm + ρgh
110,000 Pa = 100,000 Pa + (1020 kg/m³) (9.8 m/s²) h
h = 1 m
At depth of 1 m the sea would the total pressure be 110kpa.
What is pressure?The physical force applied to an object is referred to as pressure. Per unit area, a perpendicular force is delivered to the surface of the objects. F/A is the fundamental formula for pressure (Force per unit area). Pascals are a unit of pressure (Pa). Absolute, atmospheric, differential, and gauge pressures are different types of pressure.
Given that:
Atmospheric pressure at sea level has a value of p = 100 kPa = 100000 Pa.
The density of sea water is 1020kg/m-3.
Let: at depth of x m the sea would the total pressure be: P =110 kPa = 110000 Pa.
So, P = p + ρgh
⇒ 110,000 Pa = 100,000 Pa + (1020 kg/m³) (9.8 m/s²) h
⇒ h = 1 m
Hence, at depth of 1 m the sea would the total pressure be 110kpa.
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The toothpick mass was 0.14 g, its speed before entering the branch was 218 m/s, and its penetration depth was 14 mm. If its speed was decreased at a uniform rate, what was the magnitude of the force of the branch on the toothpick
Answer:
= 238N
Explanation:
mass = 0.14g = 14 × 10⁻⁵kg
initial velocity = 218m/s
final velocity = 0
penetration depth (distance) = 14mm = 14 × 10⁻³m
v²(final) = v²(initial) + 2aΔx
0² = (218)² + 2(14 × 10⁻³)a
a = -(218)² / 2(14 × 10⁻³)
a = 16.97 × 10⁵m/s²
F = ma
= (14 × 10⁻⁵)(16.97 × 10⁵)
= 237.58N
≅ 238N
Explanation:
Below is an attachment containing the solution.
What two quantities are crucial to quantifying the translational kinetic energy of an object?
Answer:
Moment of inertia and angular velocity.
Explanation:
The translational kinetic energy of an object is possessed when the object is showing rotational motion. It can be given by the formula as :
[tex]KE=\dfrac{1}{2}I\omega^2[/tex]
Here,
I is the moment of inertia of the object
[tex]\omega[/tex] is the angular velocity of the object
So, the translational kinetic energy of an object is given by moment of inertia and angular velocity of the object. Hence, this is the required solution.
The shuttles main engine provides 154,360 kg of thrust for 8 minutes. If the shuttle accelerated at 29m/s/s, and fires for at least 8 minutes, then how far does the shuttle go
Answer:
The answer to the question is
3340800 m far
Explanation:
To solve the question, we note that acceleration = 29 m/s²
Time of acceleration = 8 minutes
Then if the shuttle starts from rest, we have
S = u·t+0.5·a·t² where u = 0 m/s = initial velocity
S = distance traveled, m
a = acceleration of the motion, m/s²
t = time of travel
S = 0.5·a·t² = 0.5×29×(8×60)² = 3340800 m far
A series of pulses of amplitude 0.28 m are sent down a string that is attached to a post at one end. The pulses are reflected at the post and travel back along the string without loss of amplitude. What is the amplitude at a point on the string where two pulses cross if the string is rigidly attached to the post?
Answer:
a. The pulses cancel each other resulting in zero displacement.
b. The pulses reinforce each other, having a displacement of 2× amplitude or 0.56 m.
Explanation:
When the pulse are sent down the attached string it gets reflected and we have crossing pulses as the incident and reflected pulses cross each other.
a. If the string is rigidly attached to the post then the incident and reflected pulses will have the same amplitude but different direction. That is either the reflected will be going up and the incident down thereby resulting in a cancellation or zero displacement
Incident amplitude = 0.28 m
Reflected amplitude = -0.28 m
Sum = 0.28 m - 0.28 m = 0 m
b. If the end at which the relfection occurs is free to slide, then the incident and the reflected pulses will again have the same amplitude and in this case, the same direction. Therefore;
Incident amplitude = 0.28 m
Reflected amplitude = 0.28 m
Sum = 0.28 m + 0.28 m = 0.56 m
Answer:
The answer is zero displacement (0 m)
Explanation:
If the end of the string is not free, the reflected pulse has the same amplitude but opposite polarity to the incident pulse. Because of this, for A, the result equals zero (0 m).
The 10 kg dancer leaps into the air with an initial velocity of 5 m/s at angle of 45° from the floor. How far will she travel in the air horizontally before she lands on the ground again?
Answer:
2.55 m
Explanation:
The motion of the dancer is the motion of a projectile, which consists of 2 independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion (constant acceleration) along the vertical direction
The horizontal range of a projectile can be found by using the equations of motions along the two directions, and it is given by:
[tex]d=\frac{v^2 sin(2\theta)}{g}[/tex]
where
v is the intial velocity
[tex]\theta[/tex] is the angle of projection
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
For the dancer in this problem, we have:
v = 5 m/s
[tex]\theta=45^{\circ}[/tex]
Therefore, the horizontal range is:
[tex]d=\frac{(5)^2(sin 2\cdot 45^{\circ})}{9.8}=2.55 m[/tex]
When you lift a bowling ball with a force of 71.6 N, the ball accelerates upward with an acceleration a. If you lift with a force of 82.3 N, the ball's acceleration is 1.91a. Calculate the weight of the bowling ball.
Answer:
59.84 N
Explanation:
The net force is responsible for the upward acceleration of the bowling ball.
The net force acting on a body accelerates the body in the same direction as that in which the resultant is applied.
Net force = ma
ma = F - mg
In the first case,
F = 71.6 N, a = a m/s²
ma = 71.6 - mg (eqn 1)
In the second case,
F = 82.3 N, a = 1.91a m/s²
m(1.91a) = 82.3 - mg
1.91 ma = 82.3 - mg (eqn 2)
Substitute for ma in (eqn 2)
1.91 (71.6 - mg) = 82.3 - mg
136.756 - 1.91 mg = 82.3 - mg
136.756 - 82.3 = 1.91 mg - mg
0.91 mg = 54.456
mg = 54.456/0.91
mg = 59.84 N
Hence, the weight of the bowling ball = 59.84 N
Hope this Helps!!!
The weight of the bowling ball is 59.84 N
What is the net force?The net force should be held responsible where there is the upward acceleration of the bowling ball. When it should be acted on the body so it should be in a similar direction.
We know that
Net force = ma
So,
ma = F - mg
Now
In the first case,
F = 71.6 N, a = a m/s²
ma = 71.6 - mg (eqn 1)
And,
In the second case,
F = 82.3 N, a = 1.91a m/s²
m(1.91a) = 82.3 - mg
So,
1.91 ma = 82.3 - mg (eqn 2)
Now
1.91 (71.6 - mg) = 82.3 - mg
136.756 - 1.91 mg = 82.3 - mg
136.756 - 82.3 = 1.91 mg - mg
0.91 mg = 54.456
mg = 54.456/0.91
mg = 59.84 N
Hence, the weight of the bowling ball = 59.84 N
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Consider two conducting spheres with one having a larger radius than the other. Both spheres carry the same amount of excess charge. Which one of the following statements concerning the potential energy of the two spheres is true? O The potential energy of the larger sphere is greater than that of the smaller sphere. The potential energy of the larger sphere is the same as that of the smaller sphere. O The potential energy of the larger sphere is less than that of the smaller sphere.
Answer: Option (a) is the correct answer.
Explanation:
It is known that potential energy is the energy occupied by an object or substance due to its position is known as potential energy.
Therefore, more is the space occupied by an object more will be its position at a particular location. Hence, more will be its potential energy. On the other hand, smaller is the space occupied by an object, smaller will be the position holded by it.
Hence, smaller will be its potential energy.
Thus, we can conclude that for the given situation the statement, potential energy of the larger sphere is greater than that of the smaller sphere, is true.
Final answer:
The potential energy of the larger conducting sphere is less than that of the smaller sphere because electric potential is inversely proportional to radius and both spheres carry the same charge.
Explanation:
When two conducting spheres with the same excess charge but different radii are considered, the potential of each sphere is given by V = kQ/R, where V is the potential, k is Coulomb's constant, Q is the charge, and R is the radius of the sphere. Since the charge Q is the same for both spheres, but the radius R is larger for one of them, the electric potential energy will be higher for the smaller sphere since it has a smaller radius. This stems from the fact that electric potential is inversely proportional to the radius, meaning a smaller radius results in a higher potential for the same amount of charge.
The correct statement is: (d)he potential energy of the larger sphere is less than that of the smaller sphere.
The increase of carbon dioxide in the atmosphere has had which effect on the earth’s climate?
Collisions in two dimensions is a difficult section to read and understand. But, when we look at momentum similarly to our 'Sum of our Forces', then we can use the same concept to solve these problems. This means that the momentum in the x direction before the collision has to be as the momentum after the collision. The momentum in the y direction before the collision has to be as the momentum after the collision.1. True2. False
Answer:
2. TrueExplanation:
Since momentum is a vector, you, indeed, in two dimension collisions, you can decompose it in two components, the x-direction and the y-direction, such as you do with the force, which is a vector too.
The law of conservation of momentum states that the total momentum before and after the collision are conserved.
Let's assume a collision in one dimension: x-direction.
If object A is moving to the right, its momentum is to the right. If objcet B is at rest its momentum is zero. Then, if when object A collides with object B, the first stops, the second must move to the right with a momentum in the x-direction equal to the momentum that object A initially had.
You can apply the same reasoning if object A is moving in two dimensions, and, a similar one, if object B is not at rest: at the end the momentum in each direction before the collision has to be equal to the momentum in each direction after the collision.
The Department of Natural Resources has selected certain species that are on the verge of extinction due to a lack of prey. These species are used for game sport. In an effort to protect the species, the department has started arranging for food for these species. Which method is the department adopting?
Answer:
Habitat manipulation
Explanation:
Habitat manipulation, otherwise known as ecological engineering, is a technique of promoting natural enemies within an ecosystem by making thriving conditions more suitable for them.
In this case, thriving conditions for the species (which happens to be predators and hence, natural enemies) were promoted via artificial introduction of food.
Tevin left his house and rode his bike into town at 6 mph. Along the way he got a flat so he had to turn around and walk his bike back to his house traveling 3 mph. If the trip down and back took 12 hours, how far did he get before his tire went flat?
Answer:
Distance covered by Tevin before his tire went flat = 24 miles
Explanation:
Let x be the distance covered by Tevin before his tire went flat.
Given:
Tevin drives his bike in the town = 6 mph
Tevin back to his house = 3 mph
Total taken time by Tevin = 12 hours
We need to find the distance covered by Tevin in 12 hours.
Solution:
Using speed formula
[tex]Speed =\frac{Distance}{Time}[/tex]
We write the above formula for Time.
[tex]Time=\frac{Distance}{Speed}[/tex]-----------(1)
Time taken by Tevin when biking in town
Substitute speed = 6 mph and distance = x in equation 1.
[tex]Time=\frac{x}{6}[/tex] ----------(2)
Time taken by Tevin when he is back to his house.
Substitute speed = 3 mph and distance = x in equation 1.
[tex]Time=\frac{x}{3}[/tex] -------------(3)
Total time taken by Tevin.
Total taken time by Tevin = Time taken when biking in town + Time taken when Tevin back to his house.
Substitute time value from equation 2 and 3 in above equation and total time = 12 hours.
[tex]12=\frac{x}{6}+\frac{x}{3}[/tex]
Now, we solve the above equation for x.
[tex]12=\frac{x+2x}{6}[/tex]
[tex]12\times 6=x+2x[/tex]
[tex]72=3x[/tex]
[tex]x=\frac{72}{3}[/tex]
[tex]x=24\ mi[/tex]
Therefore, distance covered by Tevin in 12 hours is equal to 24 miles.
a 360 mile trip began on a greeway in a car traveling at 62 mph. Once the road became a 2 lane highway, the car slowed to 54 mph. If the total trip took 6 hours, find the time spent on each type of road
Answer:
Time spent on the greenway road = 4.5 hours
Time spent on the 2 lane road = 1.5 hours
Explanation:
The distance of the trip is 360 miles and the initial speed of the car is 62 miles/hr and after the road became 2 lane highway the car slowed to 54 miles/hr.
Let us divide the trip into two
Greenway
speed = distance/time
speed = 62 mph
time = a
distance = speed × time
distance = 62a
2 lane highway
speed = distance/time
speed = 54 mph
time = b
distance = speed × time
distance = 54b
Total distance
62a + 54b = 360......................(i)
Total time
a + b = 6..............................(ii)
a = 6-b
insert a in equation (i)
62(6-b) + 54b = 360
372 - 62b + 54b = 360
-8b = 360-372
-8b = - 12
b = 12/8
b = 1.5
from equation (ii)
a + 1.5 = 6
a = 6 - 1.5
a = 4.5
A sample of octane (C8H18) that has a mass of 0.750 g is burned in a bomb calorimeter. As a result, the temperature of the calorimeter increases from 21.0°C to 41.0°C. The specific heat of the calorimeter is 1.50 J/(g • °C), and its mass is 1.00 kg. How much heat is released during the combustion of this sample? Use . 22.5 J 30.0 J 31.5 J 61.5 J
Answer:
Explanation:
mass, m = 1 kg
specific heat, c = 1.5 J/g°C
rise in temperature, ΔT = 41 - 21 = 20
heat released, H = m x c x ΔT
H = 1 x 1.5 x 1000 x 20
H = 30,000 J
H = 30 kJ.
Answer:
A) 22.5 JExplanation:
on edg I got it right
How did ernest rutherford discover the nucleus of the atom
Answer:
THE ANSWER IS: Rutherford, Marsden and Geiger discovered the dense atomic nucleus by bombarding a thin gold sheet with the alpha particles emitted by radium
Explanation:
Which technique provides a smooth transition from acceleration to braking?
Answer:
Cover-braking technique
Explanation:
Cover braking involves the technique of taking off the foot from the accelerator and hovering it over the braking pedal. The foot must not be placed on the braking pedal to avoid the wear of brake unnecessarily.
It is done usually when an obstacle is expected in front of the moving car. This provides an edge in the reaction time of application of the brakes hence reducing the stopping time.
Regenerative braking and engine braking are techniques used for a smooth transition from acceleration to braking. These techniques convert kinetic and potential energy into electrical energy or use engine power to slow down the vehicle, respectively.
Explanation:The technique that provides a smooth transition from acceleration to braking is commonly recognized in physics and engineering called regenerative braking. Regenerative braking is a central component of the operation of hybrid and electric vehicles. It functions by converting a vehicle's kinetic and gravitational potential energy into electrical energy that recharges the vehicle's battery when deceleration is initiated.
Another method is engine braking, usually applied in larger vehicles like trucks to avoid overheating of brakes specially when traveling downhill.
Both methods, whether it is engine braking or regenerative, provide a smooth progression from accelerated motion to a decrease in speed, or braking. This way, they manage the transition smoothly while conserving and efficiently using energy.
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