The correct answer is Option B. If the two players were to collide, their combined momentum after the collision will be 200kgm/s.
To determine the combined momentum of the two players after the collision, we need to calculate the individual momenta of each player and then add them together.
The momentum of an object can be calculated using the equation:
Momentum = mass × velocity
For the first person:
Momentum1 = mass1 × velocity1
Momentum1 = 64 kg × 4.2 m/s
For the second person:
Momentum2 = mass2 × velocity2
Momentum2 = 59 kg × (-1.2 m/s) [since moving south indicates a negative velocity]
To find the combined momentum, we add the individual momenta:
Combined Momentum = Momentum1 + Momentum2
Substituting the values:
Combined Momentum = (64 kg × 4.2 m/s) + (59 kg × (-1.2 m/s))
After calculating the above expression, we find that the combined momentum is approximately 200 kgm/s.
Therefore, the correct answer is Option B.
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Trace the decay of U-238 to Ra-226 as shown in Figure 39.15 in the textFigure out what particles must be emitted in each step, and write the reaction for that step in terms of symbols
The reaction equations for the steps involved in the decay of U-238 to Ra-226 are;
[tex]^{238}_{92}U\ \rightarrow \ ^{234}_{90}Th \ + \ ^{4}_{2}He[/tex]
[tex]^{234}_{90}Th \ \rightarrow \ ^{230}_{88}Ra \ + \ ^{4}_{2}He[/tex]
[tex]^{230}_{88}Ra \ \rightarrow \ ^{226}_{86}Ra \ + \ ^{4}_{2}He[/tex]
What is the radioactive equation for the decay of U-238?The radioactive equation for the decay of U-238 to Ra-226 is calculated as follows;
First the uranium atom (U-238) will decay thorium by emitting alpha particle as shown in the equation below;
[tex]^{238}_{92}U\ \rightarrow \ ^{234}_{90}Th \ + \ ^{4}_{2}He[/tex]
The second stage is, the thorium will decay to radium by emitting alpha particles again as shown below;
[tex]^{234}_{90}Th \ \rightarrow \ ^{230}_{88}Ra \ + \ ^{4}_{2}He[/tex]
The third, and final stage, the radium will decay to an isotope of radium again, by emitting alpha particle as shown below;
[tex]^{230}_{88}Ra \ \rightarrow \ ^{226}_{86}Ra \ + \ ^{4}_{2}He[/tex]
Thus, the reaction equations for the steps involved in the decay of U-238 to Ra-226 are;
[tex]^{238}_{92}U\ \rightarrow \ ^{234}_{90}Th \ + \ ^{4}_{2}He[/tex]
[tex]^{234}_{90}Th \ \rightarrow \ ^{230}_{88}Ra \ + \ ^{4}_{2}He[/tex]
[tex]^{230}_{88}Ra \ \rightarrow \ ^{226}_{86}Ra \ + \ ^{4}_{2}He[/tex]
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-0,32 m - 4.2 At the instant the marble leaves the canon, the canon starts moving backwards (recoils) at a constant velocity. 4.1 State the principle of conservation of linear momentum in words. that is stationary canon is placed 0,32 m from a fixed bare After firing, the canon takes 0,33 s to collide with a barrier at a distance of 0,32 m. Calculate the speed the: W< 1 >E S 4.2.1 Canon collides with the barrier
The cannon will completely stop when it collides with the barrier.
To calculate the speed at which the cannon collides with the barrier, we can follow these step-by-step calculations:
Determine the initial momentum of the system.
Since the cannon is initially stationary, the initial momentum is zero.
Apply the conservation of linear momentum.
According to the principle of conservation of linear momentum, the initial momentum of the system (zero) is equal to the final momentum of the system. The final momentum is the momentum of the cannon after firing.
Calculate the final momentum of the system.
Let's assume the mass of the cannon is represented by 'm' and the final velocity of the cannon is represented by 'v'. The final momentum of the system is given by: final momentum = m × v.
Set up the equation.
Since the initial momentum is zero, we have: 0 = m × v.
Solve for the final velocity of the cannon.
Dividing both sides of the equation by 'm', we get: v = 0.
Interpret the result.
The calculation shows that the final velocity of the cannon is zero. This means that the cannon comes to a complete stop when it collides with the barrier.
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PLEASE HELP! THANK YOU
The International Space Station (ISS) experiences something called orbital decay, causing the ISS to get 90 m closer to earth every hour. What effect does orbital decay have on the velocity and orbital velocity of the ISS?
orbital decay causes the ISS to get closer to Earth, resulting in an increase in its velocity but not affecting its orbital velocity. The increase in velocity helps the ISS maintain its orbital stability despite the decrease in altitude.
Orbital decay of the International Space Station (ISS) leads to a decrease in its altitude, causing it to get closer to Earth.
This decrease in altitude affects the velocity and orbital velocity of the ISS.
As the ISS gets closer to Earth, the gravitational pull becomes stronger, resulting in an increase in its velocity.
This increase in velocity is a consequence of the conservation of angular momentum.
According to this principle, as the orbiting object gets closer to the central body (in this case, Earth), its velocity must increase to maintain the balance between gravitational pull and centrifugal force.
However, it's important to note that the increase in velocity due to orbital decay does not affect the orbital velocity of the ISS.
Orbital velocity is the minimum velocity required to maintain a stable orbit at a given altitude.
Even though the ISS experiences orbital decay, its orbital velocity remains constant because the decrease in altitude compensates for the increase in velocity, ensuring that the ISS continues to maintain a stable orbit.
In summary, orbital decay causes the ISS to get closer to Earth, resulting in an increase in its velocity but not affecting its orbital velocity. The increase in velocity helps the ISS maintain its orbital stability despite the decrease in altitude.
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find the electrical energy expended by an electric lamp labelled 100w in 8 seconds
Answer:800 Joules
Explanation: Energy = Power × Time = 100∗8100∗8 = 800 joules
B6. [9 Marks] 30⁰ 20140X20 DE Ofe OTO A stainless-steel orthodontic wire is applied to a tooth as shown in the diagram below. The wire has an unstretched length of 3.1 cm and a diameter of 0.22 mm. If the wire is stretched by 0.10 mm during the procedure, find the magnitude and direction of the force on the tooth. Disregard the width of the tooth and assume Young's modulus for stainless-steel is 18 × 10¹0 Nm-².
The magnitude of the force on the tooth is approximately 0.022 N.
To find the magnitude and direction of the force on the tooth, we can use Hooke's Law, which states that the force exerted on an object is directly proportional to the change in length of a material when it is stretched or compressed.
First, we need to calculate the strain (ε) of the stainless-steel wire.
Strain is defined as the change in length divided by the original length:
ε = ΔL / L₀
Given that the change in length (ΔL) is 0.10 mm [tex](0.10 \times 10^{-3} m)[/tex] and the unstretched length (L₀) is 3.1 cm [tex](3.1 \times 10^{-2} m)[/tex], we can calculate the strain:
[tex]\epsilon=(0.10 \times 10^{-3} m)/(3.1 \times 10^{-2} m)=0.003225[/tex]
Next, we can use Young's modulus (E) to calculate the stress (σ) in the wire.
Stress is defined as the force per unit area:
σ = E * ε
Given that Young's modulus (E) for stainless-steel is 18 × 10¹⁰ N/m², we can calculate the stress:
σ = (18 × 10¹⁰ N/m²) * 0.003225 = 5.805 × 10⁸ N/m²
Now, we can find the force (F) on the tooth by multiplying the stress by the cross-sectional area (A) of the wire:
F = σ * A
The cross-sectional area (A) can be calculated using the formula for the area of a circle:
A = π * (d/2)²
Given that the diameter (d) of the wire is 0.22 mm[tex](0.22 \times 10^{-3} m)[/tex], we can calculate the cross-sectional area:
[tex]A = \pi * (0.22 \times 10^-3 m / 2)^{2} = 3.802 \times 10^{-8} m^2[/tex]
Finally, we can calculate the force:
[tex]F = (5.805 \times 10^{8} N/m^{2}) * (3.802 \times 10^-8 m^{2}) \approx 2.206 \times 10^{-2} N[/tex]
Therefore, the magnitude of the force on the tooth is approximately 0.022 N.
Since the wire is stretched, the force is pulling the tooth in the direction opposite to the stretching.
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Which two layers in this model represent parts of Earth that are mostly iron
and nickel?
A. A
B. D
C. B
D. C
The correct answer is Option C. and D. The Earth consists of several layers, and the two layers that represent parts of Earth that are mostly iron and nickel are layers B and C.
The Earth's structure has been divided into several layers based on the physical and chemical properties of the materials that make up each layer.
The model is made up of four main layers: the crust, mantle, outer core, and inner core.
The outer core and inner core are the two layers that are primarily made up of iron and nickel.
Layer B, the mantle, is made up of magnesium, silicon, and iron.
Layer C, the outer core, is made up of molten iron and nickel, with some lighter elements such as sulfur and oxygen.
The inner core, layer D, is mostly composed of solid iron and nickel.
The outer core is a liquid layer located between the mantle and the inner core.
This layer is approximately 2,300 km thick and is composed primarily of iron and nickel, with some lighter elements such as sulfur and oxygen.
The outer core is responsible for generating the Earth's magnetic field.
It's also responsible for producing seismic waves that help scientists learn about the interior of the
Earth. The inner core is a solid layer located at the center of the Earth, approximately 6,371 km below the Earth's surface.
It is composed mainly of iron and nickel, with trace amounts of lighter elements such as sulfur and oxygen.
The pressure at the center of the Earth is so high that the iron and nickel in the inner core are solid, despite being at a temperature of over 5,500°C.
Therefore, The correct answer is Option C. and D.
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Part C
Increase the value of the Applied Force to 150 N. Pause the simulation and observe the magnitudes and directions of the applied force, the friction
force, and the sum of forces. Why do you think the crate moves this time?
The crate moves this time because the applied force of 150 N is greater than the maximum static friction force that opposes the motion of the crate. Static friction is a force that opposes the relative motion between two objects in contact. The maximum static friction force is determined by the normal force and the coefficient of static friction between the two surfaces in contact. The harder the surfaces are pushed together, the more force is needed to move them. When the applied force exceeds the maximum static friction force, the crate will start to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force.
A ball is allowed to fall freely from certain height it covers a distance of 1st sec equal to?
The distance covered by a ball in the first second of free fall is approximately 4.9 meters.
When an object falls freely under the influence of gravity, it experiences constant acceleration. In the case of Earth's gravity, the acceleration due to gravity is approximately 9.8 m/s². This means that the velocity of the falling object increases by 9.8 meters per second every second.
To determine the distance covered by the ball in the first second, we can use the equations of motion for uniformly accelerated motion.
The equation that relates distance (d), initial velocity (u), acceleration (a), and time (t) is:
d = ut + (1/2)at²
In this case, the initial velocity is zero (as the ball starts from rest), the acceleration is 9.8 m/s², and we want to find the distance covered in the first second (t = 1 second).
Plugging in the values:
d = 0 * 1 + (1/2) * 9.8 * (1)^2
d = 0 + (1/2) * 9.8
d = 0 + 4.9
d = 4.9 meters
Therefore, the ball covers a distance of approximately 4.9 meters in the first second of free fall.
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how many seconds makes 20 years (show all workings)
There are 630,720,000 seconds in 20 years.
To calculate the number of seconds in 20 years, we need to consider the number of seconds in a minute, minutes in an hour, hours in a day, and days in a year.
1 minute consists of 60 seconds.
1 hour contains 60 minutes (60 minutes × 60 seconds = 3600 seconds).
1 day consists of 24 hours (24 hours × 3600 seconds = 86,400 seconds).
1 year typically consists of 365 days (365 days × 86,400 seconds = 31,536,000 seconds).
To find the number of seconds in 20 years, we multiply the number of seconds in one year by 20:
20 years × 31,536,000 seconds = 630,720,000 seconds.
Therefore, there are 630,720,000 seconds in 20 years.
This calculation assumes a non-leap year with 365 days.
If the 20 years span a leap year, the total number of seconds would be slightly higher, accounting for the extra day in the leap year.
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