Work required to stop an electron which is moving with a speed of 1.10x10^6m/s is [tex]-\frac{3.971}{5^{19}\cdot \:1048576}[/tex]
What is electron?There are three main types of particles in an atom: protons, neutrons, and an electron that is bonded to an atom. For a variety of reasons, electrons are different from other particles. They possess both wave-like and particle-like properties, exist outside of the nucleus, have a substantially lower mass than protons. A particle that is not composed of smaller parts is an electron, which is also an elementary particle. They are not elementary particles because quarks are assumed to make up protons and neutrons.
We can solve the problem by using the work-energy theorem: in fact, the work done on the electron must be equal to its change in kinetic energy, therefore
[tex]W=K_{f}-K_{i}[/tex]
[tex]W=\frac{1}{2} mv^{2} - \frac{1}{2} mu^{2}[/tex]
where,
m=9.11 × 10⁻³¹kg is the mass of the electron
v is the final velocity of the electron
u is the initial velocity of the electron
In the problem, we are told that the electron is initially moving at,
v = 1.90×10⁶ m/s
Therefore, the work required to stop it as
W = -1/2 mv²
W = -1/2 (1.10×10⁻³¹)(1.90×10⁶)²
[tex]W= -\frac{3.971}{5^{19}\cdot \:1048576}\\[/tex]
and the work is negative since it is in the opposite direction to the motion of the electron.
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A concave mirror of focal length 10 cm forms an upright and diminished image of a real object placed at a distance of 5 cm from the mirror. Is this true or false?
Given,
The focal length of the concave mirror, f=10 cm
The object distance, o=5 cm
From the mirror formula, we have,
[tex]\frac{1}{f}=\frac{1}{o}+\frac{1}{i}[/tex]Where i is the image distance.
On substituting the known formula,
[tex]\begin{gathered} \frac{1}{10}=\frac{1}{5}+\frac{1}{i} \\ \Rightarrow\frac{1}{i}=\frac{1}{10}-\frac{1}{5} \\ =-\frac{1}{10} \\ \Rightarrow i=-10\text{ cm} \end{gathered}[/tex]And the magnification is given by,
[tex]m=\frac{-i}{o}[/tex]On substituting the known values,
[tex]\begin{gathered} m=\frac{-(-10)}{5} \\ =2 \end{gathered}[/tex]Thus the object is not diminished, it is magnified.
Thus the given statement is false.
A sample of silver at 10°C is heated to 90°C when 1 kcal of heat is added. What is the mass of the silver?
We can use the following formula:
[tex]q=m\cdot c\cdot(T2-T1)[/tex]Where:
q = heat = 1 kcal = 4184000J
c = specific heat (of silver) = 0.235 J °C /g
m = mass
T1 = Initial temperature = 10°C
T2 = Final temperature = 90°C
Therefore, solve for m:
[tex]\begin{gathered} m=\frac{q}{c(T2-T1)} \\ m=\frac{4184000}{0.235(90-10)} \\ m=222553.1915g \end{gathered}[/tex]Answer:
222553.1915g
You wish to lift a 720N crate of bricks to the 3rd floor of a building in a construction site. The 3rd floor is 16m high. How much work will that require?
ANSWER
[tex]11,520J[/tex]EXPLANATION
Parameters given:
Weight (force), F = 720N
Height (distance), d = 16m
To find the work required to lift the crate to the 3rd floor, we have to find the product of the force (weight of the crate) and the distance it will be lifted.
Therefore, we have:
[tex]\begin{gathered} W=F\cdot d \\ W=720\cdot16 \\ W=11,520J \end{gathered}[/tex]That is the work that it will require.
A radioactive tracer has a half-life of two hours how much of a 2500 g sample will be available after 18 hours?
ANSWER:
4.88 grams
STEP-BY-STEP EXPLANATION:
We must first calculate how many half-life there are in 18 hours, knowing that each half-life takes 2 hours.
[tex]\frac{18}{2}=9\text{ half-life}[/tex]Now, knowing this, we can calculate the number of grams applying 9 times the half-life, like this:
[tex]\frac{2500}{2^9}=4.88\text{ g}[/tex]Which means that after 18 hours there are 4.88 grams
A diagram of a student on a playground swing is shown below. At which point is the kinetic energy the greatest?A. Point 1B. Point 2C. Point 3D. Point 4
To find:
The point where the kinetic energy is the greatest.
Explanation:
The kinetic energy of an object is the energy possessed by the object due to its motion. Thus the kinetic energy of an object is proportional to the square of the velocity of the object.
The velocity of the wing will be zero at the extreme points of the path of the swing. But as the swing approaches the mean point, i.e., point 3, the velocity of the swing increases. The velocity will be maximum at the mean point.
Thus the kinetic energy will be the greatest at point 3.
Final answer:
Therefore the correct answer is option C.
You yell down a very deep well and it takes 1.5 s for your echo to return. If the speed of sound is 340 m/s then how deep is the well ?
We know that
• The time of the echo to return is 1.5 seconds.
,• The speed of the sound is 340 m/s.
It's important to consider that the sound wave has a constant speed, that is, it doesn't change its velocity. Therefore, we have to use the relation
[tex]d=v\cdot t[/tex]Where t = 1.5 sec and v = 340 m/s. Let's find d
[tex]\begin{gathered} d=340m/s\cdot1.5\sec \\ d=510m \end{gathered}[/tex]Hence, the well is 510 meters deep.Two astronauts, of masses 60 kg and 80 kg, are initially right next to each other and at rest in outer space. They suddenly push each other apart. What is their separation after the heavier astronaut has moved 12 m?24 m16 m9.0 m28 m21 m
We will have the following:
First, we recall that momentum is given by:
[tex]m_1v_0+m_2v_0=m_1v_1+m_2v_2[/tex]So:
[tex]\begin{gathered} (60kg)(0m/s)+(80kg)(0m/s)=(60kg)v_1+(80kg)v_2 \\ \\ \Rightarrow v_1=\frac{-(80kg)v_2}{(60kg)}\Rightarrow v_1=-\frac{4}{3}v_2 \end{gathered}[/tex]Then:
[Let's assume that when astronaut 2 moved 12 meters = t]
[tex]\begin{gathered} x_1=v_1t\Rightarrow x_1=-\frac{4}{3}v_2t \\ \\ x_2=v_2t\Rightarrow t=\frac{12}{v_2} \end{gathered}[/tex]Then:
[tex]x_1=-\frac{4}{3}v_2\ast\frac{12}{v_2}\Rightarrow x_1=-16m[/tex]Finally, the total distance will be:
[tex]\begin{gathered} d=|x_1|+|x_2|\Rightarrow d=16m+12m \\ \\ \Rightarrow d=28m \end{gathered}[/tex]explain how the intensity of the UV light vaires across the Earth
Some factors determine the amount of UV radiation that reach certain part of Earth's surface. They are listed and briefly explained below:
• Cloud coverage. Water molecules on clouds scatter the radiation, hence the more clouds the less UV radiation.
,• Ozone. Similarly as cloud coverage, the more concentration of ozone the less UV radiation that reaches the surface of the Earth.
,• Angle of incidence. If the angle of incidence of the UV light is oblique the light will spread in a wider area, and hence the intensity is spread across this area.
,• Aerosols. The molecules of aerosols also scatter the UV light.
,• Elevation. The more the elevation the greater the amount of UV light.
A wheel Was spinning at 2.8 rad/s. It took 3.2 seconds to stop completely. What is the acceleration of the wheel?
Given
The angular velocity is
[tex]\omega=2.8\text{ rad/s}[/tex]The time taken,
[tex]t=3.2s[/tex]To find
The acceleration of the wheel
Explanation
The acceleration is
[tex]\begin{gathered} \alpha=\frac{\omega}{t} \\ \Rightarrow\alpha=\frac{2.8}{3.2}=\frac{0.875rad}{s^2} \end{gathered}[/tex]Conclusion
The acceleration is
[tex]0.875\text{ rad/s}^2[/tex]I think this is all statements are true but I just want to make sure
Given that a bug flies into the windshield of a car going. Let's select the correct statements.
According the NEwton's third law, the force exerted on the bug by the car is equal to the force extered on the car by the bug.
To determine the acceleration, we have:
[tex]a=\frac{F}{m}[/tex]Where:
F is the force
m is the mass
The mass of the car will be greater than the mass of the bug.
Since the mass of the car is greater than the mass of the bug and they have the same force, we can say the acceleration of the bug is greater than the acceleration of the car.
Statement B is correct.
The force of impact is the same for both according to Newton's third Law.
Both the car and the bug deliver the same magnitude of impulse on each other.
Therefore, all statements are correct.
ANSWER:
All statements are true.
A deuteron particle consists of one proton and one neutron and has a mass of 3.34x10^-27 kg. A deuteron particle moving horizontally enters uniform, vertical 0.0800 T magnetic field and follows a circular arc of radius of 38.5 cm. What would be the radius of the arc followed by a proton that entered the field with the same velocity as the deuterium?
_dThe radius of curvature of a subatomic particle under a magnetic field is given by the following formula:
[tex]r=\frac{mv}{qB}[/tex]Where:
[tex]\begin{gathered} r=\text{ radius} \\ v=\text{ velocity} \\ q=\text{ charge} \\ B=\text{ magnetic field} \end{gathered}[/tex]We can determine the quotient between the velocity and the charge of the deuteron particle from the formula. First, we divide both sides by the mass:
[tex]\frac{r_d}{m_d}=\frac{v}{q_B_}[/tex]Now, we multiply both sides by the magnetic field "B":
[tex]\frac{Br_d}{m_d}=\frac{v}{q}[/tex]Since the charge of the deuterion is the same as the charge of the proton and the velocity we are considering are the same this means that the quotient between velocity and charge is the same for both particles. Therefore, we can apply the formula for the radius again, this time for the proton:
[tex]r_p=\frac{m_pv}{qB}[/tex]And substitute the quotient between velocity and charge:
[tex]r_p=\frac{m_p}{B}(\frac{Br_d}{m_d})[/tex]Now, we cancel out the magnetic field:
[tex]r_p=\frac{m_pr_d}{m_d}[/tex]Now, we substitute the values:
[tex]r_p=\frac{(1.67\times10^{-27}kg)(0.385m)}{(3.34\times10^{-27}kg)}[/tex]Solving the operations:
[tex]r_p=0.193m=19.3cm[/tex]Therefore, the radius is 19.3 cm.
Unbeknownst to most students, every time the school floors are waxed, the physics teachers get together to have a barrel of phun doing friction experiments in their socks (uhm - they do have clothes on; its just that they don't have any shoes on their feet). On one occasion, Mr. London applies a horizontal force to accelerate Mr. Schneider (mass of 84 kg) rightward at a rate of 1.2 m/s/s. If the coefficient of friction between Mr. Schneider 's socks and the freshly waxed floors is 0.35, then with what force (in Newtons) must Mr. London be pulling?
The frictional force is taken away from the net force. Mr. London will pull with a force of 388.9 N
What is Force ?Force can simply be defined as a pull or push. It is a product of mass and acceleration. It is a vector quantity.
Given that Mr. London applies a horizontal force to accelerate Mr. Schneider (mass of 84 kg) rightward at a rate of 1.2 m/s/s. If the coefficient of friction between Mr. Schneider 's socks and the freshly waxed floors is 0.35, then the force F in which Mr. London is pulling will be;
F - Fr = ma
Where
F = ?m = 84 kga = 1.2 m/s²N = ?μ = 0.35N = mg
N = 84 × 9.8
N = 823.2 N
Fr = μN
Fr = 0.35 × 823.2
Fr = 288.12 N
F - 288.12 = 84 × 1.2
F - 288.12 = 100.8
F = 100.8 + 288.12
F = 388.92 N
Therefore, Mr. London must be pulling with a force of 388.9 N
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A feather and a bowling ball are each dropped from an equal height in a vacuum and land at the same time. Which graph shows the total mechanical energy of the bowling ball as it falls?
The total mechanical energy of the bowling ball and the feather is shown by the graph in option D
What is the total mechanical energy?We know that the mechanical energy is the energy that is possessed by a body by virtue of its motion or by virtue of its staying at a place. Thus mechanical energy is possessed by an object that is moving or by an object that is at rest.
In this case, we have a feather and a bowling ball are each dropped from an equal height in a vacuum and land at the same time. We know that the mechanical energy of the two objects must be constant. This is because, the potential energy of the feathers and the ball at a height is converted to kinetic energy as the two objects begin to move.
The graph that would show the total energy must be one in which the energy axis of the graph is constant as we see in option D.
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Answer:
the answer is D
Explanation:
A 10 gram ball is rolling at 3 m/s. Calculate its kinetic energy.
ANSWER:
0.045 joules
STEP-BY-STEP EXPLANATION:
Given:
mass (m) = 10 g = 0.01 kg
velocity (v) = 3 m/s
The kinetic energy is given by the following formula:
[tex]K_E=\frac{1}{2}mv^2[/tex]We replacing:
[tex]\begin{gathered} K_E=\frac{1}{2}\cdot0.01\cdot3^2 \\ K_E=0.045\text{ J} \end{gathered}[/tex]The kinetic energy is 0.045 joules.
Objects with masses M1=12.0 kg and M2= 7.0 kg are connected by a light string that passes over a frictionless pulley as in the figure below. If, when the system starts from rest, M2 falls 1.00m in 1.33s, determine the coefficient of Kinect friction between M1 and the table
We are asked to determine the coefficient of friction in the sysmtem. First we will do a free body diagram of the first mass, like this:
Where:
[tex]\begin{gathered} N=\text{ Normal Force} \\ m_1=\text{ mass} \\ g=\text{ acceleration of gravity} \\ T=\text{ tension} \\ F_f=\text{friction force} \end{gathered}[/tex]Now, we add the forces in the horizontal direction, we get:
[tex]T-F_f=m_1a[/tex]Now, to determine the friction force we need to use the following relationship:
[tex]F_f=\mu N[/tex]To determine the normal force we will add the forces in the vertical direction. Since there is no movement in the vertical direction this sum must add up to zero:
[tex]\begin{gathered} N-m_1g=0 \\ N=m_1g \end{gathered}[/tex]Now, we substitute the value of the normal force in the equation for the friction force:
[tex]F_f=\mu m_1g[/tex]Now, we substitute the friction force in the sum of horizontal forces:
[tex]T-\mu m_1g=m_1a[/tex]Now, we turn our attention to the second mass. We add the forces in the vertical direction:
[tex]m_2g-T=m_2a[/tex]Now, since the acceleration "a" and the tension "T" is the same for both masses we will solve fot "T" in the sum of forces for the second mass, like this:
[tex]m_2g-m_2a=T[/tex]Now, we substitute the value in the sum of forces of the first mass, we get:
[tex]m_2g-m_2a-\mu m_1g=m_1a[/tex]Now, we solve for the coefficient of friction. To do that we will subtract "m2g" to both sides:
[tex]-m_2a-\mu m_1g=m_1a-m_2g[/tex]Now, we add "m2a" to both sides:
[tex]-\mu m_1g=m_1a-m_2g+m_2a[/tex]Now, we divide both sides by "-m1g":
[tex]\mu=\frac{m_1a-m_2g+m_2a}{-m_1g}[/tex]We have gotten an expression for the coefficient of friction but we need to determine the acceleration of the system. To do that we will use the fact that the second mass moves 1 meter in 1.33 seconds. Assuming constant acceleration we can use the following equation of motion:
[tex]y=v_0t+\frac{1}{2}at^2[/tex]Where:
[tex]\begin{gathered} y=\text{ distance} \\ v_0=\text{ initial velocity} \\ t=\text{ time} \\ a=\text{ acceleration} \end{gathered}[/tex]Since the mass starts from rest we have that the initial velocity is zero, therefore:
[tex]y=\frac{1}{2}at^2[/tex]Now, we solve for the acceleration. First, we multiply both sides by 2:
[tex]2y=at^2[/tex]Now, we divide both sides by the time squared:
[tex]\frac{2y}{t^2}=a[/tex]Now, we plug in the values:
[tex]\frac{2(1m)}{(1.33s)^2}=a[/tex]Solving the operations:
[tex]1.13\frac{m}{s^2}=a[/tex]Now, we substitute the values in the equation for the coefficient of friction:
[tex]\mu=\frac{(12kg)(1.13\frac{m}{s^2})-(7kg)(9.8\frac{m}{s^2})+(7kg)(1.13\frac{m}{s^2})}{-(12kg)(9.8\frac{m}{s^2})}[/tex]Solving the operations we get:
[tex]0.4=\mu[/tex]Therefore, the coefficient of friction is 0.4
A massless scaffold is held up by a wire at each end. The scaffold is 12 m long. A300-N box sits 4.0 m from the left end. What is the tension in each wire?1) left wire = 100 N; right wire = 200 N2) left wire = 200 N; right wire = 100 N3) left wire = 900 N; right wire = 2700 N4) left wire = 2700 N; right wire = 900 N
Free body diagram:
Given data:
Length of massless scaffold (end to end) L=12 m.
Weight of box m=300 N.
Length of massless scaffold (center to end) l=6 m.
As, the box sits 4.0 m from the left end, the distance of the box from the center of massless scaffold is given as,
[tex]\begin{gathered} r=6.0\text{ m}-4.0\text{ m} \\ =2.0\text{ m} \end{gathered}[/tex]Balancing force in y direction,
[tex]T_1+T_2=300\text{ N}\ldots(1)[/tex]The torque is given as,
[tex]\tau=perpendicular\text{ distance}\times force[/tex]Therefore, torque along the center of massless scaffold is given as,
[tex]\begin{gathered} \Sigma\tau=0 \\ l\times T_1+r\times(300\text{ N})-l\times\tau_2=0 \\ 6\times T_1+2\times(300\text{ N})-6\times T_2=0 \\ 6T_1+600\text{ N}-6T_2=0 \\ 6(T_1+100\text{ N}-T_2)=0 \\ T_1+100\text{ N}-T_2=0 \\ T_1-T_2=-100\text{ N}\ldots(2) \end{gathered}[/tex]Adding equation (1) and (2),
[tex]\begin{gathered} (T_1+T_2)+(T_1-T_2)=300\text{ N}-100\text{ N} \\ T_1+T_1+T_2-T_2=200\text{ N} \\ 2T_1=200\text{ N} \\ T_1=\frac{200\text{ N}}{2} \\ T_1=100\text{ N} \end{gathered}[/tex]Substituting T1 in equation (1) we get,
[tex]\begin{gathered} 100\text{ N}+T_2=300\text{ N} \\ T_2=300\text{ N}-100\text{ N} \\ T_2=200\text{ N} \end{gathered}[/tex]Therefore, tension in left wire is 100 N and tension on right wire is 200 N. Hence, option (1) is the correct choice.
How far will a bike travel if you pedal at 13 m/s for 7 seconds?
Given
[tex]\begin{gathered} v=13\text{ m/s} \\ t=7\text{ s} \end{gathered}[/tex]The distance covered is given as,
[tex]s=vt[/tex]Here, s is the distance travelled, v is the speed and t is the time .
Substituting all known values,
[tex]\begin{gathered} s=(13\text{ m/s)}\times(7\text{ s)} \\ =91\text{ m} \end{gathered}[/tex]Therefore, the bike will travel 91 m is 7 second.
I was told it was 5.886 J by another tutor on here but that was incorrect so just trying again
The potential energy of gravity is given by:
[tex]\begin{gathered} E=mgh \\ where: \\ m=0.3 \\ h=2 \\ g=9.8 \\ so: \\ E=0.3\cdot2\cdot9.8 \\ E\approx5.9J \end{gathered}[/tex]Answer:
5.9 J
For an object starting from rest and accelerating with constantacceleration, distance traveled is proportional to the square of thetime. If an object travels 2.0 furlongs in the first 2.0 s, how far willit travel in the first 4.0 s?
Since the object is accelerating with constant acceleration we can use the following formula for the position of the object:
[tex]x=x_0+v_0t+\frac{1}{2}at^2[/tex]where x0 is the initial position, v0 is the initial velocity, a is the acceleration and t is the time. In this case, the initial position and velocity are zero. Plugging the values given we have:
[tex]\begin{gathered} 2=\frac{1}{2}a(2)^2 \\ 2=2a \\ a=1 \end{gathered}[/tex]Hence, the acceleration of the object is 1 furlong per second per second.
Once we know the acceleration we can use the same formula to determine how far the object will travel in four seconds.
[tex]\begin{gathered} x=\frac{1}{2}(1)(4)^2 \\ x=\frac{16}{2} \\ x=8 \end{gathered}[/tex]Therefore, the object will travel 8 furlongs in four seconds.
I need help with this practice problem. Observe what happens in each of these three spheres with rain?Give at least two sentences each HYDROSPHERE:BIOSPHERE:GEOSPHERE:
The hydrosphere is the amount of water that there is on the planet. Water follows a cycle called the cycle of water where the heat from the sun evaporates water and is collected in clouds that eventually condensate and precipitates back to the surface in form of rain.
The biosphere is the part of the earth where there is life. When it rains this life receives the water and is used for various cycles that sustain life.
The geosphere is the soil of the earth. This soil receives the water and usually, this sinks in the soil sometimes accumulating and eventually evaporating again. This water is essential to sustain life.
I need help with #2 it off s for practice
First, find the acceleration using Newton's Second Law.
[tex]\begin{gathered} F=ma \\ a=\frac{F}{m} \end{gathered}[/tex]Where F = 8.10x10^5 N and m = 1.40x10^7 kg.
[tex]a=\frac{8.10\times10^5N}{1.40\times10^7\operatorname{kg}}=5.79\times10^{-2}\cdot\frac{m}{s^2}[/tex]Then, use a formula that relates acceleration, initial velocity, final velocity, and time.
[tex]v_f=v_0+at[/tex]Solve for t because the problem is asking to find the time.
[tex]\begin{gathered} v_f-v_0=at_{} \\ t=\frac{v_f-v_0}{a} \end{gathered}[/tex]Where vf = 64 km/h, v0 = 0, and a = 5.79x10^-2 m/s^2. Before we continue, we need to transform the final velocity to m/s.
[tex]v_f=64\cdot\frac{km}{h}\cdot\frac{1000m}{1\operatorname{km}}\cdot\frac{1h}{3600\sec}=17.78\cdot\frac{m}{s}[/tex]Once we have the velocity transformed, we are able to find t.
[tex]\begin{gathered} t=\frac{17.78\cdot\frac{m}{s}-0}{5.79\times10^{-2}\cdot\frac{m}{s^2}} \\ t=3.07\times10^2\sec \\ t=307\sec \end{gathered}[/tex]But, the answer must be in minutes.
[tex]t=307\sec \cdot\frac{1\min}{60\sec}=5.12\min [/tex]Therefore, it takes 5.12 minutes.
A spaceship travels at an enormous speed to go from one star system to another. As it reaches its destination, its “hyperdrive” deactivates, bringing the ship to an immediate halt in orbit around a planet.
Why does this defy physics? Give Equations/ Laws to justify yourself
This situation defies the laws of physics because it has no physical basis to explain the immediate stoppage of the ship.
Why does this situation defy physics?This situation defies the laws of physics because it describes an event in which an ship stops abruptly without any apparent physical explanation.
From the above, it can be inferred that this does not have scientific support. Additionally, it is defying the laws of physics because, according to Newton's second law, every body in motion continues to move at a constant speed. Therefore, if the ship goes to a hypervelocity it must continue moving, by Newton's second law. Additionally, if there were a force such as friction or gravity that causes it to stop, this would not be immediately but progressively.
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a given mass of a gas at - 73°C exerts pressure of 40 cm of mercury. what pressure will be exerted at 127 °C if the volume remains constant
The pressure that will be exerted at 127 °C if the volume remains constant is 80 cm of mercury
How do I determine the new pressure?First, we shall list out the given parameters from the question. This is shown below:
Initial temperature (T₁) = -73 °C = -73 + 273 = 200 K Initial pressure (P₁) = 40 cm of mercuryNew temperature (T₂) = 127 °C = 127 + 273 = 400 KVolume = ConstantNew pressure (P₂) = ?Thus, we can obtain the new pressure of the gas at 127 °C as illustrated below:
P₁ / T₁ = P₂ / T₂
40 / 200 = P₂ / 400
Cross multiply
200 × P₂ = 40 × 400
200 × P₂ = 16000
Divide both sides by 200
P₂ = 16000 / 200
P₂ = 80 cm of mercury
Therefore, the new pressure is 80 cm of mercury
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A can sits on a vertical wooden fencepost 1.9 meters above the ground. Billy picks up a small rock, aims at an angle ϴ = 25⁰ above the horizontal and throws the rock, releasing it 1 m above the ground with an initial speed of v0 =10 m/s. Boom! He hits the can! How far away is the fencepost?
Given,
Height of the fencepost, h=1.9 m
Angle at which the rock was thrown, θ=25°
The height at which the rock was released, a=1 m
The initial speed of the rock, v₀=10 m/s
Referring to the diagram,
[tex]\tan \theta=\frac{h-a}{d}[/tex]On rearranging the above equation,
[tex]d=\frac{h-a}{\tan \theta}[/tex]On substituting the known values,
[tex]d=\frac{1.9-1}{\tan 25^0}=1.93\text{ m}[/tex]Therefore the fencepost is at a distance of 1.93 m
Question 28 of 30The graph below shows the conservation of energy for a skydiver jumping outof a plane and landing safely on the ground. Which energy is represented byline A?сAEnergy (kJ)BDistance (m)A. Potential energyB. Kinetic energy
Answer:
Explanation:
Potential energy depends on the height of the object above the ground. As the height decreases, the potential energy decreases. Looking at line A, the energy is decreasing as the distance is increasing. Thus, the energy represented by line A is
A. Potential energy
A solenoid is wound with 259 turns per cm. An outer layer of insulated wire with 51 turns per cm is wound over the first layer of wire. The inner coil carries a current of 7.577 A, and the outer coil carries a current of 21.68 A in the opposite direction. What is the magnitude of the magnetic field, in microTeslas, at the central axis ?
Answer tab
For this question, we'll first define a number, called the "linear density of the coil" which is exactly the one the exercise gives us, in a unit of turns per length. The magnetic field can be calculated as:
[tex]B=\mu_0in[/tex]Where n is the linear density.
In our case, as the coils carry current in opposite directions, the generated magnetic fields will be opposed, and we'll have:
[tex]B=B_{outer}-B_{inner}=\mu_0*(51*10^2*21.68-259*10^2*7.577)[/tex]Please note that we had to multiply by 10^2, in order to convert turns/cm to turns/m
Then, our final magnetic field will be:
[tex]|B|=0.107664T=107664\mu T[/tex]Our final answer is B=107664uT
Describe in simple terms each of Kepler's three laws. Why were Tycho Brahe's observations so important for Kepler to develop these laws?
Kepler's first law of planetary motion states that every planet’s orbit is elliptical and has the Sun at a focus
Kepler's second law of planetary motion states that the line joining the Sun and a planet sweeps out equal areas in equal times
Kepler's third law of planetary motion states that the square of a planet’s orbital period is proportional to the cube of the semi-major axis of its orbit
Tycho Brahe provided some accurate astronomical data which makes it possible for Kepler to develop the laws of planetary motion. He devised the best instruments available before the invention of the telescope. These data and instrument provided by Tycho Brahe was able to assist Kepler in formulating these laws
Two 4.952 cm by 4.952 cm plates that form a parallel-plate capacitor are charged to +/- 0.576 nC. What is the electric field strength inside the capacitor if the spacing between the plates is 2.97 mm?
In order to solve this question, we will need to know the E-field of a capacitor
E = sigma/epsilon naught
To find sigma (charge density), we will need to divide the total charge by the area
sigma = (.576 x 10^-9)/(4.952x10^-2)^2 = 2.34 x 10^-7
E = 2.34 x 10^-7/ (8.85x10^-12) = 26541.04 V/m
If you have a convex lens whose focal length is 10.0 cm, where would you place an object in order to produce an image that is virtual?
When an object is placed between first focus and optical center of a convex lens then virtual image is produced.
Here , focal length is 10.0 cm . So object distance should be less than 10.0 cm
Final answer is : between focus and optical center of the lens
i need help asap i just need the vocab i need help with this please help
ANSWERS
0. First law of Thermodynamics:, ,B,. Thermal energy can change form and location, but it cannot be created or destroyed
,1. Thermal Energy: ,H,. Kinetic energy in transit from one object to another due to temperature difference
,2. Temperature: ,D,. the average kinetic energy of particles in an object - not the total amount of kinetic energy particles
,3. Absolute zero: ,J, occurs when all kinetic energy is removed from an object
,4. Thermal Equilibrium: ,E, obtained when touching objects within a system reach the same temperature
,5. Zeroth Law of Thermodynamics: ,C, if two systems are separately found to be in thermal equilibrium with a third system, the first two systems are in thermal equilibrium with each other.
,6. Conduction: ,F, ,the transfer of thermal energy within an object or between objects from molecule to molecule
,7. Coefficient of Heat Conductivity: ,G, is the measure of a material's ability to conduct heat.
,8. Thermal resistance of a material: ,A, the measure of a material's ability to resist heat
,9. Radiation: ,I, the process by which energy is transmitted through a medium, including empty space, as electromagnetic waves.
EXPLANATION
The zeroth and first laws of thermodynamics are known statements, as well as the definition of thermal equilibrium, so there is nothing to explain about these three.
Thermal energy is a form of kinetic energy. It is "generated" by the vibrations of the molecules or atoms that form the object, causing the object to increase its temperature. As a result, the temperature is the average kinetic energy of these particles.
Absolute zero is a temperature of 0 degrees Kelvin - approximately -273 degrees Celsius. When an object reaches this particular temperature it is considered as if the object has "no temperature" - that is why it is called absolute zero. As explained above, the temperature is the average kinetic energy of the object's particles, so if the particles have no kinetic energy, then the object has zero temperature.
Heat conductivity and thermal resistance are basically opposites. The first is the object's ability to conduct heat while the second is the ability to resist the conduction of heat.
Radiation and Conduction are two ways energy can be transferred. The first does not need a medium - that is why energy can be transmitted through empty space, and the second is by contact between two objects or an object and a medium.