(a) The work done in moving the crate along the floor is 7,050 J.
(b) The work done in moving the crate vertically is 0 J.
Work done in moving the crate horizontally
W = Fdcosθ
where;
F is total force to be appliedd is displacement of the crateθ is the horizontal angle = 0W = (1100 + 310) x 5
W = 7,050 J
Work done in moving the crate verticallyW = Fd cos(90)
W = 0
Thus, the work done in moving the crate along the floor is 7,050 J.
the work done in moving the crate vertically is 0 J.
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In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
The force required to pull the two hemispheres apart 53696.25N and the minimum number of horses required is 37 .
( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 )
(1)The contact area between the hemispheres is (π x 0.430²) = 0.5805m²
Pressure difference = (940 - 15) = 925millibars.
(925 x 100) = 92,500N/m^2.
(92500 x 0.5805) = 53696.25N. is the force required to part the hemispheres.
(2)[tex]\frac{53696.25N}{1450N}[/tex] = 37 horses for each side .
37 + 37 = 74 horses will be required.
Force is something which can change the motion of an object, stop it or move it, change its shape or size with it. There are two types of forces, contact forces and non-contact forces. Here, it is a contact force at first, then when the horses come it becomes non-contact force.
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The following table lists the speed of sound in various materials. Use this table to answer the question.
Substance Speed (m/s)
Glass 5,200
Aluminum 5,100
Iron 4,500
Copper 3,500
Salt water 1,530
Fresh water 1,500
Mercury 1,400
Hydrogen at 0°C 1,284
Ethyl Alcohol 1,125
Helium at 0°C 965
Air at 100°C 387
Air at 0°C 331
Oxygen at 0°C 316
Sound will travel fastest in air at _____.
-5°C
0°C
10°C
15°C
Sound will travel fastest in air at 15°C.
Speed of sound in air
The speed of sound in air, given in the range of 100 degrees Celsius and 0 degree Celsius include;
Air at 100°C 387 m/s
Air at 0°C 331 m/s
From the date above, the speed of sound in air increases with increases in temperature. Thus, Sound will travel fastest in air at 15°C.
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200 g of water is heated and its temperature goes from 280 K to 300 K. What was the change in temperature for this process?
A. 280 K
B. 20 F
C. 20 K
D. 300 K
The change in temperature is 20 kelvin
What is Temperature?
Temperature can simply be described as how hot or how cold an object is at a particular period in time. The unit of temperature is Kelvin.
The formula for calculating change in temperature is
Final temperature - Initial temperature
Final temperature = 300 kelvin
Initial temperature= 280 kelvin
Change in temperature= 300-280
= 20 kelvin
Thus the change in temperature for this process is 20 kelvin
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In Milikan’s experiment, a drop of radius of 1.64 μm and density 0.851 g/cm3 is suspended in the lower chamber when a downward-pointing electric field of 1.92 * 105 N/C is applied.
What is the weight of the drop?
Find the charge on the drop, in terms of e.
How many excess or deficit electrons does it have?
(a) The weight of the drop is 1.54 x 10⁻²⁵ N,
(b) The the charge on the drop, in terms of e is 5 x 10⁻¹²e and
(c) The excess electrons is 5 x 10⁻¹² electron.
Weight of the dropThe weight of the drop is calculated as follows;
Volume of the drop; V = ⁴/₃πr³
V = ⁴/₃π(1.64 x 10⁻⁶)³ = 1.845 x 10⁻¹⁷ m³
mass = density x volume
mass = 0.851 g/cm³ x 1.845 x 10⁻²³ cm³ = 1.572 x 10⁻²³ g = 1.57 x 10⁻²⁶ kg.
Weight = 1.57 x 10⁻²⁶ kg x 9.8 m/s² = 1.54 x 10⁻²⁵ N.
Charge on the dropq = F/E
q = (1.54 x 10⁻²⁵ N)/(1.92 x 10⁵)
q = 8.01 x 10⁻³¹ C
1.6 x 10⁻¹⁹ C = 1e
8.01 x 10⁻³¹ C = ?
= 5 x 10⁻¹²e
Excess electron on the drop1.6 x 10⁻¹⁹ C ------- 1 electron
8.01 x 10⁻³¹ C ------- ?
= 5 x 10⁻¹² electron
Thus, the weight of the drop is 1.54 x 10⁻²⁵ N, the the charge on the drop, in terms of e is 5 x 10⁻¹²e and the excess electrons is 5 x 10⁻¹² electron.
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. What is the potential energy of a 0.40 kg ball at a height of 9.2 m?
Explanation:
gravitational potential energy = mgh (must be in S.I. unit)
m= 0.4 kg ; g= 10m/s (gravitational acceleration occurs); h=9.2 m
hence mgh=0.4×10×9.2= 36.8J
unit for energy is joules and since the variables are in S.I. unit, we can use Joules as the final unit for measurement
Answer:
36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s²
Explanation:
Potential Energy = Mass x Gravitational Acceleration x Height
It can be expressed as [tex]\displaystyle{PE = mgh}[/tex] where PE is potential energy, m is mass, g is gravitational acceleration and h is height.
In this case, we know mass is 0.40 kg and height of 9.2 m as well as gravitational acceleration is defined to be 9.8 m/s² (You can also define g = 10 m/s²)
Therefore, substitute given information in formula:
[tex]\displaystyle{PE=0.40 \ \times \ 9.8 \ \times 9.2 }\\\\\displaystyle{PE=36.06 \ J}[/tex]
With g = 10 m/s², you'll get:
[tex]\displaystyle{PE = 0.40 \ \times 10 \ \times 9.2}\\\\\displaystyle{PE = 36.8 \ J}[/tex]
Note that J is for joule unit.
Therefore, the answer is 36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s² - both work.
The motor of a weed trimmer spins
at 9,000 rpm. The amount of time
required for the motor to reach this
speed would NOT be affected by...
A. the distribution of
trimmer line inside the
spool
B. the mass of the spool
containing the trimmer
line
C. the direction in which
the torque is applied
The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool. Option A
What is a weed trimmer?The weed trimmer is a device that is used to trim the grasses on a lawn or a field. This device has a rotating shaft that does the actual trimming of the grasses.
The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool thus it does not affect the amount of time required to reach 9,000 rpm speed.
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A 149-g baseball is dropped from a tree 15.0 m above the ground.
With what speed would it hit the ground if air resistance could be ignored?
Express your answer to three significant figures and include the appropriate units.
If it actually hits the ground with a speed of 9.00 m/s , what is the magnitude of the average force of air resistance exerted on it?
Express your answer to three significant figures and include the appropriate units.
1. The speed with which the ball hits the ground is 17.1 m/s
2. The magnitude of the average force of air resistance exerted on it is 0.77 N
1. How to determine the velocity with which the ball hits the groundInitial velocity (u) = 0 m/s Acceleration due to gravity (g) = 9.8 m/s²Height (h) = 15 m Final velocity (v) =?v² = u² + 2gh
v² = 2gh
Take the square root of both side
v = √(2 × 9.8 × 15)
v = 17.1 m/s
2. How to determine the forceWe'll begin by calculating the time to reach the ground. This is illustrated below:
Acceleration due to gravity (g) = 9.8 m/s²Height (h) = 15 m Time (t) =?h = ½gt²
15 = ½ × 9.8 × t²
15 = 4.9 × t²
Divide both side by 4.9
t² = 15 / 4.9
Take the square root of both side
t = √(15 / 4.9)
t = 1.75 s
Now we can determine the force. This can be obtained as illustrated below:
Mass (m) = 149 g = 149 / 1000 = 0.149 KgInitial velocity (u) = 0 m/sFinal velocity (v) = 9 m/sTime (t) = 5 ms = 1.75 sForce (F) = ?F = m(v –u) / t
F = 0.149(9 – 0) / 1.75
F = 0.77 N
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Forces always act in ________. A. Solitude B. Pairs C. Unpredictable ways D. Interesting ways
Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in the vein is 35.7 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?
The minimum necessary height of the IV bag above the position of the needle, if the density of the solution is 1.308kg/L, and the pressure in the vein is 35.7 mmHg is, 2.78m.
To find the answer, we need to know more about the pressure exerted by a liquid column.
How to find the height of IV bag above the position of needle?Consider a liquid of density ρ contained in a vessel of height h, the pressure exerted by the liquid column at the bottom of the vessel is given by ,[tex]P=[/tex] ρgh
In our question, it is given that,[tex]density=1.308 kg/L\\\\P=35.7 mmHg.\\[/tex]
Thus, the height of the bag h will be,[tex]h=\frac{P}{density*g} =\frac{35.7}{1.308*9.8}\\\\ h=2.78 m[/tex]
Thus, we can conclude that, the height of the Iv bag is 2.78 m above the position of needle.
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If the solution has a density of 1.308 kg/L and the vein pressure is 35.7 mmHg, the minimum height of the IV bag above the needle position is 2.78 m.
We need to learn more about the pressure that a liquid column exerts in order to determine the solution.
How can I determine the height of IV bag above where the needle is?The pressure exerted by the liquid column at the bottom of the vessel is given by for a liquid with density enclosed in a vessel of height h.[tex]P=[/tex] ρgh
In our inquiry, it is assumed that,[tex]Density=1.308kg/L\\P=35.7mmHg[/tex]
As a result, the bag's height will be,[tex]h=\frac{P}{density*g}\\\\ h=2.78m[/tex]
As a result, we may say that the I.V. bag is 2.78 m above the ground.
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1. A block of mass 0.4kg resting on the top of an inclined plane of height 20m starts to slide down on the surface of the incline to its foot, and then continues its slide horizontally. At a distance of 5m from the foot of the incline there is another block of the same mass resting on the horizontal surface to undergo an elastic collision. Next to the second block, there is a light spring of constant k = 4000N/m fixed freely against a wall. The spring is supposed to make a head-on collision with the second block. See the arrangements as in 1. Assuming all surfaces being frictionless, (a) calculate the kinetic energy of the firs block just at the foot of the incline; (b) calculate the kinetic and gravitational potential energies of the first block halfway down the incline; (c) calculate the speeds of the two blocks just after their collision; (d) compute the maximum compression of the spring resulted from its collision with the second block; (e) determine the maximum work done by the spring on the second block.
The kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.
To find the answer, we need to know about the concept of collision and kinetic energy.
How to find the kinetic energy of the first block just at the foot of the incline?Given that, the block of mass 0.4kg resting on the top of an inclined plane of height 20m.Thus, at the top of the incline it has a potential energy, and the kinetic energy will be equal to zero, or we can say that the total energy of the system is equal to the potential energy at topmost point.[tex]TE=KE+PE\\T=PE=m_1gh=(0.4*9.8*20)=78.4 J[/tex]
We have to find the kinetic energy of the first block just at the foot of the incline, and at the bottom point the PE=0, or we can say that the total energy or the potential energy is converted into kinetic energy.[tex]TE=KE=78.4J[/tex]
What is the kinetic and gravitational potential energies of the first block halfway down the incline?At the halfway, the PE will be,[tex]U'=m_1gh'=mg\frac{h}{2} \\U'=39.2J[/tex]
As we know that, the energy is conserved at each point of the motion.[tex]TE=78.4 J\\KE'+PE'=78.4J\\KE'=78-U'=78.4-39.2=39.2J[/tex]
How to find the speeds of the two blocks just after their collision?We have the KE at bottom point as, 78.4J. Thus, the velocity of first block at the bottom before collision will be,[tex]KE=\frac{1}{2} mv^2=78.4J\\v=\sqrt{\frac{2KE}{m} } =4m/s[/tex]
This is the velocity of the block 1 of mass m1 before collision, we can say, u1.As we know that, the 2 nd block of mass m2 is at rest, thus, u2=0.Given that, the collision is elastic. Thus, both the KE and the momentum will be conserved.[tex]\frac{1}{2}m_1u_1^2+ \frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/tex]
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
We have,[tex]m_1=m_2=m\\u_1=4m/s\\u_2=0\\v_1=?\\v_2=?[/tex]
Substituting this in both the equations, we get,[tex]\frac{1}{2}m*4^2=\frac{1}{2}m(v_1^2+v_2^2)\\(v_1^2+v_2^2)=16[/tex] from resolving KE equation.
[tex]4m=m(v_2+v_1)\\4=v_2+v_1\\v_1=4-v_2[/tex] From resolving momentum conservation.
solving both, we get,[tex]v_2=4m/s\\v_1=0[/tex]
Thus, we can conclude that, the kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.
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An air-filled 39.1 μH solenoid has a length 4.0 cm and a cross-sectional area 0.60 cm2. How many turns are in this solenoid?
a.21000
b.144
c.12
d.120
e.1200
No. of turns in the solenoid is an option (b) 144.
The self-inductance of a long solenoid depends only on its physical properties (such as the number of turns of wire per unit length and the volume), and not on the magnetic field or the current.
Self-inductance of solenoid = 39.1 μH
= 39.1 × [tex]10^-^6[/tex] H
Length of the solenoid = 4.0 cm
Cross-sectional area = 0.60 cm²
Expression for the self-inductance of a coil ;
L = µ₀N²A / [tex]l[/tex]
where,
L = Self- Inductance
N = No. of turns.
A = Cross-sectional area
[tex]l=[/tex] Length of the solenoid
L =( 4π × [tex]10^-^7[/tex] × N² × 0.60 ) / 4.0
39.1× 4.0 / 4π × [tex]10^-^7[/tex] × 0.60 = N²
N² = 2.07 × [tex]10^6[/tex]
N = 144
Therefore, the no. of turns of the solenoid is 144.
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Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radius of Earth = 6.37×103 km, mass of Earth = 5.98×1024 kg, G = 6.67×10-11 Nm2/kg2.)
The speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.
Speed of the satellitev = √(GM/r)
where;
G is universal gravitation constantM is mass of Earthr is radius of the satellitev = √(6.67 x 10⁻¹¹ x 5.98 x 10²⁴/3.57 x 6.37x 10³)
v = 1.32 x 10⁵ m/s
Thus, the speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.
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How does an atom of rubidium-85 a rubidium ion with a +1 charge?
A. The atom loses 1 electron to have a total of 47.
B. The atom gains proton to have a total of 38.
C. The atom loses 1 electron to have a total of 36,
D. The atom gains 1 proton to have a total of 86.
Answer:
C. The atom loses 1 electron to have a total of 36
Explanation:
The number of electrons in a Rubidium atom is 37. Since the atom loses 1 electron, it has 36 left.
A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car moves 30.0 m. Neglecting friction between car and road, find v and the horizontal force exerted on the car.
(a) the speed v
______m/s
(b) the horizontal force exerted on the car (Enter the magnitude.)
_______N
The horizontal force applied is 160 N while the velocity is 2.03 m/s.
What is the speed of the car?The work done by the car is obtained as the product of the force and the distance;
W = F x
F = ?
x = 30.0 m
W = 4,800 J
F = 4,800 J/30.0 m
F = 160 N
But F = ma
a = F/m
a = 160 N/2.30 ✕ 10^3-kg
a= 0.069 m/s
Now;
v^2 = u^2 + 2as
u = 0/ms because the car started from rest
v = √2as
v = √2 * 0.069 * 30
v = 2.03 m/s
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A truck is carrying a refrigerator as shown in the figure. The height of the refrigerator is 158.0 cm, the width is 60.0 cm. The center of gravity of the refrigerator is 67.0 cm above the midpoint of the base.
What is the maximum acceleration the truck can have so that the refrigerator doesn't tip over? (The static friction between the truck and the refrigerator is very large.)
The maximum acceleration the truck can have so that the refrigerator does not tip over is 1.86m/s².
What is acceleration?Acceleration is the change in velocity with time.
The maximum acceleration is obtained by taking moments about the tipping point of rotation.
Moment = force * perpendicular distanceClockwise moment = Anticlockwise momentF₂ * 1.58 m = F₁ * 0.67 m
where
F₂ is tipping force = mass * acceleration, a
F₁ is weight = mass * acceleration due to gravity, g
The weight acts at a distance half the width of the refrigerator = 30 cm or 0.3 m
Height of refrigerator is 158 cm 0r 1.58 cm
m * a * 1.58 = m * 9.81 * 0.30
a = 1.86 m/s²
The maximum acceleration the truck can have so that the refrigerator does not tip over is 1.86m/s².
In conclusion, if the maximum acceleration is exceeded, the refrigerator will tip over.
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Define the term work and state its unit. An ant is dragging a house-fly and the elephant is pushing a big tree which is not moving. Who is doing work, the ant or the elephant? Justify your answer. 922.5 205
Ant is performing a work
what is work?
Work is the force applied on an individual with respect to displacement.
Work = Force × displacement
Unit is Nm
Elephant is pushing bt there is no displacement occurred so the work of elephant is zero.
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The ant works, but the elephant does not.
Who works, how do find the ant and the elephant?Work done = Force × Displacement.
If there are ants and houseflies,
Ants drag the house bug, so they use specific force to move the house bug from one point to another, so we can say they work.
In the case of the elephant and the tree,
When the elephant pushes the tree (applying a force), the tree does not move, i.e., there is no displacement, so there is work.
Work done = Force × Displacement
= Force × 0
= 0
Therefore,
The ant works, but the elephant does not.
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Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.
1. What is the First Cosmic Speed i.e. the speed of a satellite on a low lying circular orbit around this planet? (Planet-X doesn't have any atmosphere.)
2. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
3. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?
(a) The speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.
(b) The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.
(c) The radius of the synchronous orbit of a satellite is 69,801 km .
Speed of the satellitev = √GM/r
where;
M is mass of the planetr is radius of the planetv = √[(6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 7,338.93 m/s
Escape velocity of the satellitev = √2GM/r
v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 10,378.82 m/s
Speed of the satellite at the given periodv = 2πr/T
r = vT/2π
r = (7,338.93 x 16.6 x 3600 s) / (2π)
r = 69,801 km
Thus, the speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.
The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.
The radius of the synchronous orbit of a satellite is 69,801 km .
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if you dip your hand in cold water after having dipped in warm water, will you feel the water colder than it actually is?
Answer:
Yes.
Explanation:
When you move your hand from the cold water to the “warmer” (room temp) water, one hand feels warm.
As you move your hand from the warm water to the “colder” (room temp) water, that hand feels colder.
Although both hands experience the last bowl of water at the same temperature, your brain senses two
separate sensations. So the water feels “warm” or “cold” relative to the water your hand was in previously.
The greater the difference in temperature, the easier it is to sense a difference.
Discuss the aspects of either the Gemini Program or the Soyuz Program.
Answer:
The Gemini program
Explanation:
The Gemini Program was the second human spaceflight program hosted by Nasa in the year 1961. Taking place between mission Mercury and Apollo, the Gemini spacecraft carried two people to space and marked the foundation to the upcoming Apollo mission to Moon. It was a series of missions into the outer orbital which took place between 1965 and 1966. Prior to the Gemini missions, NASA had little to no information about space and space traveling. It was crucial for them to get acquainted with life outside before establishing successful Moon landings. And the series of Gemini missions helped them do just that.
A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal.
The horizontal component of the force exerted by the hi.nge on the beam is 8.662x10^1 N.
What is the magnitude of the force that the beam exerts on the hi.nge?
288.51 N is the magnitude of the force that the beam exerts on the hi.nge.
Given
Mass 0f beam = 40 Kg
The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N
Angle between the beam and cable is = 90°
Angle between beam and the horizontal component = 31°
As the system of the beam, hi_nge and cable are in equilibrium.
The magnitude of the force that the beam exerts on the hi_nge can be calculated by -
F =The horizontal component of force + the vertical component of force
F = 86.62 N + 40 × 9.8 × sin 31°
F =86.62 N + 201.89 N
F = 288.51 N
Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51 N.
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I need help with my homework
C. The center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.
What is center of mass?The center of mass of an object is the unique point where the weighted relative position of the distributed mass sums to zero.
What is center of gravity?Center of gravity is the point from which the weight of a body or system may be considered to act.
Thus, the center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.
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A proton is moved from the negative to the positive plate of a parallel-plate arrangement. The plates are 1.50 cm apart, and the electric field is uniform with a magnitude of 1 500 N/C.
What is the proton’s potential energy change?
What is the potential difference between the plates?
What is the potential difference between the negative plate and a point midway between the plates?
If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate?
(a) The proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
(b) The potential difference between the negative plate and a point midway between the plates is 11.25 V.
(c) The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
Potential energy of the proton
U = qΔV
where;
q is charge of the protonΔV is potential differenceU = q(Ed)
U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)
U = 3.6 x 10⁻¹⁸ J
Potential difference between the negative plate and a point midwayΔV = E(0.5d)
ΔV = 0.5Ed
ΔV = 0.5 (1500)(1.5 x 10⁻²)
ΔV = 11.25 V
Speed of the protonU = ¹/₂mv²
U = mv²
v² = 2U/m
where;
m is mass of proton = 1.67 x 10⁻²⁷ kgv² = (2 x 3.6 x 10⁻¹⁸) / ( 1.67 x 10⁻²⁷)
v² = 4.311 x 10⁹
v = √(4.311 x 10⁹)
v = 6.57 x 10⁴ m/s
Thus, the proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
The potential difference between the negative plate and a point midway between the plates is 11.25 V.
The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
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Motorcycle safety helmet extend the time of collision hence decreasing,
a:chance of collision
b:force acting
c: velocity
c:Impulse
Answer:
D. Impulse
Explanation: Hope this helps
For a projectile launched horizontally at 650 m/s and travels for 5.75 seconds, What is the range?
The range of horizontal projectile motion is 3737.5 m.
A projectile's horizontal velocity is constant (a never changing value), Gravity causes a downward vertical acceleration with a magnitude of 9.8 m/s/s. A projectile's vertical velocity fluctuates by 9.8 m/s per second, A projectile's vertical motion is unrelated to its horizontal motion. we know that sinθ is maximum at 90°. Therefore horizontal range will maximum at 45°.
To calculate the range of horizontal projectile motion we use;
Δx = vₓ t
where , Δx = Range
vₓ = Velocity
t = Time
Initial horizontal velocity, vₓ = 650 m/s
Time = 5.75 sec
Δx = 650 × 5.75
= 3737.5 m
Therefore, the range of horizontal projectile motion is 3737.5 m.
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Object A has the same mass as Object B but Object A is traveling faster. What can we say about the momentum of Object A compared to that of Object B?
Answer:
Object A's momentum is larger
Explanation:
as the formula for momentum goes:
P = M * V
where P is momentum, M is mass, V is velocity.
so where Va (Object A's velocity) is larger than Vb (Object B's velocity) we get:
( i ) Va > Vb
as the masses of both objects are equal, we mark:
( ii) Ma = Mb = M
we multiply both sides of ( i ) by M to get:
( iii ) Va × M > Vb × M
and we finally get:
( iv ) Pa > Pb
Answer: Object A's momentum of larger than compared to that of object B.
Explanation:
How do I connect with my higher self?
Answer:
Create space
Watch your breath
Watch your thoughts
Be gentle with yourself
Affirm what you want
Explanation:
K donates, or transfers, one electron to bromine, which has 7 electrons. Both K and Br are now stable with 8 electrons. K becomes a positive ion and Br becomes a negative ion. The positive K ion and the negative Br ion attract each other to form an ionic bond.
K is cation by losing of electron whereas Br is anion due to accepting of electrons.
Is charge appears when an atom lose or accept electron?Yes, the positive ion appears on K and become cation whereas the negative ion bears on Br which make it anion because of losing and gaining of electron by these atoms. This transferring of electrons leads to formation of ionic bonds between them.
So we can conclude that K is cation by losing of electron whereas Br is anion due to accepting of electrons.
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A certain force gives mass m1, an acceleration of 12 m/s² and mass m2 an acceleration of 3.3. m/s². What acceleration will this force give to the combined mass?
Acceleration is the rate of change of velocity. Usually, acceleration means the speed is transforming, but not always. When an object moves in a circular path at a steady speed, it is still accelerating, because the focus of its velocity is changing.
a = 4,552 m / s², b) a = 2,588 m / s²
Newton's second law is
F = ma
a = F / m
in this case the force remains constant
indicate us
* for a mass m₁
a₁ = F/m₁
a₁ = 12, m/ s²
* for a mass m₂
a₂= 3.3 m / s²
acceleration m = m₂-m₁
substitute
[tex]$\begin{aligned}&a=\frac{F}{m_{2}-m_{1}} \\&1 / a=\frac{m_{2}}{F}-\frac{m_{1}}{F}\end{aligned}$[/tex]
let's calculate
1/a=1/3.3 - 1/12 = 0.21969
a = 4,552 m / s²
a= 4,552m/s²
What is speed and acceleration?Speed estimates the rate of movement of an object, that is, the distance traveled per unit of time. Acceleration calculates the rate of change of velocity, that is, the change in velocity between two different moments
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A 72.5-kg hiker starts at an elevation of 1230 m and climbs to the top of a peak 2660 m high.
What is the hiker's change in potential energy?
Express your answer to three significant figures and include the appropriate units.
What is the minimum work required of the hiker?
Express your answer to three significant figures and include the appropriate units.
Change in potential energy = 1.02 * 10⁶ J
Minimum work required of the hiker = 1.89 * 10⁶ J
What is the change in potential energy of the hiker?The potential energy of a body is calculated as follows:
Potential energy = mghChange in potential energy = Final PE - Initial PE
Change in potential energy = mg(H - h)
Change in potential energy = 72.5 * 9.81 * (2660 - 1230)
Change in potential energy = 1.02 * 10⁶ J
The minimum work required of the hiker is the potential energy at the highest point.
Minimum work = mgH
Minimum work required of the hiker = 1.89 * 10⁶ J
In conclusion, potential energy is energy due to state or position of a body.
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The three displacement vectors have magnitude of A=5.00m,B=5.00m and C=4.00m.Find the resultant (magnitude and directional angle) of the three vectors?
3.00 m is the magnitude of the resultant vector
137.1° is the directional angle of the resultant vector.
1) To find the magnitude of the resultant
Discover each vector's individual components first.
We must consider the signs of the components because the angles in the figure are measured in various ways.
In this case, both the y component of vector C and the x component of vector A are negative.
The vectors' components are as follows, using a little trigonometry:
Magnitude of A = 5 m
[tex]A_{x}[/tex] = - (5.00m) cos20° = -5 ×0.408 = -4.698 m
[tex]A_{y}[/tex] = + (5.00m)sin20° = +5 × 0.342 = +1.710 m
Magnitude of B = 5m
[tex]B_{x}[/tex] = +(5.00m)cos60° = 5 × 0.5 = +2.5m
[tex]B_{y}[/tex] = +(5.00m)sin60° = 5 × √3/2 = +4.33 m
Cx = 0
Cy = -4.00m
The sum of all three vectors, which we refer to as R, produces components.
Rₓ = Aₓ + Bₓ + Cₓ
= -4.698 + 2.5 + 0
= -2.198 m
[tex]R_{y}[/tex] = [tex]A_{y} + B_{y} + C_{y}[/tex]
= +1.71 +4.33 - 4.00
= 2.040 m
R = [tex]\sqrt{R_{x^{2} }+R_{y^{2} }[/tex]
= [tex]\sqrt{(-2.198)^{2 } + (2.040)^{2} }[/tex]
= 3.00 m
2) To find the directional angle of resultant
tan θ = 2.040/-2.198 = -0.928
θ = -42.9°
Such a vector would be in the so-called "fourth quadrant," as it is well known. However, we discovered that R has a negative x component and a positive y component, indicating that such a vector must reside in the "second quadrant.
The calculator accidentally returned an angle that is 180 degrees off, thus we must add 180 degrees to the naïve angle in order to get the right angle.
Therefore, R's actual direction is determined by-
θ = -42.9° + 180° = 137.1°
Hence, the magnitude of resultant vector is 3.00 m and directional angle is 137.1°
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