the answer is: magnitude = 5, direction = positive y-axis.
To determine the magnitude and direction of the revitant force, we need to break down the given vectors. We have a vector F1 and a vector F2, and we know that the sum of these vectors is in the positive y-axis direction.
We can start by graphing these vectors to visualize the problem. Then we can use trigonometry to determine the magnitude and direction of the revitant force.
Step 1: Graph F1 and F2Since we know that the sum of the vectors is in the positive y-axis direction, we can assume that F1 and F2 point in opposite directions. Therefore, we can graph F1 pointing up and F2 pointing down, both starting at the origin.
Here is a graph of the two vectors:graph{F1+F2 [-5, 5, -5, 5]}
Step 2: Break down the vectors into componentsTo determine the revitant force, we need to break down F1 and F2 into their x- and y-components. We can use trigonometry to do this. Here are the components of each vector: F1:x-component: 0y-component: 15F2:x-component: 0y-component: -10
Step 3: Add up the components of the vectorsNow that we have the components of each vector, we can add them up to find the components of the revitant force. The x-component is 0 since there is no horizontal component. The y-component is: 15 - 10 = 5So the revitant force has a magnitude of 5 and points in the positive y-axis direction.
Therefore, the answer is: magnitude = 5, direction = positive y-axis.
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you lifted up a snowball to a height of 0.20 m at constant velocity. how much is the mass of the snowball when the potential energy of the snowball is 0.089 joules?
the magnitude of gravitational acceleration is 9.8 m/s^2.
You lifted up a snowball to a height of 0.20 m at constant velocity. The mass of the snowball is 0.0454 kg when the potential energy of the snowball is 0.089 joules.
To find the mass of the snowball, we can use the formula for gravitational potential energy:
PE = mgh
Where:
PE is the potential energy,
m is the mass of the object,
g is the gravitational acceleration, and
h is the height.
Given:
PE = 0.089 J
g = 9.8 [tex]m/s^2[/tex]
h = 0.20 m
Rearranging the formula, we have:
m = PE / (gh)
Substituting the given values:
m = 0.089 J / (9.8 [tex]m/s^2[/tex] * 0.20 m)
m = 0.089 J / 1.96 J
m ≈ 0.0454 kg
Therefore, the mass of the snowball is approximately 0.0454 kg.
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Two blocks of masses 20 kg and 8.0 kg are connected together by a light string and rest on a frictionless level surface. Attached to the 8.0 kg mass is a second light string, which a person uses to pull both blocks horizontally with a force of 14 N. What is the tension in the connecting string between the two blocks?
The tension in the connecting string between the two blocks is 196 N.
In this scenario, there are two blocks. The masses of the two blocks are 20 kg and 8.0 kg, respectively. The two blocks are connected by a light string and are placed on a smooth, horizontal surface with no friction. There is also a second light string that is attached to the 8.0 kg mass. The person uses this string to pull both blocks horizontally with a force of 14 N.
Since the surface is frictionless, there is no frictional force acting on the block. Now let's draw the free-body diagram for the 8.0 kg block.
Now let's apply Newton's second law to each block. For the 20 kg block: ΣF = maT - W = 20a
Since the block is not accelerating, we know that ΣF = 0. Therefore, we can say that:
T = W = mgT = 20(9.8)T = 196 N
For the 8.0 kg block: ΣF = maF - T = 8a + 14
Since the block is accelerating, we need to solve for the acceleration first. We can use the equation for the net force to do this:
ΣF = maF - T = 8a + 14a = (F - T)/8 - 14/8
Now we can plug in the values for F and T: a = (14 - T)/8 - 14/8a = (14 - T - 14)/8a = (T - 28)/8
Now we can plug this value for a into the equation for the net force:
ΣF = maF - T = 8a + 14F - T = 8[(T - 28)/8] + 14F - T = T - 28 + 14F = 2TF = T/2 + 14
Now we can solve for T by substituting this expression for F into the equation we found for the tension in the 20 kg block: T = 20g = 196 N
Thus, the tension in the connecting string between the two blocks is 196 N.
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is the difference in the water pressure between the top and bottom of the pipe? \( \mathrm{Pa} \)
The difference in water pressure between the top and bottom of a pipe is determined by the height of the water column and the density of the water. The pressure at the bottom is higher than the pressure at the top due to the weight of the water column.
The difference in water pressure between the top and bottom of a pipe is determined by the height of the column of water in the pipe and the gravitational force acting on it. This is known as hydrostatic pressure.
The pressure at any point in a fluid is given by the formula P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the column of fluid above the point.
In this case, the water pressure at the bottom of the pipe is higher than the pressure at the top due to the weight of the water column above it. The difference in pressure is equal to the weight of the water column divided by the cross-sectional area of the pipe.
For example, if the height of the water column is 10 meters and the density of water is 1000 kg/m^3, the pressure difference between the top and bottom of the pipe would be approximately 100,000 Pa (Pascals).
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Water travels at 3.15 m/s in a horizontal pipe with cross-sectional area 0.45 m2. The pipe rises by 4.8 m, while the cross sectional area narrows by half. If the pressure in the lower part of the pipe is 143,319, what is the pressure in the upper part of the pipe?
Therefore, the pressure in the upper part of the pipe is 127,541.81 Pa. Answer: 127,541.81 Pa.
Given, Water travels at 3.15 m/s in a horizontal pipe with cross-sectional area 0.45 m2. The pipe rises by 4.8 m, while the cross-sectional area narrows by half. If the pressure in the lower part of the pipe is 143,319 Pa, we need to find the pressure in the upper part of the pipe.
Let's use Bernoulli's equation to solve the problem.
Bernoulli's equation states that the sum of the pressure, kinetic energy and potential energy per unit volume of a fluid is constant at all points along a streamline.
Here, the fluid flow is horizontal, so the potential energy per unit volume is the same at all points.
Therefore, we can ignore it while using the Bernoulli equation. Let p1 and p2 be the pressures at the lower and upper points respectively and A1 and A2 be the cross-sectional areas at these points respectively. Using Bernoulli's equation, we get:[tex]p1 + (1/2)ρv1^2 = p2 + (1/2)ρv2^2[/tex] .....................(1)
We know that the fluid flow is horizontal.
Therefore, the height difference (Δh) between the two points is 4.8 m.
We can find the velocity of the fluid at each point using the continuity equation. The continuity equation states that the mass flow rate is constant along a streamline.
Mathematically, it can be written as A1v1 = A2v2...................(2)
Here, the cross-sectional area at the upper point (A2) is half of that at the lower point (A1).
Therefore, we have A2 = (1/2)A1.
Substituting this value of A2 in equation (2), we get:
v2 = 2v1
Now, we can substitute the values of v1, v2, p1, ρ, A1, and A2 in equation (1) to get the value of p2.
[tex]Let's do that: p1 + (1/2)ρv1^2[/tex]
[tex]= p2 + (1/2)ρv2^2p2[/tex]
[tex]= p1 + (1/2)ρ(v1^2 - v2^2)[/tex]
[tex]p2 = 143,319 + (1/2) × 1000 × (3.15^2 - 2 × 3.15^2)[/tex]
[tex]p2 = 143,319 - (1/2) × 1000 × 3.15^2[/tex]
p2 = 143,319 - 15,777.19
p2 = 127,541.81 Pa
Therefore, the pressure in the upper part of the pipe is 127,541.81 Pa. Answer: 127,541.81 Pa.
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Find the net force of two forces F1 = 20 N and F2 = 30 N when they are forming an angle of 120 between them.
The net force of the two forces with magnitudes 20 N and 30 N, forming an angle of 120° between them, is approximately 51.1 N directed at an angle of -60.9° with the horizontal axis.
To find the net force of two forces when they are forming an angle between them, we can use the concept of vector addition. The net force is the vector sum of the individual forces.
Given:
F1 = 20 N (magnitude of force 1)
F2 = 30 N (magnitude of force 2)
θ = 120° (angle between the forces)
To find the net force, we need to break down the forces into their horizontal and vertical components. The horizontal component is given by F * cos(θ), and the vertical component is given by F * sin(θ).
For Force 1 (F1):
F1x = F1 * cos(θ) = 20 N * cos(120°) = -10 N (horizontal component)
F1y = F1 * sin(θ) = 20 N * sin(120°) = 17.32 N (vertical component)
For Force 2 (F2): F2 is acting along positive x-axis.
F2x = F2 * cos(θ) = 30 N * cos(120°) = -15 N (horizontal component)
F2y = F2 * sin(θ) = 30 N * sin(120°) = 25.98 N (vertical component)
Now, we can find the net horizontal and vertical components by adding the corresponding components of the forces:
Net horizontal component (Fx) = F1x + F2x = -10 N + (-15 N) = -25 N
Net vertical component (Fy) = F1y + F2y = 17.32 N + 25.98 N = 43.3 N
Finally, we can calculate the magnitude and direction of the net force using the net horizontal and vertical components:
Magnitude of the net force[tex]$(F_{net}) = \sqrt{(F_x)^2 + (F_y)^2} = \sqrt{(-25 \text{ N})^2 + (43.3 \text{ N})^2} \approx 51.1 \text{ N}$[/tex]
Angle (θnet) of the net force with the horizontal axis can be found using the inverse tangent (arctan) function:
[tex]\theta_{net} = \arctan\left(\frac{F_y}{F_x}\right) = \arctan\left(\frac{43.3 \, \text{N}}{-25 \, \text{N}}\right) \approx -60.9^{\circ}[/tex]
The negative sign indicates that the net force is directed in the fourth quadrant, below the negative x-axis.
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A car has a velocity vector with components 24 m/s east and 24 m/s north. What is the speed of the car?
The speed of the car can be calculated using the Pythagorean theorem as the magnitude of the velocity vector, which is [tex]\sqrt((24 m/s)^2 + (24 m/s)^2) = 33.94 m/s.[/tex]
For finding the speed of the car, calculate the magnitude of the velocity vector. Given that the car's velocity vector has components of 24 m/s east and 24 m/s north, can represent it as a right-angled triangle. The magnitude of the velocity vector (speed) can be found using the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Using this theorem, calculate the speed as follows:
[tex]speed = \sqrt((24 m/s)^2 + (24 m/s)^2)\\= \sqrt(576 m^2/s^2 + 576 m^2/s^2)\\= \sqrt(1152 m^2/s^2)\\\approx 33.94 m/s.[/tex]
Therefore, the speed of the car is approximately 33.94 m/s.
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A hill that has a 17.4% grade is one that rises 17.4 m vertically for every 100.0 m of distance in the horizontal direction. At what angle is such a hill inclined above the horizontal? Number Units
The hill is inclined at an angle of approximately 9.96 degrees above the horizontal.
To find the angle at which a hill is inclined above the horizontal, we can use trigonometry. The angle can be determined by taking the inverse tangent of the ratio of the vertical rise to the horizontal distance.
Given:
Grade of the hill = 17.4% = 17.4/100 = 0.174
Vertical rise = 17.4 m
Horizontal distance = 100.0 m
Let θ represent the angle of inclination.
θ = arctan(vertical rise / horizontal distance)
θ = arctan(17.4 / 100.0)
Using a calculator or trigonometric tables, we can find the inverse tangent of the ratio to determine the angle:
θ ≈ 9.96 degrees
Therefore, the hill is inclined at an angle of approximately 9.96 degrees above the horizontal.
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A child on a bob sled with a combined mass of 67 kg is on a horizontal surface. What is the minimum force the child needs to apply to start sliding given a static coefficient of friction between the ice and the sled is 0.26 ? Use g=9.81 ms −2where needed Answer to 2 decimal places. Remember to include the unit in your answer. The kinetic coefficient of friction is 0.04.
The minimum force the child needs to apply to start sliding is approximately 170.28 N (Newtons).
To determine the minimum force the child needs to apply to start sliding, we need to consider the force of friction acting on the sled. The force of friction can be calculated using the equation:
Frictional force = coefficient of friction * Normal force
The normal force is equal to the weight of the child and the sled, which can be calculated as:
Normal force = mass * gravitational acceleration
Given that the combined mass of the child and sled is 67 kg and the gravitational acceleration is 9.81 m/s^2, we have:
Normal force = 67 kg * 9.81 m/s^2 = 656.27 N
Now we can calculate the minimum force required to overcome static friction:
Minimum force = coefficient of friction * Normal force
Minimum force = 0.26 * 656.27 N = 170.28 N
Therefore, The minimum force the child needs to apply to start sliding is approximately 170.28 N (Newtons).
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Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is 6.98 m. The stones are thrown with the same speed of 8.57 m/s. Find the location (above the base of the cliff) of the point where the stones cross paths. D=
The location where the stones cross paths is approximately 3.07 meters above the base of the cliff.
The downward throw (starting from the top of the cliff) is y1(t) = 6.98 m - (1/2)gt².
The upward throw (starting from the base of the cliff) is y2(t) = 1.5 m + (1/2)gt².
Where g is the acceleration due to gravity, which is -9.81 m/s².
Based on the time t, let's say that the two stones cross paths at a height of D meters above the base of the cliff.
Therefore, we can :6.98 - (1/2)gt² = 1.5 + (1/2)gt²6.98 - 1.5 = gt²g = 4.1 m/s²
We can plug g into the first equation:y1(t) = 6.98 - (1/2)(4.1)t²
We can plug g into the second equation:y2(t) = 1.5 + (1/2)(4.1)t²
Now let's use the fact that both stones are at the same height D when they cross paths:
y1(t) = y2(t)6.98 - (1/2)(4.1)t² = 1.5 + (1/2)(4.1)t²5.48 = (4.1)t²t = sqrt(5.48/4.1) = 1.15 s
We can plug this time back into either equation to find D:
D = y1(1.15) = 6.98 - (1/2)(4.1)(1.15)²D ≈ 3.07 m
Therefore, the location where the stones cross paths is approximately 3.07 meters above the base of the cliff.
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Two 1.8 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has +10μC of charge. What is the magnitude of the electric force on one of the masses? What is the initial acceleration of the mass if it is released and allowed to move?
The initial acceleration of the mass if it is released and allowed to move is 83.33 m/s².
Given data;Masses = 2 * 1.8 kg = 3.6 kg
Distance between masses = 1.1 mCharge on each mass = +10 μC
Friction is frictionlessWe know that Coulomb's Law states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. That is,F = k q₁ q₂ / r²where k is Coulomb's constant = 9 × 10⁹ N m² C⁻², q₁ and q₂ are the charges in Coulombs, and r is the distance between the two charges in meters.
Let us calculate the force between the two masses.
Since both the masses have same charge, and they are equidistant, force on either of the two masses can be found using this formula.F = k q₁ q₂ / r²F = (9 * 10^9) * (10 * 10^-6)^2 / (1.1/2)^2F = 150 N
So the magnitude of the electric force on one of the masses is 150 N.The initial acceleration of the mass if it is released and allowed to move is given by the formula;
a = F/mwhere F is force, and m is the mass of the object.a = 150 N / 1.8 kg = 83.33 m/s²Thus, the initial acceleration of the mass, if it is released and allowed to move, is 83.33 m/s².
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A small turbo-prop commuter airplane, starting from rest on a Lansing airport runway, accelerates for 19.0 s before taking off. Its speed at takeoff is 50.0 m/s. Calculate the acceleration of the plane, in g′s, assuming it remains constant. Acceleration is the change in velocity divided by the time taken to change the velocity. Divide your answer by 9.81 to get the number of g′s. Tries 3/12 Previous Tries In the problem above, how far did the plane move while acceleratina. A Boeing 727-200 jet taking off from Lansing airport accelerates from rest for 29.0 s before leaving the ground. Its acceleration is 2.19 m/s2. Assuming that the acceleration is constant, calculate the plane's speed at take off. Tries 0/12 What is the take off speed in mi/hr ?
The acceleration of the small turbo-prop commuter airplane is approximately 0.268 g's, and the takeoff speed of the Boeing 727-200 jet is approximately 142.07 mi/hr.
To calculate the acceleration of the plane in g's, we can use the formula:
acceleration = (change in velocity) / (time taken)
The speed at takeoff is 50.0 m/s and the time taken to accelerate is 19.0 s, we can calculate the acceleration:
acceleration = (50.0 m/s - 0 m/s) / 19.0 s
acceleration = 50.0 m/s / 19.0 s
acceleration ≈ 2.63 m/s²
To convert this acceleration to g's, divide by the acceleration due to gravity (9.81 m/s²):
acceleration in g's = 2.63 m/s² / 9.81 m/s²
acceleration in g's ≈ 0.268 g's
Therefore, the acceleration of the plane is approximately 0.268 g's.
In the second problem, we are given that a Boeing 727-200 jet accelerates at a constant rate of 2.19 m/s² for 29.0 s before taking off. We need to calculate the speed at takeoff.
Using the formula:
final velocity = initial velocity + (acceleration * time)
Since the initial velocity is 0 m/s, we can calculate the final velocity:
final velocity = 0 m/s + (2.19 m/s² * 29.0 s)
final velocity = 63.51 m/s
To convert the speed from meters per second to miles per hour, we can use the conversion factor: 1 m/s = 2.237 mi/hr
takeoff speed in mi/hr = 63.51 m/s * 2.237 mi/hr
takeoff speed in mi/hr ≈ 142.07 mi/hr
Therefore, the takeoff speed of the Boeing 727-200 jet is approximately 142.07 mi/hr.
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How do the delays of the AND gates compare with the delays in the data sheet for the 74LS08 chip? (b) Why did we put in a square wave at one input of the AND gate and a 1 in the other? (c) Are the delays of all the not gates the same. If so, could they have been different? What may be the cause for different delays for gates in the chip? (d) A NAND gate has the functionality of an AND gate followed by a NOT gate. Compare the sum of the delays of an AND gate and one NOT gate (that you determined), with that of a NAND gate (obtained from the data sheet for 74LS00). What can you conclude about how the NAND gate has been constructed? (e) Draw a diagram of the circuit for the ring oscillator. Put in a logic 0 at the αβ=y5 input. Let this logic value propagate through the inverters 1,2,3,4,5, until it comes back to where it started. What is the new value. How long do you think it takes for this new value to be generated at αβ. (f) How is the time-period of the ring oscillator related to the sum of the gate delays of inverters 1−5 ?
In summary, the delays of the AND gates in the 74LS08 chip can be found in the datasheet and may vary due to factors such as temperature and voltage. A square wave and a constant 1 are used to test the behavior of the AND gate.
The delays of the NOT gates in the chip are typically the same, but can have slight variations. The delay of a NAND gate can be compared to the sum of the delays of an AND gate and a NOT gate to determine its construction. The time-period of a ring oscillator is determined by the sum of the gate delays of the inverters.
(a) The delays of the AND gates in the 74LS08 chip can be found in the datasheet. These delays are typically measured in terms of propagation delay, which is the time it takes for the output of the gate to respond to a change in input. The datasheet provides the maximum and minimum propagation delays for the chip, as well as the typical value. The delays of the AND gates can vary depending on factors such as temperature and voltage.
(b) A square wave is used at one input of the AND gate and a 1 is used at the other to test the behavior of the gate under different input conditions. By using a square wave, we can observe how the gate responds to changing inputs and whether it introduces any delays or distortions in the output signal. The use of a 1 at the other input allows us to determine how the gate performs when one input is held constant.
(c) The delays of all the NOT gates in the chip are typically the same. However, it is possible for them to have slightly different delays due to variations in manufacturing or environmental factors. The cause for different delays among gates in the chip can be attributed to factors such as process variations, transistor mismatch, and parasitic capacitance. These factors can affect the performance of individual gates and result in slight variations in their delays.
(d) A NAND gate is constructed by combining an AND gate followed by a NOT gate. The sum of the delays of an AND gate and a NOT gate can be compared with the delay of a NAND gate from the datasheet to determine the construction of the NAND gate. If the sum of the delays of the individual gates is similar to the delay of the NAND gate, it suggests that the NAND gate is implemented using an AND gate and a NOT gate in series.
(e) Unfortunately, the information provided does not specify the circuit diagram or provide sufficient information to determine the new value and the time it takes for it to be generated at αβ.
(f) The time-period of the ring oscillator is related to the sum of the gate delays of the inverters 1-5. Each inverter in the ring oscillator introduces a delay, and the time taken for the signal to propagate through each inverter contributes to the overall time-period of the oscillator. The sum of the gate delays determines the time it takes for the ring oscillator to complete one full cycle or oscillation.
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Most long period comets have orbits that are and have orbital inclinations. prograde and retrograde - low retrograde - high prograde - low prograde and retrograde - high
The most long period comets have orbits that are high retrograde and have high orbital inclinations.
Long-period comets are defined as comets that take more than 200 years to orbit around the sun once.
Since they spend more time in outer space, they are also known as Oort cloud comets. Oort cloud is a massive, spherical cloud of comets that surrounds our solar system.
It stretches out as far as halfway to the nearest star. Alpha Centauri, which is 4.3 light-years away from our solar system. Orbital inclinations refer to the angle at which a comet's orbit is inclined to the plane of the ecliptic.
Ecliptic is the imaginary plane where the Earth and other planets in our solar system orbit around the Sun. Most long-period comets have highly elliptical orbits, which means their orbits can take them far out of the solar system.
Some even travel beyond the Kuiper belt, which is a region of space beyond Neptune that contains objects such as Pluto. The inclination of a comet's orbit is either prograde or retrograde.
Prograde orbits refer to orbits that are in the same direction as the planets in our solar system. Retrograde orbits, on the other hand, are in the opposite direction.
Most long-period comets have high orbital inclinations, which means their orbits are inclined at steep angles to the plane of the ecliptic.
These comets have highly elliptical orbits, which means they travel to great distances from the Sun.
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An aluminum bar has the desired length when at 20 ∘C. How much stress is required to keep it at this length if the temperature increases to 31 ∘C ? Young's modulus for the aluminum is E = 70 ×109 N/m2 and the coefficient of thermal expansion of aluminium is 25 ×10-6 1/C∘.
The stress required to keep it at this length if the temperature increases to 31 ∘C is approximately 19,250 N/m^2.
To calculate the stress required to keep the aluminum bar at its original length when the temperature increases, we can use the formula:
Stress = Young's modulus * Coefficient of thermal expansion * Temperature change
Young's modulus (E) = 70 × 10^9 N/m^2
Coefficient of thermal expansion (α) = 25 × 10^-6 1/°C
Temperature change (ΔT) = (31 °C - 20 °C) = 11 °C
Calculate the stress:
Stress = E * α * ΔT
Stress = (70 × 10^9 N/m^2) * (25 × 10^-6 1/°C) * (11 °C)
Stress ≈ 19,250 N/m^2
Therefore, the stress required to keep the aluminum bar at its original length when the temperature increases to 31 °C is approximately 19,250 N/m^2.
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A shot putter can throw the shot with an initial velocity of 20 m/s from a height of 1.8 m at an angle of 30 degrees to the horizontal. With what speed does the shot hit the ground?
The shot hits the ground with a speed of approximately 10.54 m/s.
To find the speed at which the shot hits the ground, we can analyze the horizontal and vertical components of its motion.
Given:
Initial velocity (v₀) = 20 m/s
Height (h) = 1.8 m
Launch angle (θ) = 30 degrees
Acceleration due to gravity (g) = 9.8 m/s²
First, we can find the time it takes for the shot to hit the ground by analyzing its vertical motion. Using the equation of motion:
h = v₀ᵧ * t - (1/2) * g * t²
where v₀ᵧ is the initial vertical component of velocity. Since the shot is launched horizontally, v₀ᵧ = 0. Therefore, we can simplify the equation to:
h = -(1/2) * g * t²
Solving for t:
t² = (2h) / g
t = √((2 * 1.8 m) / 9.8 m/s²)
t ≈ 0.608 s
Next, we can find the horizontal component of velocity (vₓ) using the initial velocity and launch angle:
vₓ = v₀ * cos(θ)
vₓ = 20 m/s * cos(30°)
vₓ ≈ 17.32 m/s
Finally, we can find the speed at which the shot hits the ground by multiplying the horizontal component of velocity by the time of flight:
Speed = vₓ * t
Speed = 17.32 m/s * 0.608 s
Speed ≈ 10.54 m/s
Therefore, the shot hits the ground with a speed of approximately 10.54 m/s.
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2)[30] A \( 4 \mathrm{~kg} \) mass in free space (no gravity) is seen moving with a constant acceleration of \( 2 \mathrm{~m} / \mathrm{s}^{2} \) in the \( +\mathrm{j} \) direction because of four for
A \(4 \mathrm{~kg}\) mass in free space (no gravity) is seen moving with a constant acceleration of \(2 \mathrm{~m}/\mathrm{s}^2\) in the \(+\mathrm{j}\) direction because of four forces.
The forces acting on the mass are:
1. The gravitational force (Fg) acting on the mass is zero because there is no gravity.
2. The normal force (N) is the force exerted by any surface on the object in contact. In this case, since the mass is in free space, the normal force acting on the mass is also zero.
3. The frictional force (Ff) is also zero as there is no friction in free space.
4. The external force (Fe) is the only force acting on the mass.
The external force is responsible for the motion of the mass. In this case, the external force is \(F_e = ma = 4 \mathrm{~kg} \times 2 \mathrm{~m}/\mathrm{s}^2 = 8 \mathrm{~N}\).
Hence the force is in the \(+\mathrm{j}\) direction.
The only force acting on the mass is the external force of \(8 \mathrm{~N}\) acting in the \(+\mathrm{j}\) direction.
The mass will continue to move with a constant acceleration of \(2 \mathrm{~m}/\mathrm{s}^2\) in the \(+\mathrm{j}\) direction as long as the external force remains constant.
If the external force changes, the acceleration of the mass will also change.
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Describe how the behavior of the velocity and acceleration vectors over time, and how they are affected by air resistance.
As the object moves through the air, air resistance opposes its motion. The velocity of the object decreases as a result of the force of air resistance.
When the object stops accelerating, it reaches its maximum speed, or terminal velocity. The direction of the velocity vector is in the direction of the object's motion. This direction remains constant unless the object is redirected. Its magnitude, on the other hand, is determined by the object's speed. It changes over time as a result of air resistance. When air resistance opposes the motion of the object, the acceleration vector is impacted. The acceleration vector is in the same direction as the net force acting on the object. The force of air resistance opposes the object's direction of motion, reducing the net force acting on it. As a result, the acceleration vector decreases over time.
When an object travels through the air, the air resistance force opposes its motion, causing the object's velocity to decrease. The direction of the velocity vector is always in the direction of the object's motion. The magnitude of the velocity vector changes over time as a result of air resistance.When air resistance opposes the object's motion, the acceleration vector is affected. The acceleration vector is in the same direction as the net force acting on the object. Air resistance opposes the object's direction of motion, reducing the net force acting on it, thus causing the acceleration vector to decrease over time. Finally, the object will achieve its maximum speed or terminal velocity because the air resistance force balances the force of gravity.
Air resistance has a significant impact on the behavior of velocity and acceleration vectors over time. Air resistance causes the object's velocity to decrease over time, while the direction of the velocity vector remains constant. Air resistance also opposes the motion of the object, causing the acceleration vector to decrease over time. As a result of air resistance, the object eventually reaches its maximum speed or terminal velocity.
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In the figure below, m = 1 and the coefficient of kinetic friction is = 0.1. Determine the tension in the rope and the acceleration of the blocks.
In the given scenario, the mass of the block is 1 kg, and the coefficient of kinetic friction is 0.1. We need to determine the tension in the rope and the acceleration of the blocks.
The tension in the rope can be found by considering the forces acting on the block. There are two forces acting on the block: the force of gravity (mg) pulling it downwards and the tension in the rope pulling it upwards. Since the block is in equilibrium vertically, the tension in the rope must be equal to the force of gravity, which is given by T = mg.
To find the acceleration of the blocks, we need to consider the net force acting on the system. The only horizontal force acting on the block is the force of friction (μkN), where N is the normal force. The normal force is equal to the weight of the block (N = mg). The net force can be calculated as the difference between the tension force and the force of friction: net force = T - μkN.
Using Newton's second law (F = ma), we can equate the net force to the mass of the block (1 kg) multiplied by its acceleration (a). Therefore, we have the equation T - μkN = ma.
Substituting the values, T - 0.1 * (1 kg * 9.8 m/s^2) = 1 kg * a. Solving for T and a, we can find the tension in the rope and the acceleration of the blocks.
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When a certain traffic light turns green, a waiting car starts off with constant acceleration of 3.0 m/s2 . At the same instant a truck with constant speed 12 m/s passes by the car in the next lane. (a) How far must the car travel in order to catch up to the truck (and then pass)? (b) How fast will the car be moving as it passes the truck?
The car must travel a distance of 48 meters to catch up to the truck and pass it. When the car passes the truck, its speed will be 27 m/s.
(a) To find the distance the car must travel to catch up to the truck, we can use the equation of motion: s = ut + (1/2)[tex]at^2[/tex], where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. The truck is already at a constant speed, so its acceleration is zero. The car's initial velocity, u, is also zero because it starts from rest when the light turns green. We need to determine the time it takes for the car to catch up to the truck. Let's assume this time is t. Since the truck is moving at a constant speed, its distance traveled in time t is given by d = vt, where v is the truck's speed. Equating the distances traveled by the car and the truck, we have: 3[tex]t^2[/tex]/2 = 12t. Simplifying the equation, we find t = 8 seconds. Now, we can calculate the distance traveled by the car using s = (1/2)[tex]at^2[/tex]. Plugging in the values, we get s = (1/2) * 3 * [tex](8)^2[/tex] = 48 meters.
(b) When the car catches up to the truck, it will be moving at the same speed as the truck. Therefore, the speed of the car as it passes the truck will be equal to the truck's speed, which is 12 m/s. However, the car continues to accelerate at a rate of 3 [tex]m/s^2[/tex] even after passing the truck. Therefore, its speed will increase. To find the final speed of the car, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we have v = 12 + 3 * 8 = 36 m/s. So, the car will be moving at a speed of 36 m/s (or 27 m/s faster than the truck) as it passes the truck.
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A volume of 1.5 L of water at 15 °C is placed in an electric kettle. If it takes 5 minutes for the kettle to boil, estimate (a) its wattage and (b) the current that it draws if the mains voltage is 230 V.
Converting 5 minutes to seconds, we have t = 5 minutes * 60 seconds/minute = 300 seconds. Therefore, P = 535.05 kJ / 300 s = 1783.5 W.
To estimate the wattage of the electric kettle, we can use the equation P = (m * c * ΔT) / t, where P is the power in watts, m is the mass of water in kilograms, c is the specific heat capacity of water, ΔT is the change in temperature, and t is the time taken.
Given that the kettle boils 1.5 liters (or 1.5 kg) of water from 15°C to 100°C in 5 minutes, we can calculate the wattage. The current drawn by the kettle can be determined using the equation P = IV, where P is the power in watts, I is the current in amperes, and V is the voltage in volts. Assuming a mains voltage of 230 V, we can find the current drawn by the kettle.
The specific heat capacity of water is approximately 4.18 J/g°C. The change in temperature is ΔT = 100°C - 15°C = 85°C. Using the equation Q = mcΔT, where Q is the energy in joules, m is the mass of water in kilograms, c is the specific heat capacity, and ΔT is the change in temperature, we have Q = 1.5 kg * 4.18 J/g°C * 85°C = 535.05 kJ.
Next, we can determine the power (wattage) of the kettle using the equation P = Q/t, where P is the power in watts, Q is the energy in joules, and t is the time taken in seconds. Converting 5 minutes to seconds, we have t = 5 minutes * 60 seconds/minute = 300 seconds. Therefore, P = 535.05 kJ / 300 s = 1783.5 W.
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The driver of a 840.0 kg car decides to double the speed from 20.0 m/s to 40.0 m/s. What effect would this have on the amount of work required to stop the car, that is, on the kinetic energy of the car?
The kinetic energy increases from 168000 J to 672000 J. Therefore, more work would be required to stop the car.
The driver should be cautious when doubling the speed, as it significantly increases the amount of work required to bring the car to a stop.
Given:
Mass of the car, m = 840.0 kg
Initial velocity, v₁ = 20.0 m/s
The initial kinetic energy (KE₁):
KE₁ = (1/2)mv₁²
= (1/2)(840.0 kg)(20.0 m/s)²
= 168000 J
The final velocity after doubling the speed:
Final velocity, v₂ = 2 * v₁
= 2 * 20.0 m/s
= 40.0 m/s
The final kinetic energy (KE₂):
KE₂ = (1/2)mv₂²
= (1/2)(840.0 kg)(40.0 m/s)²
= 672000 J
Analyze the change in kinetic energy:
Change in kinetic energy = KE₂ - KE₁
= 672000 J - 168000 J
= 504000 J
Doubling the speed of the car results in a quadrupling of its kinetic energy.
The kinetic energy increases from 168000 J to 672000 J.
Therefore, more work would be required to stop the car.
The driver should be cautious when doubling the speed, as it significantly increases the amount of work required to bring the car to a stop.
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After solving a difficult physics problem, an excited student throws his book straight up. It leaves his hand at 3.9 m/s from 1.5 m above the ground a) Calculate the maximum height reached by the book. b) How much time does it take until the book hits the floor? c) What's its velocity, then?
The answer is that a) the maximum height reached by the book is 0.75 meters; b) it takes 0.39 sec until the book hits the floor; c) the velocity of the book is 7.58 m/s just before it hits the ground. Initial velocity, u = 3.9 m/s; Distance above the ground level, h = 1.5 m; Acceleration due to gravity, g = 9.8 m/s²
a) Calculation of maximum height reached by the book: For finding the maximum height, we can use the formula:
h = (u²/2g) + u²/2g; since the book is thrown upwards, the final velocity is zero.
Substituting the given values, we get: h = (3.9)²/(2×9.8) = 0.75 meters
Therefore, the maximum height reached by the book is 0.75 meters.
b) Calculation of time taken until the book hits the floor: We know that the distance covered by the book, s = h = 0.75 meters
Let’s use the formula s = ut + 1/2gt², which can be rewritten as: t = √(2s/g)where g = 9.8 m/s² and s = 0.75 m
Substituting the values in the equation above we get: t = √(2 × 0.75 / 9.8) = 0.39 sec
Therefore, it takes 0.39 sec until the book hits the floor.
c) Calculation of its velocity: We can use the formula, v = u + gt, to find the velocity of the book just before it hits the ground. Substituting the values of u, g and t, we get: v = 3.9 + (9.8 × 0.39) = 7.58 m/s
Therefore, the velocity of the book is 7.58 m/s just before it hits the ground.
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)An electric charge of 2×10
−6
C has a speed of 1×10
6
m/s enters a uniform magn field of 1.5 T with angle of 30
∘
. What is the magnitude of the magnetic force acting field of 1.5 T with
The magnitude of the magnetic force is 3 × 10⁻¹ N.
Charge, q = 2 × 10⁻⁶ C
Velocity, v = 1 × 10⁶ m/s
Magnetic field, B = 1.5 T
Angle, θ = 30°
The magnitude of the magnetic force acting on a moving charged particle in a magnetic field is given by:
F = qvBsinθ
Where,
F = Magnitude of the magnetic force
q = Charge
v = Velocity
B = Magnetic field
θ = Angle between the magnetic field vector and the velocity vector
Substitute the given values in the above equation,
F = qvBsinθ
= 2 × 10⁻⁶ × 1 × 10⁶ × 1.5 × sin 30°
= 3 × 10⁻¹ N
Therefore, the magnitude of the magnetic force acting on the electric charge of 2×10⁻⁶ C that has a speed of 1×10⁶ m/s and enters a uniform magnetic field of 1.5 T with an angle of 30° is 3 × 10⁻¹ N.
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an object is launched in parabolic motion from the ground and reaches the ground again after 20 seconds. The horizontal component of its velocity is 12 m/s. How far from its original position did it travel in the X direction?
The object traveled a horizontal distance of 240 meters from its original position in the x-direction.
In parabolic motion, the horizontal and vertical motions are independent of each other. The horizontal component of the velocity remains constant throughout the motion.
Given:
Time of flight (time taken to reach the ground) = 20 seconds
Horizontal component of velocity = 12 m/s
To find the horizontal distance traveled (Δx), we can use the formula:
Δx = (horizontal velocity) * (time of flight)
Δx = 12 m/s * 20 s
Δx = 240 meters.
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A 1.00−kg piece of iron, initially at 800 ∘C, is dropped into 0.20 kg of water initially at 20 ∘C. How much water boils away? The specific heat of iron is 449 J/kg.K. A) 0.11 kg B) 0.20 kg C) 0.042 kg D) 0.32 kg E) 0.17 kg
When a 1.00 kg piece of iron at 800 °C is dropped into 0.20 kg of water at 20 °C, approximately 0.11 kg (option A) of water boils away due to the heat transferred from the iron.
To determine how much water boils away, we need to calculate the heat transferred from the iron to the water, taking into account the specific heat capacities and temperature changes.
The heat transferred can be calculated using the formula:
Q = mcΔT
Where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
For the iron:
m_iron = 1.00 kg
c_iron = 449 J/kg.K
ΔT_iron = final temperature - initial temperature = 100 - 800 = -700 °C
For the water:
m_water = 0.20 kg
c_water = 4186 J/kg.K
ΔT_water = final temperature - initial temperature = 100 - 20 = 80 °C
Now, let's calculate the heat transferred:
Q_iron = m_iron * c_iron * ΔT_iron
Q_water = m_water * c_water * ΔT_water
The negative sign for ΔT_iron indicates a decrease in temperature.
To find the amount of water that boils away, we need to consider the heat of vaporization of water, which is approximately 2260 kJ/kg.
The amount of water boiled away (Δm_water) can be calculated using the formula:
Δm_water = Q_water / heat of vaporization
Plugging in the values:
Δm_water = Q_water / 2260 kJ/kg
After calculating Q_iron, Q_water, and Δm_water, we find that the amount of water boiled away is approximately 0.11 kg (option A).
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Suppose you have a 107−kg wooden crate resting on a wood floor. (For each answer, enter a number. μ
k
=0.3 and μ
s
=0.5 ) (a) What maximum force (in N) can you exert horizontally on the crate without moving it? N (b) If you continue to exert this force (in m/s
2
) once the crate starts to slip, what will the magnitude of its acceleration (in m/s
2
) then be? m/s
2
Part B.) A contestant in a winter games event pushes a 30.0 kg block of ice across a frozen lake as shown in the figure below. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03. (For each answer, enter a number.) (a) Calculate the minimum force F (in N ) he must exert to get the block moving. N (b) What is its acceleration (in m/s
2
) once it starts to move, if that force is maintained? m/s
2
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of a 93.0 kg skydiver falling in a pike (headfirst) position with a surface area of 0.145 m
2
. (Assume that the density of air is 1.21 kg/m
3
and the drag coefficient of a skydiver in a pike position is 0.7. For each answer, enter a number.) terminal velocity in m/s m/s terminal velocity in km/h km/h
Part A.)
For a wooden crate resting on a wood floor, let μk = 0.3 and μs = 0.5.
(a) The maximum force you can exert horizontally on the crate without moving it is given by the formula `F_s <= μ_s* F_N = μ_s * m * g`. Here, `F_N` = m * g is the normal force that the floor exerts on the crate. F_s is the maximum static frictional force that opposes the applied force to keep the crate at rest. Thus, the maximum force that can be exerted on the crate without moving it is given by `F_s = μ_s * F_N = μ_s * m * g = 0.5 * 107 * 9.8 ≈ 524.9 N`.
Hence, the maximum force that can be exerted horizontally on the crate without moving it is 524.9 N.
(b) If you continue to exert this force once the crate starts to slip, the magnitude of its acceleration will then be equal to the kinetic frictional force acting on the crate, divided by the mass of the crate. The kinetic frictional force acting on the crate is given by `F_k = μ_k * F_N = μ_k * m * g`.
So, the magnitude of the acceleration of the crate is given by `a = F_k / m = μ_k * g = 0.3 * 9.8 ≈ 2.94 m/s^2`. Thus, the magnitude of the acceleration of the crate will be 2.94 m/s^2.
Part B.)
Let the coefficient of static friction be μs = 0.1 and the coefficient of kinetic friction be μk = 0.03.
(a) The minimum force that the contestant must exert to get the block moving is equal to the product of the coefficient of static friction and the normal force acting on the block. This force is given by `F_s = μ_s * F_N = μ_s * m * g`, where `F_N` = m * g is the normal force acting on the block. Here, `F_N` = m * g = 30.0 kg * 9.8 m/s^2 = 294.0 N.
Therefore, the minimum force that the contestant must exert to get the block moving is `F_s = μ_s * F_N = 0.1 * 294.0 ≈ 29.4 N`. Hence, the minimum force that the contestant must exert to get the block moving is 29.4 N.
(b) Once the force required to start the block in motion is exceeded and the block is set in motion, the force of kinetic friction acts on the block, which is given by `F_k = μ_k * F_N = μ_k * m * g`. Here, `F_k = 0.03 * 294.0 = 8.82 N`. Therefore, the net force acting on the block is given by `F_net = F - F_k = 29.4 - 8.82 = 20.58 N`.
Thus, the acceleration of the block is given by `a = F_net / m = 20.58 / 30.0 ≈ 0.686 m/s^2`. Hence, the acceleration of the block once it starts to move is 0.686 m/s^2. The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid.
Find the terminal velocity (in meters per second and kilometers per hour) of a 93.0 kg skydiver falling in a pike (headfirst) position with a surface area of 0.145 m2.
The terminal velocity is given by the formula `v_t = sqrt((2 * m * g) / (ρ * A * C_d)))`, where `m` is the mass of the skydiver, `g` is the acceleration due to gravity, `ρ` is the density of air, `A` is the cross-sectional area of the skydiver facing the fluid, and `C_d` is the drag coefficient of the skydiver in the pike position. Here, `m` = 93.0 kg, `g` = 9.8 m/s^2, `ρ` = 1.21 kg/m^3, `A` = 0.145 m^2, and `C_d` = 0.7. Hence, the terminal velocity of the skydiver is `v_t = sqrt((2 * m * g) / (ρ * A * C_d))) = sqrt((2 * 93.0 * 9.8) / (1.21 * 0.145 * 0.7))) ≈ 56.5 m/s`.
Thus, the terminal velocity of the skydiver is 56.5 m/s or 203.5 km/h (rounded to one decimal place). Therefore, the terminal velocity of the skydiver is 56.5 m/s and 203.5 km/h.
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Two cars drive on a straight highway. At time t=0, car 1 passes mile marker 0 traveling due east with a speed of 190 m/s. At the same time, car 2 is 1.1 km east of mile marker 0 traveling at 27.0 m/s due west. Car 1 is speeding up with an acceleration of magnitude 0.20 m/s
2
, and car 2 is slowing down with an acceleration of magnitude 0.30 m/s
2
. You may want to review (Pages 40−43 ) . Part A Write x-versus-t equations of motion for both cars, taking east as the positive direction.
x
1
=−1100 m+(19.0 m/s)t−(0.20 m/s
2
)t
2
;x
2
=(27.0 m/s)t+(0.30 m/s
2
)t
2
x
1
=(19.0 m/s)t−(0.1)t
2
;x
2
=1100 m−(27.0 m/s)t−(0.15)t
2
x
1
=(19.0 m/s)t+(0.20 m/s
2
)t
2
;x
2
=1100 m−(27.0 m/s)t−(0.30 m/s
2
)t
2
x
1
=(19.0 m/s)t+(0.1)t
2
;x
2
=1100 m−(27.0 m/s)t+(0.15)t
2
x
1
=(19.0 m/s)t+(0.1)t
2
;x
2
=1100 m+(27.0 m/s)t+(0.15)t
2
x
1
=(19.0 m/s)t+(0.20 m/s
2
)t
2
;x
2
=1100 m+(27.0 m/s)t−(0.30 m/s
2
)t
2
Part B At what time do the cars pass next to one another? Express your answer using two significant figures.
The cars pass next to one another at t = 33.2 s. Hence, the correct option is (A) 33.
The two cars are:
x1=−1100 m+(19.0 m/s)t−(0.20 m/s^2)t^2;
x2=(27.0 m/s)t+(0.30 m/s^2)t^2
We have to calculate the time at which the cars pass next to one another.
−1100 m+(19.0 m/s)t−(0.20 m/s^2)t^2 = (27.0 m/s)t+(0.30 m/s^2)t^2 ... (i)
We can rearrange the above equation as follows:(0.20 m/s^2 + 0.30 m/s^2)t^2 − (19.0 m/s + 27.0 m/s)t - 1100 m = 0
Substitute the given values in the above equation as follows:
a = 0.20 m/s^2 + 0.30 m/s^2 = 0.50 m/s^2;b = -(19.0 m/s + 27.0 m/s) = -46.0 m/s;c = -1100 m
By substituting these values, the above equation becomes:0.50t^2 - 46.0t - 1100 = 0
The value of t, as shown below:t = [46 ± √(46^2 - 4 × 0.50 × (-1100))] / (2 × 0.50)
We get two values of t:t = 147.1 s (taking negative sign before square root) and t = 33.2 s (taking positive sign before square root)
Since both cars are on the same side of mile marker 0 at time t = 0, the next time the cars pass next to one another is after the time t = 33.2 s (time t = 147.1 s is not possible).
Therefore, the cars pass next to one another at t = 33.2 s.
Hence, the correct option is (A) 33.
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nove through this displactirent. Nin Sle choise (b) What aralrue motei a moet aspepriate for desirrbing this situation? pertide Lader constare aveed particie unger coinetart acreieration pertigin in echutibum v
y =vh = ar A. =w 1+ 21ae 3Ny 2=w 2=7a4 mot
The situation of the ball rolling off the platform and hitting the ground can be accurately described using the principles of projectile motion, considering both the horizontal and vertical components of motion and the effects of gravity on the vertical trajectory of the ball.
To properly describe the situation of the ball rolling off a platform and hitting the ground, the most appropriate motion model to use is projectile motion. Projectile motion is the motion of an object that is launched into the air and moves along a curved path under the influence of gravity. In projectile motion, the object has both horizontal and vertical components of motion. The horizontal motion is constant and unaffected by gravity, while the vertical motion is influenced by gravity and follows a parabolic trajectory. In this case, as the ball rolls off the platform, its horizontal velocity remains constant at 5 m/3. The vertical motion of the ball is influenced by the acceleration due to gravity, which is approximately 10 m/s^2. The ball follows a parabolic trajectory as it moves through the air, reaching a maximum height before descending and eventually hitting the ground. The time it takes for the ball to hit the ground is determined by the vertical motion and can be calculated using equations of motion.
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Two forcos,
F
1
and
F
2
, act at a point, as shown in (Figure 1).
F
1
has a magnitude of 8.20 N and is directed at an angle of α=58.0
∘
above the negative x axis in the second quadrant.
F
2
has a magnitude Part A of 6.60 N and is difected at an angle of β=52.8
∘
below the negative x axis in the third quadrant. What is the x component F
x
of the resultant force? Express your answer in newtons. What is the y component F
y
of the resultant force? Express your answer in newtons. What is the magnitude F of the resultant force? Express your answer in newtons. What is the angle γ that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis. Express your answer in degrees.
The x component of the resultant force is approximately 0.96 N, the y component is approximately 1.31 N, the magnitude of the resultant force is approximately 1.61 N, and the angle γ formed with the negative x axis is approximately 52.9°
To find the x and y components of the resultant force, as well as the magnitude and angle formed by the resultant force, we can use vector addition. The x and y components can be found using trigonometry, and then the magnitude and angle can be calculated.
First, let's find the x component of the resultant force, Fx. The x component is given by:
Fx = F1x + F2x
To find F1x and F2x, we can use the cosine function:
F1x = F1 * cos(α)
F2x = F2 * cos(180° - β)
Plugging in the given values, we get:
F1x = 8.20 N * cos(58.0°)
F2x = 6.60 N * cos(180° - 52.8°)
Calculating these expressions, we find:
F1x ≈ 4.33 N
F2x ≈ -3.37 N (note the negative sign indicating the direction)
Therefore, the x component of the resultant force, Fx, is:
Fx = F1x + F2x ≈ 4.33 N - 3.37 N ≈ 0.96 N
Next, let's find the y component of the resultant force, Fy. The y component is given by:
Fy = F1y + F2y
To find F1y and F2y, we can use the sine function:
F1y = F1 * sin(α)
F2y = F2 * sin(180° - β)
Plugging in the given values, we get:
F1y = 8.20 N * sin(58.0°)
F2y = 6.60 N * sin(180° - 52.8°)
Calculating these expressions, we find:
F1y ≈ 6.96 N
F2y ≈ -5.65 N (note the negative sign indicating the direction)
Therefore, the y component of the resultant force, Fy, is:
Fy = F1y + F2y ≈ 6.96 N - 5.65 N ≈ 1.31 N
Now, we can calculate the magnitude of the resultant force, F:
F = √(Fx^2 + Fy^2)
Plugging in the values, we get:
F = √((0.96 N)^2 + (1.31 N)^2) ≈ 1.61 N
Finally, we can find the angle γ that the resultant force forms with the negative x axis. The angle can be found using the arctangent function:
γ = atan(Fy / Fx)
Plugging in the values, we get:
γ = atan(1.31 N / 0.96 N) ≈ 52.9°
Therefore, the x component of the resultant force is approximately 0.96 N, the y component is approximately 1.31 N, the magnitude of the resultant force is approximately 1.61 N, and the angle γ formed with the negative x axis is approximately 52.9°.
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Find the capacitance of a parallel-plate capacitor consisting of circular plates 20 cm in radius separated by 1.8 mm Express your answer in picofarads.
The capacitance of the parallel-plate capacitor is approximately 0.306 picofarads (pF).
The capacitance (C) of a parallel-plate capacitor can be calculated using the formula:
C = (ε0 * A) / d
where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.
Given:
Radius of circular plates (r) = 20 cm = 0.2 m
Distance between the plates (d) = 1.8 mm = 0.0018 m
The area of each plate (A) can be calculated using the formula:
A = π * r^2
Plugging in the values, we have:
A = π * (0.2 m)^2
A = π * 0.04 m^2
Now, we can calculate the capacitance using the formula:
C = (ε0 * A) / d
The value of ε0 is approximately 8.854 × 10^(-12) F/m.
Plugging in the values, we have:
C = (8.854 × 10^(-12) F/m) * (π * 0.04 m^2) / (0.0018 m)
C ≈ 0.306 pF
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