Before the development of quantum theory, Ernest Rutherford's experiments with gold atoms led him to propose the so-called Rutherford Model of atomic structure. The basic idea is that the nucleus of the atom is a very dense concentration of positive charge, and that negatively charged electrons orbit the nucleus in much the same manner as planets orbit a star. His experiments appeared to show that the average radius of an electron orbit around the gold nucleus must be about 10−1010−10 m. Stable gold has 79 protons and 118 neutrons in its nucleus.
What is the strength of the nucleus' electric field at the orbital radius of the electrons?
What is the kinetic energy of an electron in a circular orbit around the gold nucleus?
Answer:
1. [tex] E = 1.14 \cdot 10^{13} N/C [/tex]
2. [tex]E_{k} = 9.1 \cdot 10^{-17} J[/tex]
Explanation:
1. The strength of the nucleus' electric field (E):
[tex]E = \frac{kq}{r^{2}}[/tex]
Where:
k: is the Coulomb constant = 9x10⁹ Nm²/C²
q: is the proton charge = 1.6x10⁻¹⁹ C
r: is the radius = 10⁻¹⁰ m
[tex]E = \frac{kq}{r^{2}} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*79*1.6 \cdot 10^{-19} C}{(10^{-10} m)^{2}} = 1.14 \cdot 10^{13} N/C[/tex]
2. The kinetic energy (Ek) of an electron is the following:
[tex] E_{k} = \frac{1}{2}mv^{2} [/tex]
Where:
m is the electron's mass = 9.1x10⁻³¹ kg
v: is the speed of the electron
We can find the speed of the electron by equaling the centripetal force (Fc) and the electrostatic force (Fe):
[tex] F_{c} = F_{e} [/tex]
[tex] \frac{mv^{2}}{r} = \frac{kq^{2}}{r^{2}} = qE [/tex]
[tex] v^{2} = \frac{qEr}{m} = \frac{1.6 \cdot 10^{-19} C*1.14 \cdot 10^{13} N/C*10^{-10} m}{9.1 \cdot 10^{-31} kg} = 2.00 \cdot 10^{14} m^{2}/s^{2} [/tex]
Now, we can find the kinetic energy:
[tex] E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}9.1 \cdot 10^{-31} kg*2.00 \cdot 10^{14} m^{2}/s^{2} = 9.1 \cdot 10^{-17} J [/tex]
I hope it helps you!
please answer this question with an explanation
will mark brainiest
Answer:
jenny
Explanation:
A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will
Complete Question:
A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will
A. be behind the package.
B. be over the package.
C. be in front of the package.
D. depend on the speed of the plane when the package was released.
Answer:
B.
Explanation:
As no other horizontal forces are present, due to the horizontal movement and the vertical one are independent each other (as they are perpendicular), the plane and the package continue moving horizontally at the same speed, so when the package hits the ground (due to the action of gravity in the vertical direction only) the plane will be exactly over the package.
The package hits the ground in such a way that the plane will be exactly over the package.
The given problem is based on the effect of air resistance on the motion of object. The resistive force exerted on any object on account of dust, and other air particles is known as air resistance or simply as air drag.
In the given problem, as no other horizontal forces are present, due to the horizontal movement and the vertical one are independent each other (as they are perpendicular), the plane and the package continue moving horizontally at the same speed.
So when the package hits the ground (due to the action of gravity in the vertical direction only) the plane will be exactly over the package.
Thus, we can conclude that the package hits the ground in such a way that the plane will be exactly over the package.
Learn more about the air drag here:
https://brainly.com/question/14755232
True.or false A railroad track runs southwest to northeast.
Answer:
ns for high-speed rail in the United States date back to the High Speed Ground Transportation Act of 1965. Various state and federal proposals have followed. Despite being one of the world's first countries to get high-speed trains (the Metroliner service in 1969), it failed to spread. Definitions of what constitutes high-speed rail vary, including a range of speeds over 110 mph (180 km/h) and dedicated rail lines. Inter-city railwith top speeds between 90 and 125 mph (140 and 200 km/h) is sometimes referred to in the United States as higher-speed rail.[1]
Amtrak's Acela Express (reaching 150 mph, 240 km/h), Silver Star, Northeast Regional, Keystone Service, Vermonter and certain MARC Penn Line express trains (all five reaching 125 mph, 201 km/h) are the only high-speed services in the country.
As of 2020, the California High-Speed Rail Authority is working on the California High-Speed Rail project and construction is under way on sections traversing the Central Valley. The Central Valley section is planned to open in 2029 and Phase I is planned for completion in 2031.[2]
Contents
1 Definitions in American context
2 History
2.1 Faster inter-city trains: 1920–1941
2.2 Post-war period: 1945–1960
2.3 First attempts: 1960–1992
2.4 Renewed interest: 1993–2008
2.5 Plans for 2008–2013
3 Current state and regional efforts
3.1 The Northeast
3.1.1 Northeast Corridor: Next Generation High-Speed Rail
3.1.1.1 Proposed routes
3.1.2 Northeast Maglev proposal
3.1.3 New Jersey–New York City upgrades
3.1.4 New York
3.1.5 Pennsylvania
3.2 Western States
3.2.1 California
3.2.2 Pacific Northwest
3.2.3 Arizona
3.3 Mid-Atlantic and the South
3.3.1 Florida
3.3.2 Southeast
3.3.3 Texas
3.4 Midwest
3.4.1 Illinois and the Midwest
3.5 The Southwest
4 Federal high-speed rail initiatives
4.1 American Recovery and Reinvestment Act of 2009
4.1.1 Strategic plan
4.2 2009 federal grant funding
4.3 2010 allocation
4.3.1 Cancellation of funds for Wisconsin, Ohio, and Florida
4.4 2011 and 2012 proposals and rejections of funding
5 See also
6 Notes
7 Further reading
8 External links
Explanation:
How much work is done lifting a 5 kg ball from a height of 2 m to a height of 6 m? (Use 10 m/s2 for the acceleration of gravity.)
A) 100 J B) 200 J C) 300 J D) 400 J
Answer:
B) 200 [J]
Explanation:
In order to solve this problem we must remember the definition of work which tells us that it is equal to the product of force by a distance, in this case, the force is the weight of the ball. The distance traveled is 4 [m] since 6-2 = 4[m]
F = m*g
where:
m = mass = 5 [kg]
g = gravity acceleration = 10 [m/s^2]
F = 5*10 = 50 [N]
w = F*d
where:
F = force = 50 [N]
d = 4 [m]
w = 50*4 = 200 [J]
what is the meaning of the word physics
Answer:
the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.
Explanation:
mark as brainliest
Bird A, with a mass of 2.2 kg, is stationary while Bird B, with a mass of 1.7 kg, is moving due north from Bird A at 3 m/s. What is the velocity of the center of mass for this system of two birds
Answer:
1.3 m/s
Explanation:
It is given that,
Mass of bird A, [tex]m_A=2.2\ kg[/tex]
Mass of bird B, [tex]m_B=1.7\ kg[/tex]
Initial speed of bird A is 0 as it was at rest
Initial speed of bird B is 3 m/s
We need to find the velocity of the center of mass for this system of two birds. Let it is V. so,
[tex]v_{cm}=\dfrac{m_Au_A+m_Bu_B}{m_A+m_B}\\\\v_{cm}=\dfrac{2.2\times 0+1.7\times 3}{2.2+1.7}\\\\v_{cm}=1.3\ m/s[/tex]
So, the center of mass for this system is 1.3 m/s.
Two identical items, object 1 and object 2, are dropped from the top of a 50.0m50.0m building. Object 1 is dropped with an initial velocity of 0m/s0m/s, while object 2 is thrown straight downward with an initial velocity of 13.0m/s13.0m/s. What is the difference in time, in seconds rounded to the nearest tenth, between when the two objects hit the ground
Answer:
Δt = 1.1 s
Explanation:
Given information:
H= 50.0 m
g= 9.8 m/s²
Object 1v₀ = 0[tex]H = \frac{1}{2} * g* t^{2}[/tex]
Solving for t, we get:
t₁= 3.2 s
Object 2v₀ = 13 m/sWe can find the final velocity for the object when it hits the ground, using the following expression:
[tex]v_{f}^{2} - v_{o}^{2} = 2*g*H[/tex]
Solving for vf, we get:
vf = 33.9 m/s
Applying the definition of acceleration, being this acceleration the one due to gravity (g), we can write the following equation:
[tex]v_{f} = v_{o} + g*t[/tex]
(Assuming the downward direction to be positive).
Solving for t, we get:
t₂ = 2.1 s
So the difference in time when both objects hit the ground, it's simply
Δt = t₂ - t₁ = 3.2 s - 2.1 s = 1.1 s
It takes 3.8 x 10^-5 for a pulse of the radio waves from a radar to reach a plane and bounce back. How far is the plane from the radar?
Answer: 11400 m
Explanation:
Given:
t = 3.8 x 10^-5 s
v = 3 x 10^8 m/s
d = ?
Formula:
d = vt
= (3.8 x 10^-5 s)(3 x 10^8 m/s)
= 11400 m
hope this helps :)
The uniform movement allows to find the results for the distance from the radar to the plane is: 5.7 10³ m or 5.7 km
Kinematics studies the motion of objects looking for relationships between position, velocity and acceleration, in the special case that the acceleration is zero is called uniform motion and is described by the expression
[tex]v = \frac{d}{t}[/tex]
d = v t
Where v is the velocity, d the displacement and t the time.
Radar waves are electromagnetic waves with constant velocity
v = 3 10⁸ m/s
They indicate that the time of the waves to go to the plane and return is 3.8 10⁻⁵ s, therefore if the speed is constant, the time to reach the plane is half of the total time.
t = [tex]\frac{t_{total} }{ 2}[/tex]
t = [tex]\frac{3.8 \ 10^{-5}}{2}[/tex]
t = 1.9 10⁻⁵ s
Let's calculate
d = 3 10⁸ 1.9 10⁻⁵
d = 5.7 10³
In conclusion with the uniform movement we can find the results for the distance from the radar to the plane is: 5.7 10³ m or 5.7 km
Learn more here: brainly.com/question/20369552
If vector A = 6i - 2j + 3k, determine
(a) A vector in the same direction as A with magnitude 2A
(b) A unit vector in the direction of A
(c) a vector opposite to A with magnitude of 4 m
Answer:
(a) [tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]
(b) [tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]
(c) [tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]
Explanation:
Vectors
Given a vector
[tex]\vec A=6\hat i-2\hat j+3\hat k[/tex]
We must determine the following:
a) A vector in the same direction as A with double magnitude 2A.
If the vector goes in the same direction but has a different magnitude, we only need to multiply each component by a common factor, in this case, by 2. Thus, the required vector is:
[tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]
b) A unit vector in the same direction of A.
The unit vector needs to compute the magnitude of the vector:
[tex]\mid A\mid=\sqrt{6^2+2^2+3^2}[/tex]
[tex]\mid A\mid=\sqrt{36+4+9}=\sqrt{49}=7[/tex]
[tex]\mid A\mid=7[/tex]
The unit vector is:
[tex]\displaystyle \vec{U_A}=\frac{\vec A}{\mid \vec A\mid}[/tex]
[tex]\displaystyle \vec{U_A}=\frac{12\hat i-4\hat j+6\hat k}{7}[/tex]
[tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]
c) A vector opposite to A with magnitude 4 m. We assume the original vector is also expressed in m.
The opposite vector to A is obtained simply by multiplying the unit vector by -1. To make its magnitude equal to 4, also multiply by 4. In all, we multiply the unit vector by -4:
[tex]-4\vec{U_A}=-4(12/7\hat i-4/7\hat j+6/7\hat k)[/tex]
[tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]
The equation that governs the period of a pendulum’s swinging. T=2π√L/g
Where T is the period, L is the length of the pendulum and g is a constant, equal to 9.8 m/s2. The symbol g is a measure of the strength of Earth’s gravity, and has a different value on other planets and moons.
On our Moon, the strength of earth’s gravity is only 1/6th of the normal value. If a pendulum on Earth has a period of 4.9 seconds, what is the period of that same pendulum on the moon?
Answer:
The period of that same pendulum on the moon is 12.0 seconds.
Explanation:
To determine the period of that same pendulum on the moon,
First, we will determine the value of g (which is a measure of the strength of Earth's gravity) on the Moon. Let the value of g on the Moon be [tex]g_{M}[/tex].
From the question, the strength of earth’s gravity is only 1/6th of the normal value. The normal value of g is 9.8 m/s²
∴ [tex]g_{M}[/tex] = [tex]\frac{1}{6} \times 9.8 m/s^{2}[/tex]
[tex]g_{M}[/tex] = 1.63 m/s²
From the question, T=2π√L/g
[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]
We can write that,
[tex]T_{E} = 2\pi \sqrt{\frac{L}{g_{E} } }[/tex] .......... (1)
Where [tex]T_{E}[/tex] is the period of the pendulum on Earth and [tex]g_{E}[/tex] is the measure of the strength of Earth's gravity
and
[tex]T_{M} = 2\pi \sqrt{\frac{L}{g_{M} } }[/tex] .......... (2)
Where [tex]T_{M}[/tex] is the period of the pendulum on Moon and [tex]g_{M}[/tex] is the measure of the strength of Earth's gravity on the Moon.
Since we are to determine the period of the same pendulum on the moon, then, [tex]2\pi[/tex] and [tex]L[/tex] are constants.
Dividing equation (1) by (2), we get
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
From the question,
[tex]T_{E} = 4.9secs[/tex]
[tex]g_{E}[/tex] = 9.8 m/s²
[tex]g_{M}[/tex] = 1.63 m/s²
[tex]T_{M}[/tex] = ??
From,
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
[tex]\frac{4.9}{T_{M} } = \sqrt{\frac{1.63}{9.8} }[/tex]
[tex]\frac{4.9}{T_{M} } = 0.40783[/tex]
[tex]T_{M} =\frac{4.9}{0.40783 }[/tex]
[tex]T_{M} = 12.01 secs[/tex]
∴ [tex]T_{M} = 12.0secs[/tex]
Hence, the period of that same pendulum on the moon is 12.0 seconds.
Answer:
The period of that same pendulum on the moon is 12.0 s
Explanation:
Given;
period of a pendulum’s swinging, T=2π√L/g
the strength of earth’s gravity on moon, g₂ = ¹/₆(g₁)
period of pendulum on Earth, T₁ = 4.9 s
period of pendulum on moon, T₂ = ?
The length of the pendulum is constant, make it the subject of the formula;
[tex]T = 2\pi \sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}}\\\\(\frac{T}{2\pi} )^2 =\frac{L}{g}\\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\ L = \frac{gT^2}{4\pi^2}\\\\L_1 = L_2\\\\\frac{g_1T_1^2}{4\pi^2}= \frac{g_2T_2^2}{4\pi^2}\\\\g_1T_1^2 = g_2T_2^2\\\\T_2^2 = \frac{g_1T_1^2}{g_2} \\\\T_2 = \sqrt{\frac{g_1T_1^2}{g_2}}\\\\ T_2 = \sqrt{\frac{g_1T_1^2}{g_1/6}}\\\\ T_2 = \sqrt{\frac{6*g_1T_1^2}{g_1}}\\\\T_2 = \sqrt{6T_1^2}\\\\ T_2 = T_1\sqrt{6} \\\\T_2 = (4.9)\sqrt{6}\\\\ T_2 = 12.0 \ s[/tex]
Therefore, the period of that same pendulum on the moon is 12.0 s
Converting compound units
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram/meter3 and the density of silicon in other units: 2.33 gram/centimeter3. You decide to convert the density of silicon into units of kilogram/meter3 to perform the comparison. By which combination of conversion factors will you multiply 2.33 gram/centimeter3 to perform the unit conversion?
Answer:
Dividing the silicon density by 1000 and then multiply it by 1000000.
Explanation:
A kilogram equals 1000 grams and a cubic meter equals 1000000 cubic centimeters. Hence, we must divide the silicon density by 1000 and then multiply itby 1000000 to convert the value into kilograms per cubic centimeter. That is:
[tex]x = 2.33\,\frac{g}{cm^{3}}\times \frac{1\,kg}{1000\,g}\times \frac{1000000\,cm^{3}}{1\,m^{3}}[/tex]
[tex]x = 2330\,\frac{kg}{m^{3}}[/tex]
In a nutshell, we must multiply the density of silicon by 1000 to obtains its value in kilograms per cubic meter.
What is the volume of an object if it has a mass of 10 grams and a density of 87 g/ml
Answer:
The answer is 0.115 mLExplanation:
The volume of a substance when given the density and mass can be found by using the formula
[tex]volume = \frac{mass}{density} \\[/tex]
From the question
mass = 10 g
density = 87 g/ml
We have
[tex]volume = \frac{10}{87} \\ = 0.114942528...[/tex]
We have the final answer as
0.115 mLHope this helps you
i need help, for physics
how far will a brick starting from rest fall freely in 3.0 seconds?
Answer: It will be about 44.1m
Explanation:
There are 5.5 L of a gas present at -38.0 C. What is the temperature if the volume of the gas has changed to 1.30 L?
Answer:
We are given:
V1 = 5.5L T1 = -38 C or 235 k
V2 = 1.3L T2 = T
From the gas equation:
PV = nRT
Since the pressure (P) , number of moles (n) and the universal gas constant (R) are constants, we can write the same equation as:
V / T = k (where k is a constant)
so a bit more insight, since the values noted above are constant, when multiplied by each other, they will provide us with a constant number irrespective of the value of the variables
Changing the variables for the first case:
V1 / T1 = k (where k is the same constant) ----------------(1)
Similarly,
V2 / T2 = k (again, k has the same value)----------------(2)
From (1) and (2):
k is the common value
V1 / T1 = V2 / T2
Replacing the variables
5.5 / 235 = 1.3 / T
T = 1.3 * 235 / 5.5
T = 55.54 k
Therefore, at 55.54 K the gas will have a volume of 1.3L
HELP PLS7. A steel ball is dropped from a height of 100 meters. Which velocity-time graph best describes the
motion of the ball?
Answer:
Option C.
Explanation:
To know which velocity-time graph best describes the motion of the ball, let us calculate the velocity of the ball and the time taken for the ball to get the ground. This can be obtained as follow:
1. Determination of the velocity.
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 100 m
Final velocity (v) =.?
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 100)
v² = 0 + 1960
v² = 1960
Take the square root of both side.
v = √(1960)
v = 44.27 m/s
2. Determination of the time taken.
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 100 m
Time (t) =.?
h = ½gt²
100 = ½ × 9.8 × t²
100 = 4.9 × t²
Divide both side by 4.9
t² = 100 / 4.9
Take the square root of both side
t = √(100 / 4.9)
t = 4.52 s
From the above illustration,
Initial time (t1) = 0 s
Final time (t2) = 4.52 s
Initial velocity (u) = 0 m/s
Final velocity (v) = 44.27 m/s
Thus, we can see that as the time increase, the velocity also increase. Therefore, option C gives the correct answer to the question.
The diagram shows two forces acting on the dog. What are these two forces
Answer:
kenietic and potential i guess
Explanation:
How long does it take a P-wave to travel 7,000 km? ______ minutes b) How long does it take an S-wave to travel 7,000 km? ______ minutes c) How long does it take a Love wave to travel 7,000 km? ______ minutes d) How long does it take a Rayleigh wave to travel 7,000 km? ______minutes
Answer:
A. 8.64 secs.
B. 14.58 secs.
C. 26.002 secs.
D. 33.46secs.
Explanation:
A. P wave would travel 7000km
p-wave travels on a speed of 13.5km/s
= 7000km/13.5km/s
= 8.64 secs.
B. S-wave time to travel 7000km
s-wave travels on a speed of 8km/s
= 7000km/8km/s
= 14.58 secs.
C Love wave travels at a speed of 10,000m/s ( 2.7778 m/s ).
= 7000km to miles
= 4349.598m/2.788m/s
= 26.002 secs.
D. Rayleigh wave to travel 7,000 km
10,000m/s ( 2.1667 m/s ).
= 7000km to miles
= 4349.598m/2.1667m/s
= 33.46secs.
What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.9 A current in a 1.60 T field
Answer:
The maximum torque on the loop is 395.80 N.m.
Explanation:
Given;
number of turns of the wire, N = 150 turns
length of the square loop, L = 18.0 cm = 0.18 m
current in the wire, I = 50.9 A
Magnetic field, B = 1.6 T
Maximum torque on the loop is given by;
τ = NIAB
τ = (150)(50.9)(0.18²)(1.6)
τ = 395.80 N.m
Therefore, the maximum torque on the loop is 395.80 N.m.
An airplane flies with a constant speed of 600 km/h. to the west. How far can it travel in 1/4 hour?
Answer:
d = 150 km
Explanation:
Speed of an airplane is 600 km/h
We need to find how far it travel in 1/4 hour.
We know that the speed of an object is given by distance travelled divided by the time taken. Let d is the distance covered in 1/4 hour. So,
[tex]v=\dfrac{d}{t}\\\\d=v\times t\\\\d=600\ km/h \times \dfrac{1}{4}\ h\\\\d=150\ km[/tex]
So, it will cover 150 km.
Which of the organisms in the food web above is the top level carnivore
Answer:
apex consumers
Explanation:
they are top
what is the volume of an object that has a density of 65g/cm3 and a mass of 130g.
Density ρ is mass m per unit volume v, or
ρ = m / v
Solving for v gives
v = m / ρ
So the given object has a volume of
v = (130 g) / (65 g/cm³) = 2 cm³
If the loudness drops to 90 % of its original value in 5.0 s , what is the time constant of the damped oscillation
This question is incomplete, the complete question is;
A gong makes a loud noise when struck. The noise gradually gets less and less loud until it fades below the sensitivity of the human ear. The simplest model of how the gong produces the sound we hear treats the gong as a damped harmonic oscillator. The tone we hear is related to the frequency f of the oscillation, and its loudness is proportional to the energy of the oscillation.
If the loudness drops to 90 % of its original value in 5.0 s , what is the time constant of the damped oscillation
Answer: the time constant of the damped oscillation is 47.44s
Explanation:
Given that;
t = 5.0s
Lets say Ao is the amplitude of initial loudness and later A(t) = 0.9 Ao
EXPRESSION for amplitude is A(t) = Ao e^-t / T
t is time while T is time constant
so
0.9Ao = Ao e^-t / T
0.9 = e^ -t/T
So we take the natural log of both the sides
ln (0.9) = -t/T
-0.1054 = -t/T
0.1054 = t/T
WE now substitute our value of t
0.1054 = t/T
0.1054T = 5.0
T = 5 / 0.1054
T = 47.44s
therefore the time constant of the damped oscillation is 47.44s
Pleeeease Help!!
How did the parallel circuit respond differently to these changes than a series circuit?
Adding Bulbs: __________
Removing Bulbs: ________
Answer:
adding bulbs:bulbs will glow in parallel circuit.
but the brightness of each bulb differs in series circuit.
removing bulbs:if u remove a bulb from parallel circuit,the other bulbs continue glowing in the same brightness
but if u remove a bulb in series circuit,the brightness of each bulb increases.
Answer:
The person above is correct.
Explanation:
In a parallel sequence, the light bulbs have electrons that flow equally. Though, in a series circuit, it's mainly connected to one wire, and the electrons get weaker.
correct me if im wrong
why do some athletes get injuries before and after the game?
Answer:
they don't strech so they tear a muscle when they perform
Explanation:
Newton's first law states that objects do not change their motion unless acted upon by a net force. What does the word 'net' mean in this context?
A woven net, such as a fishing net or basketball net
B To catch or ensnare
C Remaining or left over after everything has been accounted for
D To cover, such as with mosquito netting
An object, initially at rest, is subject to an acceleration of 45 m/s^2. How long will it take that object to travel 1000m? Round to one decimal place.
Answer:
6.7 seconds
Explanation:
d=(1/2)at^2
equation
1000=(1/2)45t^2.
substitute
2000=45t^2.
multiply by 2 for both sides
44.44=t^2.
divide both sides by 45
6.7=t
take the square root of both sides
Please provide an explanation.
Thank you!!
Answer:
(a) 22 kN
(b) 36 kN, 29 kN
(c) left will decrease, right will increase
(d) 43 kN
Explanation:
(a) When the truck is off the bridge, there are 3 forces on the bridge.
Reaction force F₁ pushing up at the first support,
reaction force F₂ pushing up at the second support,
and weight force Mg pulling down at the middle of the bridge.
Sum the torques about the second support. (Remember that the magnitude of torque is force times the perpendicular distance. Take counterclockwise to be positive.)
∑τ = Iα
(Mg) (0.3 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L)
F₁ = ½ Mg
F₁ = ½ (44.0 kN)
F₁ = 22.0 kN
(b) This time, we have the added force of the truck's weight.
Using the same logic as part (a), we sum the torques about the second support:
∑τ = Iα
(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)
F₁ = ½ Mg + ⅔ mg
F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)
F₁ = 36.0 kN
Now sum the torques about the first support:
∑τ = Iα
-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)
F₂ = ½ Mg + ⅓ mg
F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)
F₂ = 29.0 kN
Alternatively, sum the forces in the y direction.
∑F = ma
F₁ + F₂ − Mg − mg = 0
F₂ = Mg + mg − F₁
F₂ = 44.0 kN + 21.0 kN − 36.0 kN
F₂ = 29.0 kN
(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)
As x increases, F₁ decreases and F₂ increases.
(d) Using our equation from part (c), when x = 0.6 L, F₂ is:
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)
F₂ = ½ Mg + mg
F₂ = ½ (44.0 kN) + 21.0 kN
F₂ = 43.0 kN
Answer:
a. Left support = Right support = 22 kNb. Left support = 36 kN Right support = 29 kNc. Left support force will decrease Right support force will increase.d. Right support = 43 kNExplanation:
given:
weight of bridge = 44 kN
weight of truck = 21 kN
a) truck is off the bridge
since the bridge is symmetrical, left support is equal to right support.
Left support = Right support = 44/2
Left support = Right support = 22 kN
b) truck is positioned as shown.
to get the reaction at left support, take moment from right support = 0
∑M at Right support = 0
Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0
Left support = 44 (0.3) + 21 (0.4)
0.6
Left support = 36 kN
Right support = weight of bridge + weight of truck - Left support
Right support = 44 + 21 - 36
Right support = 29 kN
c)
as the truck continues to drive to the right, Left support will decrease
as the truck get closer to the right support, Right support will increase.
d) truck is directly under the right support, find reaction at Right support?
∑M at Left support = 0
Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0
Right support = 44 (0.3) + 21 (0.6)
0.6
Right support = 43 kN