Answer:
Explanation:
The deflection of a charged particle by a magnetic field is proportional to its electric charge and to its velocity. The deflection is also inversely proportional to its mass. So given a proton and an electron going at the same velocity in a magnetic field and having equal (but opposite) electric charge the electron will deflect much more since the ratio of the masses is 1836.
what would happen if gravity were to stop everywhere?
Answer:
everything will float up and go up to space and die
Explanation:
gravity keeps us down and once it stops everything will float up. And if it were to stop everywhere everything and everyone will die and everything will be destroyed.
The 6 strings on a guitar all have about the same length and are stretched with about the same tension. The highest string vibrates with a frequency that is 4 times that of the lowest string. 1)If the strings are made of the same material, how would you expect the diameters of the lowest and highest strings to compare
Answer:
not sure i need points
Explanation:
After passing point 2 the hill becomes frictionless and the ring's rotational velocity remains constant. What is the linear velocity of the ring at point 3 in m/s
The energy in the system is given by the initial potential energy at the point 1.
The linear velocity at point 3, is approximately 33.59 m/s.
Reasons:
The parameters are;
Height at point 1, h₁ = 83 m
Radius of the ring = 8 cm
Mass of the ring, M = 8 kg
Height at point 2, h₂ = 32 m
At point 2, we have;
Change in potential energy = Kinetic energy
Which gives;
(83 - 32) × 9.81 × 8 = 0.5 × 8 × v² + 0.5 × 8 × 0.08² × (v/0.08)²
Which gives;
v ≈ 22.37 m/s
At point 3, the rotational kinetic energy remains constant while the
translational kinetic energy increases as follows;
K.E. at point 3 = Initial kinetic energy + Change in potential energy
Which gives;
K.E. at point 3 = 0.5 × 8 × v₃³ ≈ 0.5×8×22.37² + 32×9.81×8
[tex]v_3^2 = \dfrac{0.5 \times 8 \times 22.37^2 + 32 \times 9.81 \times 8}{0.5 \times 8} = 1128.15[/tex]
v₃ ≈ √(1128.15) ≈ 33.59
The linear velocity at point 3, v₃ ≈ 33.59 m/s
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The probable question parameters as obtained from a similar question online are;
Height at point 1, h₁ = 83 m
Radius of the ring = 8 cm
Mass of the ring, M = 8 kg
Height at point 2, h₂ = 32 m
Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.120 rev/s . As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight.
Answer:
Explanation:
0.120 rev/s(2π rad/rev) = 0.24π rad/s
At the highest point of the arc, gravity must supply the required centripetal acceleration. As the normal force is 1/4 of his normal weight, then 3/4 of gravity acceleration must be used as centripetal acceleration
0.75g = ω²R
R = 0.75(9.81) / (0.24π)²
R = 12.942198...
R = 12.9 m
The radius of the circle is equal to 12.95m which is rotating with an angular velocity of 0.120 rev/s.
What is vertical circular motion?A body spins in a vertical circle so that its motion at different points is different from the motion of the body is said to be vertical circular motion.
The velocity and tension vary in maximum magnitude from the lowest to the highest position because of the effect of the gravitational force of the earth.
Given, the angular velocity of the Ferris wheel, ω = 0.120 rev/s
ω = 0.120 rev/s × 2π rad/rev
ω= 0.7536 rad/s
If r is the radius of the circle and 'm' is the mass of the jack.
From newton's second law of motion, the net force will be equal to
mg - N = mrω²
mg - (mg/4) = mrω²
r = 3g/4ω²
r = 3×9.81 / (4× 0.7536)
r = 12.95 m
Therefore, the radius of the circle in which the jack travels is equal to 12.95m.
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Your question is incomplete, most probably the complete question was,
Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.120 rev/s. As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight. What is the radius of the circle in which Jack travels?
A man walks 30m to the west, then 5m to the East in 45 seconds. What is the total distance walked?
Answer:
In this problem basically we will use directions to find displacement and distance .
(Keep in mind that displacement is vector while distance is scalar quantity)
And then we will use displacement and distance to find average velocity and average speed.
Explanation:
A car is driving 12m/sec, has to stop suddenly because a pedestrian dashes out in front of the car. If the coefficient of kinetic friction between the tires and parking lot is ∪=60
what is the time, after the breaks are applied, before the car comes to a stop? Sketch the velocity time graph for the car's motion from the instant the breaks are applied until the car comes to a stop.
Answer:
Approximately [tex]2\; \rm s[/tex], assuming that the floor of this parking lot is level, [tex]\mu_{\rm k} = 0.60[/tex], and [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].
Explanation:
Let [tex]m[/tex] denote the mass of this vehicle. Weight of this vehicle: [tex]m\, g[/tex].
If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight: [tex]N = m \, g[/tex].
Given that [tex]\mu_{\rm k}[/tex], the kinetic friction between this vehicle and the ground would be consistently [tex]\mu_{\rm k} \, N = \mu_{\rm k} \, m \, g[/tex] until the vehicle comes to a stop.
Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be [tex](-\mu_{\rm k} \, m \, g)[/tex] (negative because this force is opposite to the direction of the motion.)
By Newton's second law of motion, the acceleration of this vehicle would be:
[tex]\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= \frac{-\mu_{\rm k} \, m \, g}{m} \\ &= -\mu_{\rm k}\, g \\ &= -0.60 \times 9.81\; \rm m\cdot s^{-2} \\ &= -5.886\; \rm m\cdot s^{-2}\end{aligned}[/tex].
In other words, braking would reduce the velocity of this vehicle by a constant [tex]5.886\; \rm m\cdot s^{-1}[/tex] every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from [tex]v_{0} = 12\; \rm m\cdot s^{-1}[/tex] to [tex]v_{1} = 0\; \rm m\cdot s^{-1}[/tex]:
[tex]\begin{aligned}t &= \frac{v_{1} - v_{0}}{a} \\ &= \frac{0\; \rm m\cdot s^{-1} - 12\; \rm m\cdot s^{-1}}{-5.886\; \rm m \cdot s^{-2}} \\ &\approx 2.0\; \rm s \end{aligned}[/tex].
Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.
A ball is dropped from an 80.0 m building. What is the ball's velocity after 3.00 s? Use an order-of-magnitude estimation to identify the correct choice.
A. -2.9 m/s
B. -29.4 m/s
C -8.8 m/s
D. -88.3 m/s
Answer:b
Explanation:
-29.4 m/s
The velocity of the ball dropped from 80 m if it reaches the ground within 3 seconds is 26.6 m/s. If it is in midway within this time, then the velocity will be 29.4 m/s.
What is velocity ?Velocity of a moving object is the measure of its distance travelled per unit time. Velocity is a vector quantity having both magnitude and direction. Acceleration is the rate of change in velocity.
Given that, height of the building = 80 m
the ball is moving downwards by acceleration due to gravity g = 9.8 m/s².
Then after 3 seconds, the velocity of the ball is calculated as follows:
velocity = acceleration × time
v = 9.8 m/s² × 3 s = -29.4 m/s
If the ball reaches the ground within the time of 3 s, then, the velocity is:
v = 80 m/3s = 26.6 m/s.
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Anita Knapp needs to get hay to cows in a frozen field using an airplane flying
80.0 m/s, at a height of 300,m. If at the last minute, how far from the cow would
she have to release the hay in order to hit the cow?*
756 m
626m
700m
575 m
Other:
Answer:
626m
Explanation:
Read the sentence from the text. “They are as glossy as satin or sunlight reflecting off water!" What does the word glossy mean in the sentence? O A. pointed o B. shiny O C. small O D. strong
Answer:
b Shiny
Explanation:
Trust me it's right
True or False: The basketball should be dribbled below the waist.
Elevations on the tongue are called
sulci
taste buds
papillae
gyri
Answer:
Papillae is correct
Explanation:
hope it helps you
Answer:
Papillae is the correct answer of this question
Acceleration of a Car A car traveling along a straight road at accelerated to a speed of over a distance of ft. What was the acceleration of the car, assuming that it was constant
Answer:
how many feet?
Explanation:
Conservation of Energy Roller Coaster A roller coaster cart of mass 100kg travels on a track with one loop. Fill in blanks A-H. А. KE=OJ PE=120000J h= А. V= B B KE=___CE PE=60000J h= _D V= E KE=__F PE=40000J h=__G_ V= KE= PE= h=Om v= K D E F G H K
(a) The height of the roller coaster at 120,000 potential energy is 122.45 m.
(b) The velocity of the roller coaster at 0 J kinetic energy is 0.
(c) The height of the roller coaster at 60,000 potential energy is 61.23 m.
(d) The velocity of the roller coaster at 60,000 J kinetic energy is 34.64 m/s.
(e) The height of the roller coaster at 40,000 potential energy is 40.82 m.
(f) The velocity of the roller coaster at 80,000 J kinetic energy is 40 m/s.
The given parameters:
mass of the roller coaster, m = 100 kgWhen the kinetic energy = 0 and potential energy = 120,000 J
The height of the roller coaster is calculated as follows;
P.E = mgh
[tex]h = \frac{P.E}{mg}\\\\h = \frac{120,000}{100 \times 9.8} \\\\h = 122.45 \ m[/tex]
Since the kinetic energy = 0, the velocity of the roller coaster = 0
When the potential energy, P.E = 60,000 J, the kinetic energy, K.E is calculated as;
P.E + K.E = M.A
P.E + K.E = 120,000
60,000 + K.E = 120,000
K.E = 120,000 - 60,000
K.E = 60,000 J
The height of the roller coaster at 60,000 potential energy is calculated as follows;
[tex]h = \frac{P.E}{mg} \\\\h = \frac{60,000}{100 \times 9.8} \\\\h =61.23 \ m[/tex]
The velocity of the roller coaster at 60,000 J kinetic energy is calculated as follows;
[tex]K.E = \frac{1}{2} mv^2\\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{ \frac{2K.E}{m}} \\\\v = \sqrt{ \frac{2\times 60,000}{100}}\\\\v = 34.64 \ m/s[/tex]
When the potential energy, P.E = 40,000 J, the kinetic energy, K.E is calculated as;
P.E + K.E = M.A
40,000 + K.E = 120,000
K.E = 120,000 - 40,000
K.E = 80,000
The height of the roller coaster at 40,000 potential energy is calculated as follows;
[tex]h = \frac{P.E}{mg} \\\\h = \frac{40,000}{100 \times 9.8} \\\\h = 40.82 \ m[/tex]
The velocity of the roller coaster at 80,000 J kinetic energy is calculated as follows;
[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 80,000}{100} } \\\\v = 40 \ m/s[/tex]
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Match the sport/physical activity in column B with the primary physical fitness component needed to perform it in column A . Write the letters of your answer in your activity notebook.
1. Power
A. Patintero
2. Speed
B. Marathon
3. Balance
C. dodgeball / tamaan bata
4. Coordination
D. 100m sprint
5. Flexibility
E. badminton and table tennis
6. Muscular Strength
F. exercise and proper diet
7. Agility
G. hopscotch/piko
8. Cardiorespiratory Endurance
H. Shotput
9. Reaction time
I. Archery
J. Leg Splits and yoga poseso
sagutan po plss
. Power
Patintero
, also known as harangang-taga or tubigan, (Intl. Translate: Escape from the hell or Block the runner) is a traditional Filipino children's game. Along with tumbang preso, it is one of the most popular outdoor games played by children in the Philippines.[1]
2. Speed The Barkley Marathonsis an ultramarathon trail race held in Frozen Head State Park near Wartburg, Tennessee. If runners complete 60 miles (97 km) this is known as a "fun run." The full course is about 100 miles (160 km). The race is limited to a 60-hour period and takes place in late March or early April of each year.
Which of the following is a fundamental unit of the SI system of units?
A) kilometer
B) joule
C) kilogram
D) gram
E) newton
The answer is C) kilogram
The SI unit in the given options is kilogram, the correct option is C.
What is SI unit?The most common unit of measurement is the standard unit. These units are standardized, which means that everyone understands an object's size, weight, and other properties.
Standard Units of Measurement have fixed values that cannot be changed.
The International System of Units, abbreviated SI in all languages and sometimes pleonastically as the SI system, is the modern version of the metric system and the world's most widely used measurement system.
The SI is a coherent system of units of measurement that begins with seven base units: the second (symbol s, the unit of time), the metre (m, length), the kilogram (kg, mass), the ampere (A, electric current), and so on.
Thus, the correct option is C.
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what are saturated and non saturated vapour?
Answer:
A saturated vapour is one that is in balance with its own liquid. Space is considered to be unsaturated if it contains fewer vapours than the maximum amount it can retain at a given temperature. Saturated vapour pressure is unaffected by volume, although it rises as temperature rises.
Explanation:
Lab report on velocity of sound
Rachel drops a ball from a hot–air balloon while her friend Lisa is watching her from the ground. Which statement about the ball's motion is true from Lisa's point of view?
Assume that there is no air resistance and the hot–air balloon is moving horizontally.
A. The ball drops to the ground along a straight–line path.
B.When the ball lands, the hot–air balloon will be ahead of it.
C. When the ball lands, the hot–air balloon will be behind it.
D. When the ball lands, the hot–air balloon will be directly above it.
Answer:
According Lisa, both the ball and the balloon have the same forward velocity of Vx.
(D) is correct
Two blocks are set in a pully system as shown in fig below. Block A sits on the frictionless table while block B hags freely. The pully is light and frictionless towards the light string that runs over it. If the Block A has mass of 3.4 kg and Block has 3.5 kg, what would be the magnitude of the acceleration (in ms-2) of the blocks? [g = 9.8 ms=2]
Answer:
Explanation:
F = ma
a = F/m
a = mBg / (mB + mA)
a = 3.5(9.8)/(3.5 + 3.4)
a = 4.971014...
a = 5.0 m/s²
If you want to use individual Free Body Diagrams
mass A will have downward weight and upward normal forces equal at mAg
and a horizontal force of string tension T
F = ma
T = mAa
mass B will have a downward force of mBg and an upward force of T
mBg - T = mBa
substitute for T
mBg - mAa = mBa
mBg = a(mB + mA)
a = mBg / (mB + mA) which is identical to the above answer.
Water has higher specific heat than aluminum. This is all what you should know to answer following questions. An aluminum rod of mass 1 kg at temperature of 80^0C is placed into 1l of water of temperature 10^0C . there is no heat exchange with surroundings. Which material experiences greater change in temperature while system is reaching the thermal equilibrium
Answer:
The Aluminum
Explanation:
With a larger specific heat, water requires more heat to raise its temperature by a temperature degree.
In this system, with equal masses of water and aluminum, the heat moving from the aluminum lowering its temperature by one degree is not sufficient to raise the water temperature by one degree.
Use a trigonometric equation to
determine the leg of this triangle.
10 m
[?]
m
30°
Answer:
5m
Explanation:
x=?
[tex]sin30^{0} =\frac{x}{10}[/tex]
[tex]x=10sin30^{0} =10(0.5)=5[/tex]
Hope this helps
Uranus (mass = 8.68 x 1025 kg) and its moon Miranda (mass = 6.59 x 1019 kg) exert a gravitational force of 2.28 x 1019 N on each other. How far apart are they? cs [?] x 10?'m Coefficient (green) Exponent (yellow) Enter
Answer:
Explanation:
F = GMm/d²
d = √(GMm/F)
d = √(6.674e-11(8.68e25)(6.59e19) / 2.28e19)
d = 1.29398e8 = 1.29 x 10^8 m center to center
Answer:
1.29 x 10^8 m apart
Explanation:
Works in Acellus!
tell types of instruments
Answer:
Instrument Use
Accelerometer Measures acceleration
Altimeter Measures altitude of an aircraft
Ammeter Measures electric current in ampere
Anemometer Measures wind speed
A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so small that it can be ignored. A sandbag of mass m is dropped onto the merry-go-round, at a position designated by r. The sandbag does not slip or roll upon contact with the merry-go-round.
The figure shows a top view of a merry-go-round of radius capital R rotating counterclockwise. A sandbag is located on the merry-go-round at a distance lowercase r from the center.
Rank the following different combinations of m and r on the basis of the angular speed of the merry-go-round after the sandbag "sticks" to the merry-go-round.
The angular speed of the merry-go-round reduced more as the sandbag is
placed further from the axis than increasing the mass of the sandbag.
The rank from largest to smallest angular speed is presented as follows;
[m = 10 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 20 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 0.5·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 1.0·R]
Reasons:
The given combination in the question as obtained from a similar question online are;
1: m = 20 kg, r = 0.25·R
2: m = 10 kg, r = 1.0·R
3: m = 10 kg, r = 0.25·R
4: m = 15 kg, r = 0.75·R
5: m = 10 kg, r = 0.5·R
6: m = 40 kg, r = 0.25·R
According to the principle of conservation of angular momentum, we have;
[tex]I_i \cdot \omega _i = I_f \cdot \omega _f[/tex]
The moment of inertia of the merry-go-round, [tex]I_m[/tex] = 0.5·M·R²
Moment of inertia of the sandbag = m·r²
Therefore;
0.5·M·R²·[tex]\omega _i[/tex] = (0.5·M·R² + m·r²)·[tex]\omega _f[/tex]
Given that 0.5·M·R²·[tex]\omega _i[/tex] is constant, as the value of m·r² increases, the value of [tex]\omega _f[/tex] decreases.
The values of m·r² for each combination are;
Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²
Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²
Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²
Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²
Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²
Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²
Therefore, the rank from largest to smallest angular speed is as follows;
Combination 3 > Combination 1 > Combination 5 = Combination 6 >
Combination 2
Which gives;
[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =
10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].
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Find the self-inductance of a 1700-turn solenoid 55 cm long and 4.0 cm in diameter. Express your answer with the appropriate units.
The self-inductance of the solenoid is 8.25 mH.
The given parameters;
number of turns, N = 1700 turnlength of the solenoid, l = 55 cm = 0.55 mdiameter of solenoid, d = 4 cmradius of the solenoid, r = 2 cm = 0.02 mThe area of the solenoid is calculated as follows;
[tex]A = \pi r^2\\\\A = \pi \times (0.02)^2\\\\A = 0.00125 \ m^2[/tex]
The self-inductance of the solenoid is calculated as follows;
[tex]L = \frac{\mu_o N^2 A }{l} \\\\L = \frac{(4\pi \times 10^{-7}) \times 1700^2 \times 0.00125}{0.55} \\\\L = 0.00825 \ H\\\\L = 8.25\ mH[/tex]
Thus, the self-inductance of the solenoid is 8.25 mH.
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give with an example a cause where the velocity of an object is zero but its acceleration is not zero .
Answer:
At the highest point when you toss a ball into the air.
Explanation:
At the higest point of a trajectory of a ball, the velocity is zero for a split second and there is no speed and direction. However, there still is acceleration of -10 m/s^2 because the force of gravity is still acting upon it at that point.
Hi there!
An example of this could be when a ball is thrown vertically into the air and reaches the TOP of its trajectory.
When an object is thrown with a vertical velocity, the acceleration due to gravity results in a decrease in its positive (upward) velocity until it reaches its highest point, where the instantaneous velocity = 0 m/s and the object begins to fall back down (negative velocity).
Additionally, throughout its entire trajectory, the ball experiences an acceleration due to gravity of g = 9.8 m/s², even at its highest point where there is a velocity = 0 m/s.
6) An object is released from rest at the top of a ramp inclined at 30. degrees up from the horizontal. Due to friction, the ramp is only 20. % efficient. What is the object's speed after it slides down ALONG the ramp for 2.0 m? *
Answer:
Explanation:
I've been doing these types of problems for many years and I don't think I've ever seen an "efficiency" rating on a ramp.
I'm going to ASSUME that 20% efficient means that 80% of the Potential energy that gets converted becomes system internal heat energy.
Potential energy at the start of a 2.0 m slide
PE = mgh = mg2sin30 = mg2(½) = mg J
0.8mg J gets converted to heat and 0.2mg converts to kinetic energy
0.2mg = ½mv²
v² = 0.4g
v = √(0.4(9.8)) = 1.979898... ≈ 2.0 m/s
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According to "Patterns of Change," select the ways that rocks are changed naturally over time. Choose three answers. O A. water O B. location O C. animals D. wind O E. people O F temperature
Answer: A , D & F
Explanation:
A baseball player notices the ball when it is 3.4 m above the
ground, traveling at 4.4 m/s. He wants to make the catch when
the ball is 1.5 m above the ground, how long does it take to reach
his glove?
Find the distance the ball travels:
3.4 meters - 1.5 meters = 1.9 meters
Now divide the distance the ball travels by the speed:
1.9 meters / 4.4 m/s = 0.43 seconds
Answer:
Explanation:
s = s₀ + v₀t + ½at²
There are an infinite number of solutions to this question as posed because we are not told the direction of the initial velocity.
Assuming ground is level and origin and UP the positive direction
The shortest amount of time possible is when the initial velocity is straight down
1.5 = 3.4 - 4.4t + ½(-9.8)t²
0 = -4.9t² - 4.4t + 1.9
t = (4.4 ±√(4.4² - 4(-4.9)(1.9))) / (2(-4.9))
positive answer is
t = 0.32 s
The longest amount of time possible is when the initial velocity is straight up.
1.5 = 3.4 + 4.4t + ½(-9.8)t²
0 = -4.9t² + 4.4t + 1.9
t = (-4.4 ±√(4.4² - 4(-4.9)(1.9))) / (2(-4.9))
positive answer
t = 1.22 s
If the initial velocity is horizontal, meaning no vertical velocity
1.5 = 3.4 + 0t + ½(-9.8)t²
-4.9t² = -1.9
t² = 0.38775...
t = 0.62 s
Any angle between UP and Down will have a different initial vertical velocity and result in a different time to catch height.
It appears from the comments on the other answer, that I have shown you how to arrive at three of the four possible solutions. The initial direction is very important.
A Pump discharges water at 1MPaa and 165 deg C. Determine the specific volume and internal energy of the water at the discharge point.
This problem is describing a pump from which water is discharged at 1 MPa and 165 °C and is asking for the specific volume and internal energy at those conditions, thus, we can use the steam tables for resolving this requirement.
First of all, we need to remember that water can be a saturated liquid, vapour or liquid-vapour mixture, and this is determined for the temperature and pressure it is at.
In this case, we find that at 165 °C the saturation pressure is about 0.6178 MPa; this means we are referring to a saturated liquid so that both the specific volume and internal energy can be simply read from the steam tables as vf and uf as follows:
[tex]v=0.001127\frac{m^3}{kg}\\\\u=761.67 \frac{kJ}{kg}[/tex]
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