Answer:
The gravitational force on the ball in newtons is 4.9 N.
The gravitational force on a 500 gram ball that falls is 4.9 N (newtons).
Mass is the quantity that describes the amount of matter in an object.
The kilogram is the metric unit of mass.
The weight is the force exerted on an object due to gravity, and it varies depending on the object's mass and the acceleration due to gravity.
Force = mass × acceleration due to gravity (F=ma)
We can use the formula to calculate the gravitational force that acts on a 500-gram ball when it falls.
We know the ball's mass (m = 500 grams) and the acceleration due to gravity (g = 9.8 m/s²).
F = m × gF
= (0.5 kg) × (9.8 m/s²)
F = 4.9 N
Thus, the gravitational force on the ball in newtons is 4.9 N.
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A cylindrical water-storage tank has a height of 7.9 m and a radius of 5.7 m. The storage tank is full of water but is vented to the atmosphere. The bottom of the tank is 25 m off the ground. A 10 cm diameter pipe runs vertically down from the tank and goes 1.3 m underground before turning horizontal. The water flow in the horizontal pipe is 83 L/s.
A- What is the water pressure (gauge) in the horizontal pipe underground?
(include units with answer)
B- How fast does the water level in the tank drop?
(include units with answer)
C-Rick Barnes drills a 5.7 mm diameter hole near the bottom of the tank. How fast does the water shoot out?
(include units with answer)
D-What volume flow of water is lost out the hole?
(include units with answer)
The water pressure (gauge) in the horizontal pipe underground is 3,407,229 Pa, the rate at which the water level in the tank drops is 0.02226 m/s, the speed of efflux is 9.89 m/s, and the volume flow rate of water lost through the hole is 8.008 x 10⁻⁵ m³/s.
Water pressure (gauge) in the horizontal pipe underground:
To determine the water pressure (gauge) in the horizontal pipe underground, the formula is:
P2 = P1 + (density x g x h),
P1 is the pressure at the surface,
density = 1000 kg/m³,
g = 9.8 m/s² and
h = 1.3 + 7.9 + 25 = 34.2 m.
Then, substituting the values we get:
P2 = 101,325 + (1000 x 9.8 x 34.2)
= 3,407,229 Pa
B. The rate at which the water level in the tank drops:
We must first calculate the volume of water in the tank, which is given by the formula:
V = πr²hwhere r = 5.7 m and h = 7.9 m.
Substituting the values we get:
V = π (5.7)² (7.9) = 1253.2 m³
Then we need to determine the rate at which the water level in the tank drops using the formula:
dV/dt = A x v, Where A is the cross-sectional area of the tank, which is equal to πr², r = 5.7 m, v is the speed at which the water level drops, and dV/dt is the rate at which the water level drops.
Substituting the values, we get:
dV/dt = π (5.7)² x v
= 162.396v
Since the flow rate is 83 L/s, the volume flow rate can be calculated as: 83 L/s = 83 x 10⁻³ m³/s. Therefore, the rate at which the water level in the tank drops is:
v = dV/dt / A = (83 x 10⁻³) / π (5.7)²
= 0.02226 m/sC.
Therefore, the water pressure (gauge) in the horizontal pipe underground is 3,407,229 Pa, the rate at which the water level in the tank drops is 0.02226 m/s, the speed of efflux is 9.89 m/s, and the volume flow rate of water lost through the hole is 8.008 x 10⁻⁵ m³/s.
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sam little sister is 60 pound (77.2 kg) but can't push her 40 pound (18.1 kg) . So sam father makes her pull the sled, with her little sister sitting inside, up a 22 m tall hill, How much work did sam do?
a) 2,700J
b)-2700J
C)-3400J
d)3400J
The work done by Sam is approximately 9950 J or 9950 Joules.
The work that Sam did to pull the sled up the hill can be determined by using the formula W = F.d
where W is the work done,
F is the force applied, and
d is the distance covered.
Here,
the force applied by Sam is equal to the weight of the sled and his little sister, which is given as:
= (60 + 40)pounds
= 100 pounds
= 100 * 0.4536 = 45.36 kg.
The distance covered is the height of the hill, which is 22 meters.
Therefore, W = F.d = 45.36 * 9.81 * 22
= 9949.96 J ≈ 9950 J
Hence, the work done by Sam is approximately 9950 J or 9950 Joules.
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Predict/Calculate Problem 2.56 - Part B Arer braking hal the time found in part A, hyour speed 60 m/h, greater than 60 m/h, or less than 6.0 m/s ? \begin{tabular}{|l|} \hline 60 m/s \\ peaterthan 60 m/s \\ 10s Han 60 m/s \\ \hline \end{tabular}
The speed of 60 m/h is equivalent to approximately 26.84 m/s. It's is greater than both 60 m/h and 6.0 m/s.
To answer the question, we need to consider the units and conversion between meters per second (m/s) and miles per hour (m/h).
Given that the speed is 60 m/h, we can convert it to m/s by multiplying it by the conversion factor 1 m/2.237 m/h (since 1 m/h = 2.237 m/s). Performing the calculation, we find:
60 m/h * (1 m/2.237 m/h) ≈ 26.84 m/s
Comparing the obtained value of 26.84 m/s to the given options:
- Is it greater than 60 m/h? No, 26.84 m/s is less than 60 m/h.
- Is it less than 6.0 m/s? No, 26.84 m/s is greater than 6.0 m/s.
Therefore, neither of the given options (greater than 60 m/h or less than 6.0 m/s) accurately represents the calculated value. The speed of 26.84 m/s is greater than both 60 m/h and 6.0 m/s.
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A small rock is throwm vertically upward with a speed of 13.0 m/s from the edge of the roof of a 40.0 m tall building. The rock doesn't hit the building on its way back down and lands in the What is the speed of the rock just beiore it hits the street? strect below. Air resistance can be neglected. Express your answer with the appropriate units. Part B How much time elapses from when the rock is thrown unlil it hits the street? Express your answer with the appropriate units.
(A) The speed of the rock just before it hits the street is 30.9 m/s
(B) Time taken by the rock to hit the street is 16.332 s.
Part A: final velocity (v) = 0 m/s, acceleration (a) = g = -9.81 m/s², displacement (s) = -40 m (negative sign indicates downward direction), initial velocity (u) = 13.0 m/s.
Now, we can use the third equation of motion, v² = u² + 2as
v² = (13.0 m/s)² + 2(-9.81 m/s²)(-40 m)
v² = 169.0 m²/s² + 784.0 m²/s².
v² = 953.0 m²/s²
v = √953.0 m²/s². v = 30.9 m/s.
Therefore, the speed of the rock just before it hits the street is 30.9 m/s. Answer: 30.9 m/s.
Part B: acceleration (a) = g = -9.81 m/s², initial velocity (u) = 13.0 m/s, displacement (s) = -40 m (negative sign indicates downward direction). Now, we can use the first equation of motion, s = ut + 1/2 at²
-40 m = (13.0 m/s)t + 1/2 (-9.81 m/s²)t²
-40 m = 13.0 mt - 4.905 t²
t² - 13.0 m/s t - 40.0 m / (-4.905 m/s²) = 0
This is a quadratic equation in t. We can solve this using the quadratic formula, t = [-(-13.0 m/s) ± √[(-13.0 m/s)² - 4(1/2)(-4.905 m/s²)(-40.0 m)]] / [2(1/2)(-4.905 m/s²)]
t = [13.0 m/s ± √(169.0 + 3924.48)] / (-4.905 m/s²)
t = [13.0 m/s ± 67.1668 m/s] / (-4.905 m/s²)
t = (-80.1668 m/s) / (-4.905 m/s²) or (-53.8332 m/s) / (-4.905 m/s²)t = 16.332 s or 10.962 s,
However, time cannot be negative. Therefore, time taken by the rock to hit the street is 16.332 s.
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The drawing shows two perpendicular, long, straight wires, both of which lie in the plane of the paper. The current in each of the wires is I = 4.8 A. In the drawing dH = 0.23 m and dV = 0.30 m. Find the magnitudes of the net magnetic fields at points A and B.
The current in each of the wires is I = 4.8 A. In the drawing dH = 0.23 m and dV = 0.30 m. The magnitude of wires A and B is 0.104 T and 0.032 T, respectively.
To find the magnitudes of the net magnetic fields at points A and B, we can use the formula for the magnetic field produced by a long straight wire:
B = (μ₀ * I) / (2π * r),
where B is the magnetic field, μ₀ is the permeability of free space (μ₀ ≈ 4π × [tex]10^{(-7)[/tex] T·m/A), I is the current in the wire, and r is the distance from the wire.
For point A:
The wire with current I is perpendicular to the page, and the distance from the wire to point A is dH. Therefore, the magnetic field at point A is:
[tex]B_A[/tex]= (μ₀ * I) / (2π * dH).
For point B:
The wire with current I is parallel to the page, and the distance from the wire to point B is dV. Therefore, the magnetic field at point B is:
[tex]B_B[/tex]= (μ₀ * I) / (2π * dV).
Now we can substitute the given values and calculate the magnitudes of the magnetic fields:
[tex]B_A[/tex]= (4π × [tex]10^{(-7)[/tex] T·m/A * 4.8 A) / (2π * 0.23 m) ≈ 0.104 T,
[tex]B_B[/tex]= (4π × [tex]10^{(-7)[/tex] T·m/A * 4.8 A) / (2π * 0.30 m) ≈ 0.032 T.
Therefore, the magnitudes of the net magnetic fields at points A and B are approximately 0.104 T and 0.032 T, respectively.
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A woman rides a carnival Ferris wheel at radius 16 m, completing 5.9 turns about its horizontal axis every minute. What are (a) the period of the motion, and the magnitude of her centripetal acceleration at (b) the highest point and (c) the lowest point? (a) Number Units (b) Number Units (c) Number Units
Given: Radius of Ferris wheel, r = 16 m Angular speed, ω = 5.9 revolution/min = 2π× 5.9 rad/min(a) The period of motion is the time taken for one complete revolution of the Ferris wheel.
The time for one revolution, T = 1/ω = 1/ (2π× 5.9) = 0.1767 min(b) At the highest point, the centripetal acceleration is equal to the gravitational acceleration and the direction is towards the center of the circular motion.The centripetal acceleration at the highest point, a = rω² = 16 × (2π× 5.9)²= 905.95 m/min²(c)
At the lowest point, the centripetal acceleration is equal to the gravitational acceleration and the direction is towards the center of the circular motion.The centripetal acceleration at the lowest point, a = rω² = 16 × (2π× 5.9)²= 905.95 m/min²Therefore, the period of motion is 0.1767 min, and the magnitude of her centripetal acceleration at the highest and lowest points is 905.95 m/min².
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Why did the normal force not enter into your solution? Frictionless ramp does not exert normal force on the crate. Normal force is perpendicular to the displacement and does no work. Normal for
In the case of a frictionless ramp, the normal force does not enter into the solution as it does no work and is not relevant for determining the height of the ramp.
Apologies for the oversight. You are correct that in the case of a frictionless ramp, the normal force does not enter into the solution because it does no work on the crate. Here's a step-by-step explanation without considering the normal force:
Step 1: Identify the forces acting on the crate.
In this case, we only need to consider the force of gravity (mg), acting vertically downwards.
Step 2: Analyze the motion of the crate.
The crate is moving along the ramp, so we need to consider the component of gravity that acts parallel to the ramp.
Step 3: Determine the parallel component of gravity.
The parallel component of gravity is given by F_parallel = mg sin(θ), where θ is the angle of the ramp.
Step 4: Apply Newton's second law in the direction of motion.
F_parallel = ma, where a is the acceleration of the crate along the ramp.
Step 5: Solve for acceleration.
a = F_parallel / m = (mg sin(θ)) / m = g sin(θ), where g is the acceleration due to gravity.
Step 6: Calculate the displacement of the crate.
We can use the kinematic equation s = ut + (1/2)at², where u is the initial velocity (assumed to be zero), t is the time, and a is the acceleration.
Step 7: Simplify the equation.
s = (1/2)g sin(θ) t²
Step 8: Determine the height of the ramp.
The height (h) of the ramp can be found using the equation s = h sin(θ).
Therefore, the height of the ramp can be calculated as h = s / sin(θ), where s is the displacement of the crate.
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A spaceship moves past Earth with a high speed. As it is passing a student on Earth measures the spaceship's length to be 76.7 m. If the proper length is 133 m how fast is the ship going? 0.65c 0.818c 0.72c Question 14 10 pts An alarm clock is set to sound in 12 h. At t=0, the clock is placed in a spaceship moving with a speed of 0.85c (relative to Earth). What distance, as determined by an Earth observer, does the snaceship travel before the alarm clock sounds?
d = 1.10 * 10¹⁴ m is the distance travelled by the spaceship before the alarm clock sounds, as determined by an Earth observer.
A spaceship moves past Earth with a high speed. As it is passing a student on Earth measures the spaceship's length to be 76.7 m.
If the proper length is 133 m how fast is the ship going?The formula to find the speed is given as:v = √(c² - l²) / √(c² - l0²)
Where, v = velocityc = speed of lightl = measured lengthl0 = proper length
After substituting the given values, we get:v = √(299,792,458² - 76.7²) / √(299,792,458² - 133²)
Therefore, v = 0.818c0.818c is the velocity of the spaceship. 100 word answerAn alarm clock is set to sound in 12 h. At t=0, the clock is placed in a spaceship moving with a speed of 0.85c (relative to Earth).
The distance travelled by the spaceship can be calculated using the formula:d = v * td = distance travelle
dv = velocity
t = timeThe time is given as 12 hours or 12 * 60 * 60 = 43200 seconds.v = 0.85c = 0.85 * 299,792,458 = 254824089.3 m/st = 43200 seconds
By substituting these values in the above formula, we get:d = 254824089.3 * 43200
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For a lens of focal length f, where should the object be
located to produce a real image that is
the same size as the object itself? Again, careful about
signs.
When the object is placed at a distance of 2f from the lens, the image is formed at a distance of 2f on the other side of the lens. This configuration is known as the "2f - 2f" configuration.
The object should be placed to a distance of 2 times the total length of the lens's objective (2f) from the lens to produce a true image the same size like the object itself.
Distances measured beyond the lens to the object are regarded negative in terms or sign convention for lens formula.
Positive distances are those measured from the lens to the picture.
A converging lens' focal length (f) is positive.
By positioning the object at this distance, the lens's image will be actual, on the other side of the lens, and the same size as the object.
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A vector A has components Ax=−5.8 m and Ay=−4.2 m. a) What is the magnitude of the vecotor A ? Unit b) What is the angle made by the vector A relative to the +x-axis? Unit
The magnitude of the vector A is approximately [tex]7.16 m[/tex]. The angle made by the vector A relative to the +x-axis is approximately [tex]36.74\°[/tex]
a) The magnitude of the vector A is given by:
[tex]|A| = \sqrt{Ax^2 + Ay^2}[/tex]
Substituting the given values:
[tex]|A| =\sqrt{(-5.8 m)^2 + (-4.2 m)^2)}[/tex]
[tex]|A| = \sqrt{(33.64 + 17.64)}[/tex]
[tex]|A| = \sqrt{51.28}[/tex]
[tex]= 7.16 m[/tex]
Therefore, the magnitude of the vector A is approximately [tex]7.16 m[/tex]
b) The angle made by the vector A relative to the +x-axis is given by:
[tex]\theta = tan^-^1(A_y/A_x)[/tex]
Substituting the given values:
[tex]\theta = tan^-^1(-4.2 m/-5.8 m)[/tex]
[tex]\theta = 36.74\°[/tex]
Therefore, the angle made by the vector A relative to the +x-axis is approximately [tex]36.74\°[/tex]
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A dolphin wants to swim directly back to its home bay, which is 0.800 km due west. It can swim at a speed of 6.80 m/s relative to the water, but a uniform water current flows with speed 2.83 m/s in the southeast direction.
What direction should the dolphin head? Enter the angle in degrees where positive indicates south of west and negative indicates north of west.
The dolphin should head at an angle of -17.6 degrees, north of west, to reach its home bay.
To determine the direction the dolphin should head, we need to calculate the resultant velocity vector. The dolphin's velocity relative to the water is 6.80 m/s due west, while the water current flows at 2.83 m/s in the southeast direction. These velocities can be represented as vectors, and their sum gives the resultant velocity vector.
Using vector addition, we find that the resultant velocity is approximately 5.63 m/s at an angle of -17.6 degrees with respect to west. Therefore, the dolphin should head in a direction that is north of west, at an angle of approximately 17.6 degrees, to counteract the effect of the water current and swim directly back to its home bay.
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PROBLEM (2) 4 marks A lucky student has a chance to do an experiment at the top of the Aspire tower. which is 300.0 m high above the ground. He kicked a ball verlically up. When he measured the initial velocity of the ball it was 28.2 m/s. After some lime the ball hils the ground. Neglect air resistence and the rotation of the ball. A) Calculate the maximum height the ball can reach with respect to the ground. B) Calculate the time required for the ball to reach a height of 200.0 m above the gyround. C) Calculate the total time of the trip. D) Calculate the speed of the ball when it hils the ground.
The maximum height reached by the ball is approximately 40.29 m.The time required for the ball to reach a height of 200.0 m is approximately 8.16 s. The total time of the trip is approximately 16.32 s. The speed of the ball when it hits the ground is approximately 40.07 m/s in the downward direction.
To solve the given problem, we can use the equations of motion under constant acceleration. Considering the motion of the ball in the vertical direction, we can use the following equations:
(a) The maximum height reached by the ball can be calculated using the equation:
h_max = (v_initial^2) / (2 * g)
where v_initial is the initial velocity of the ball and g is the acceleration due to gravity (approximately 9.8 m/s^2).
(b) The time required for the ball to reach a height of 200.0 m can be found using the equation:
h = v_initial * t - (1/2) * g * t^2
Rearranging the equation, we get a quadratic equation in terms of t, which can be solved to find the time.
(c) The total time of the trip is the time taken for the ball to reach the maximum height and the time taken for it to descend back to the ground. Since the motion is symmetric, the total time is twice the time taken to reach the maximum height.
(d) The speed of the ball when it hits the ground can be found using the equation:
v_final = v_initial - g * t
where v_final is the final velocity of the ball when it hits the ground.
Now, let's calculate the values:
(a) h_max = (28.2^2) / (2 * 9.8) ≈ 40.29 m
(b) Using the quadratic equation, we find that the time to reach a height of 200.0 m is approximately 8.16 s.
(c) The total time of the trip is 2 * 8.16 s = 16.32 s.
(d) The final velocity of the ball can be calculated as:
v_final = 28.2 - 9.8 * 8.16 ≈ -40.07 m/s (negative sign indicates downward direction)
Therefore, the answers are:
(a) The maximum height reached by the ball is approximately 40.29 m.
(b) The time required for the ball to reach a height of 200.0 m is approximately 8.16 s.
(c) The total time of the trip is approximately 16.32 s.
(d) The speed of the ball when it hits the ground is approximately 40.07 m/s in the downward direction.
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If a magnet is pushed in a coil, an electromotive force (voltage) is induced across the coil. If the same magnet is pushed into the width coil twice the number of loops, what do you think will happen to the induced emf?
According to Faraday's Law of Electromagnetic Induction, when a magnet is pushed into a coil, an electromotive force (EMF) or voltage is induced across the coil. The amount of induced EMF depends on the rate of change of the magnetic field lines passing through the coil.
When the same magnet is pushed into a coil with twice the number of loops, the induced EMF will increase.Increasing the number of loops of a coil increases the amount of magnetic flux passing through the coil. This, in turn, increases the rate of change of the magnetic field passing through the coil.
According to Faraday's Law, this increase in the rate of change of magnetic field lines passing through the coil will cause an increase in the induced EMF.In other words, increasing the number of loops of the coil increases the amount of induced EMF. This means that if the same magnet is pushed into a coil with twice the number of loops, the induced EMF will increase by a factor of 2.
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3.18 The plots shown in 떤 Figure P3.18 are the voltage across and the current through an ideal capacitor. Determine its capacitance.
To find the capacitance, we can use the formula: C = Q / V, where C represents the capacitance, Q represents the charge stored on the capacitor, and V represents the voltage across the capacitor.
The capacitance of an ideal capacitor can be determined by analyzing the voltage and current plots. In this case, the voltage across the capacitor is given by the y-axis of the graph, and the current through the capacitor is given by the x-axis of the graph.
To find the capacitance, we can use the formula:
C = Q / V
where C represents the capacitance, Q represents the charge stored on the capacitor, and V represents the voltage across the capacitor.
To find the charge, we can integrate the current over time. By examining the graph of the current, we can see that it is a straight line. The area under this straight line represents the charge stored on the capacitor.
To find the voltage, we can examine the graph and determine the maximum voltage reached by the capacitor.
Once we have the values for charge and voltage, we can substitute them into the formula to find the capacitance.
It is important to note that the scale of the graph should be taken into consideration when determining the charge and voltage values. Make sure to convert the values to the appropriate units if necessary.
By following these steps and analyzing the given plots, you will be able to determine the capacitance of the ideal capacitor.
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A ball (Ball A) moving at 8.00 m/s collides head on with a ball (Ball B) at rest. If the collision is elastic, and the mass of ball B is three times the mass of ball A, find the speeds of both balls after the collision.
The speed of ball A after the collision is 2.67 m/s, and the speed of ball B after the collision is 5.33 m/s.According to the Law of Conservation of Energy and Momentum, if the collision is elastic, then the total momentum of the system before the collision is equal to the total momentum of the system after the collision. It also means that the kinetic energy of the system is conserved.
The momentum before the collision can be calculated as:m1v1 + m2v2 = (m1 + m2)v where,m1 = mass of ball A = m2/3m2 = mass of ball Bv1 = initial velocity of ball A = 8.00 m/sv2 = initial velocity of ball B = 0v = final velocity of both the ballsAfter the collision, the balls will move in opposite directions, and their velocities will be v and -v. Therefore, the momentum after the collision can be calculated as:m1v + m2(-v) = (m1 + m2)0m1v - m2v = 0v(m1 - m2) = 0v = 0 or m2/m1As the collision is elastic, the kinetic energy of the system is conserved.
Therefore, the kinetic energy before the collision is equal to the kinetic energy after the collision. The kinetic energy before the collision can be calculated as:1/2m1v1² = 1/2(1/3m2)v2²After the collision, the balls will move in opposite directions, and their velocities will be v and -v.
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basketball player, standing near the basket to grab a rebound jumos 86 cm vertically.
a.) How much (total) time does the player spend in the top 20 cm of jump?
b.) How much (total) time does the player spend in the bottom of the 20 cm of this jump?
A basketball player, standing near the basket to grab a rebound, jumps 86 cm vertically. Let's find out the time the player spends in the top 20 cm of jump and the bottom of the 20 cm of this jump. The total time spent in the bottom 20 cm of jump is also 0.404 seconds.
a) The player spends the total time in the top 20 cm of jump Time taken for the player to go up = ?u = 0 (initial velocity) Acceleration = a = g = 9.8 m/s² (downwards)Distance covered = s = 20 cm = 0.2 m From the first equation of motion, we know that: s = ut + 1/2at²Where t is the time taken for the player to reach the top of the jump.
By substituting the above values, we get: 0.2 = 0 + 1/2 × 9.8 × t²0.2 = 4.9t²t² = 0.2/4.9t² = 0.04081632653061224t = √(0.04081632653061224)t = 0.202 seconds So, the player takes 0.202 seconds to reach the top of the jump. Time taken for the player to come down = ?u = 0 (initial velocity) Acceleration = a = g = 9.8 m/s² (downwards) Distance covered = s = 20 cm = 0.2 m (distance covered in coming down is equal to the distance covered in going up) From the first equation of motion,
we know that: s = ut + 1/2at²Where t is the time taken for the player to reach the ground after reaching the top of the jump. By substituting the above values, we get: 0.2 = 0 + 1/2 × 9.8 × t²0.2 = 4.9t²t² = 0.2/4.9t² = 0.04081632653061224t = √(0.04081632653061224)t = 0.202 seconds Therefore, the player takes 0.202 seconds to reach the ground after reaching the top of the jump.
Total time spent in the top 20 cm of jump = Time taken to reach the top of the jump + Time taken to reach the ground after reaching the top of the jump= 0.202 + 0.202= 0.404 seconds. Therefore, the player spends a total of 0.404 seconds in the top 20 cm of jump. b) The player spends the total time in the bottom of the 20 cm of jump.
The player spends the same amount of time in the bottom of the jump as in the top of the jump. Hence, the total time spent in the bottom 20 cm of jump is also 0.404 seconds.
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You are driving to the grocery store at 19 m/s. You are 110 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.70 s and that your car brakes with constant acceleration. Part B What magnitude braking acceleration will bring you to rest right Express your answer with the appropriate units. ✓ Correct Correct answer is shown. Your answer 1.86
s
2
m
was eith significant figures than required for this part. Part C How long does it take you to stop? Express your answer with the appropriate units.
To come to a stop at the intersection, the driver needs a braking acceleration of 5.4 m/s². It will take approximately 7.9 seconds for the car to come to a complete stop.
To find the magnitude of the braking acceleration needed to bring the car to rest, we can use the following kinematic equation:
[tex]v^2 = u^2 + 2as[/tex]
where:
v = final velocity (0 m/s, as the car comes to a stop)
u = initial velocity (19 m/s)
a = acceleration
s = distance traveled (110 m)
Rearranging the equation to solve for acceleration, we get:
[tex]a = (v^2 - u^2) / (2s)[/tex]
Substituting the given values, we have:
[tex]a = (0^2 - 19^2) / (2 * 110)[/tex]
a ≈ -5.4 m/s²
The negative sign indicates that the acceleration is in the opposite direction of the car's initial motion, which corresponds to braking. Therefore, the magnitude of the braking acceleration needed to bring the car to rest is approximately 5.4 m/s².
To determine the time it takes to stop, we can use the equation:
v = u + at
where:
v = final velocity (0 m/s)
u = initial velocity (19 m/s)
a = acceleration (-5.4 m/s²)
t = time
Rearranging the equation to solve for time, we have:
t = (v - u) / a
Substituting the given values, we have:
t = (0 - 19) / -5.4
t ≈ 7.9 seconds
Therefore, it will take approximately 7.9 seconds for the car to come to a complete stop after the traffic light turns red.
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A particle of mass m is constrained to lie along a frictionless, horizontal plane subject to a force given by the expression F(x)=−kx. It is projected from x=0 to the right along the positive x direction with initial kinetic energy T
0
=1/2kA
2
.k and A are positive constants. Find (a) the potential energy function V(x) for this force; (b) the kinetic energy, and (c) the total energy of the particle as a function of its position. (d) Find the turning points of the motion. (e) Sketch the potential, kinetic, and total energy functions. (Optional: Use Mathcad or Mathematica to plot these functions. Set k and A each equal to 1.)
(a) The potential energy function is V(x) = (1/2)kx^2. (b) The kinetic energy is T = (1/2)kA^2. (c) The total energy is constant and equal to T. (d) The turning points occur at x = ±A. (e) The potential energy increases quadratically, while the kinetic energy remains constant, resulting in a constant total energy.
(a) The potential energy function, V(x), is derived from the force expression F(x) = -kx using the definition of potential energy. Integrating F(x) with respect to x yields V(x) = (1/2)kx^2.
(b) The kinetic energy, T, is given as T₀ = (1/2)kA^2, representing the initial kinetic energy when the particle is projected.
(c) The total energy, E, of the particle is constant and equal to the initial kinetic energy, E = T₀. This is due to the absence of external forces and the conservation of mechanical energy.
(d) The turning points of the motion occur when the particle reaches its maximum displacement. This happens at x = ±A, where the potential energy is maximum, and the kinetic energy is zero.
(e) The potential energy function, V(x), increases quadratically as the particle moves away from the origin, while the kinetic energy remains constant throughout. Consequently, the total energy, E, remains constant, representing a balance between potential and kinetic energies.
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What is the coefficient of volume expansion of the container? Density of water at 60
∘
C is 0.98324 g/mL Express your answer using two significant figures. 2 Incorrect; Try Again; 10 attempts remaining Part B What is the most likely material of the container? iron copper aluminum glass What is the net force on each side of the box? Express your answer to two significant figures and include the appropriate units. Xง Incorrect; Try Again; 12 attempts remaining A cubic box of volume 6.1×10
−2
m
3
is filled with air at atmospheric pressure at 20
∘
C. The box is closed and heated to 200
∘
C. Part A What is the net force on each side of the box? Express your answer to two significant figures and include the appropriate units What is the coefficient of volume expansion of the container? Density of water at 60
∘
C is 0.98324 g/mL Express your answer using two significant figures. 2 Incorrect; Try Again; 10 attempts remaining Part B What is the most likely material of the container? iron copper aluminum glass It is observed that 54.90 mL of water at 20
∘
C completely fills a container to the brim. When the container and the water are heated to 60
∘
C,0.35 g of water is lost. Part A What is the coefficient of volume expansion of the container? Density of water at 60
∘
C is 0.98324 g/mL. Express your answer using two significant figures.
The coefficient of volume expansion of the container is approximately -0.0006 cm³/°C.
To find the coefficient of volume expansion of the container, we can use the formula:
β = (V_f - V_i) / (V_i * ΔT),
Where β is the coefficient of volume expansion, V_f is the final volume, V_i is the initial volume, and ΔT is the temperature change.
Given:
The density of water at 60°C is 0.98324 g/mL.
Change in temperature, ΔT = 60°C - 20°C = 40°C.
To find the initial volume, we can use the density of water at 60°C. Since the density of water is given in g/mL, we can convert it to g/cm³:
Density of water at 60°C = 0.98324 g/mL = 0.98324 g/cm³.
Let's assume the initial volume of the container is Vi. Therefore, the mass of the water filled in the container at 20°C is:
Mass = Density * Volume
0.35 g = 0.98324 g/cm³ * Vi.
Solving for Vi:
Vi = 0.35 g / 0.98324 g/cm³ ≈ 0.356 cm³.
Now, using the formula for the coefficient of volume expansion:
β = (V_f - V_i) / (V_i * ΔT)
= (0 - 0.356 cm³) / (0.356 cm³ * 40°C).
Simplifying:
β ≈ -0.0089 / 14.24
≈ -0.0006 cm³/°C.
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If a flea can jump straight up to a height of 0.460 m, what is its initial speed as it leaves the ground? Part B How long is it in the air?
a)The initial speed of a flea as it leaves the ground is 2.999 m/s.
b) It is in the air for approximately 0.611 seconds.
a) The initial speed of a flea as it leaves the ground can be determined using the formula:
v_f = \sqrt{2gh} ,
where v_f is the final velocity, g is the acceleration due to gravity, and h is the height.
Given h = 0.460 m, and g = 9.8 m/s^2, we can calculate v_f:
v_f = \sqrt{2gh} ,= /sqrt{2 * 9.8} m/s^2 * 0.460 m)
=sqrt(9.016)
=2.999 m/s
Therefore, the initial speed of the flea as it leaves the ground is 2.999 m/s.
b) In order to find how long it is in the air, we can use the formula t = 2 * v_f / g. So, t = 2 * 2.999 m/s / 9.8 m/s^2 = 0.611 s.
Therefore, it is in the air for approximately 0.611 seconds.
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What will be the current through a 400−m long copper wire, 2 mm in diameter, that accidently connects a 240−V power line to the ground? Show your work. For a full credit cite numbers of relevant formulas and problems from the notes. 4. The electric company charges $0.50 per kilowatt hour. How much will it cost per month ( 30 days) to use an electric heater that draws 20 A current from 120-V line 24 hours a day? Show your work. For a full credit cite numbers of relevant formulas and problems from the notes.
(a) The current through the copper wire is approximately 111.11 A.
(b) The cost per month to use the electric heater is $864.
(a) To find the current through the copper wire, we can use Ohm's Law:
I = V / R
Where:
I is the current
V is the voltage
R is the resistance
Length of the wire, L = 400 m
Diameter of the wire, d = 2 mm = 0.002 m
Voltage, V = 240 V
First, we need to calculate the resistance of the wire using the formula:
R = ρ * (L / A)
Where:
ρ is the resistivity of copper (given in the notes)
L is the length of the wire
A is the cross-sectional area of the wire
The cross-sectional area of the wire can be calculated using the formula:
A = π * (d/2)^2
Now, let's calculate the resistance:
A = π * (0.002 m / 2)^2
A ≈ 3.14 × 10^(-6) m^2
R = (resistivity of copper) * (L / A)
Now, substitute the given values:
R = (1.7 × 10^(-8) Ω·m) * (400 m / 3.14 × 10^(-6) m^2)
Calculating:
R ≈ 2.16 Ω
Now, we can find the current:
I = V / R
I = 240 V / 2.16 Ω
Calculating:
I ≈ 111.11 A
Therefore, the current through the copper wire is approximately 111.11 A.
(b) To calculate the cost per month to use the electric heater, we can use the formula:
Cost = (Power * Time * Cost per kWh) / 1000
Current, I = 20 A
Voltage, V = 120 V
Time, T = 24 hours * 30 days = 720 hours
Cost per kWh = $0.50
First, we need to calculate the power consumed by the electric heater:
Power = V * I
Now, substitute the given values:
Power = 120 V * 20 A
Calculating:
Power = 2400 W = 2.4 kW
Now, we can find the cost per month:
Cost = (2.4 kW * 720 hours * $0.50) / 1000
Calculating:
Cost = $864
Therefore, it will cost $864 per month to use the electric heater.
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Two stones lie on top of each other in a lift. Both have a mass of 1.0 kg. The elevator accelerates downwards with a = 0.71 m/s2. Draw in and calculate all the forces acting on the lower one the rock during acceleration.
To solve this problem, we need to consider the forces acting on the lower stone in the lift during acceleration. Let's break it down step by step:
1. Identify the forces acting on the lower stone:
- Gravitational force (weight): The weight of the stone acts vertically downward and can be calculated using the formula F = m * g, where m is the mass of the stone (1.0 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²).
- Normal force: The normal force acts perpendicular to the contact surface between the stone and the lift. In this case, since the lift is accelerating downwards, the normal force will be less than the weight of the stone.
2. Calculate the gravitational force:
F_gravity = m * g
= 1.0 kg * 9.8 m/s²
= 9.8 N
3. Calculate the net force acting on the lower stone:
Since the elevator is accelerating downwards, the net force acting on the stone will be the difference between the gravitational force and the normal force.
F_net = F_gravity - F_normal
4. Calculate the normal force:
To find the normal force, we need to consider the acceleration of the elevator.
F_net = m * a
F_net = 1.0 kg * (-0.71 m/s²) [Negative sign because the elevator is accelerating downwards]
F_net = -0.71 N
F_normal = F_gravity - F_net
= 9.8 N - (-0.71 N)
= 9.8 N + 0.71 N
= 10.51 N
5. Summarizing the forces acting on the lower stone during acceleration:
- Gravitational force (weight): 9.8 N (acting downward)
- Normal force: 10.51 N (acting upward)
- Net force: -0.71 N (acting downward)
It's important to note that the net force is the force responsible for the stone's acceleration.
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How does the terminal voltage compare to the load voltage? Why? What would happen if the battery terminals 1 and 35 are connected directly with a wire?
The terminal voltage of a battery is equal to the load voltage when there is no internal resistance or voltage drop within the battery. If the battery terminals 1 and 35 are connected directly with a wire, it would create a short circuit.
In an ideal scenario, where the battery has no internal resistance, the terminal voltage and the load voltage would be the same. However, in practical situations, batteries have some internal resistance due to factors like the resistance of the electrolyte and the material used in the battery construction. When a load is connected to a battery, the current flows through both the load resistance and the internal resistance of the battery. As the current passes through the internal resistance, there is a voltage drop within the battery. This voltage drop causes the terminal voltage to be lower than the load voltage. In other words, the terminal voltage decreases compared to the open circuit voltage of the battery.
If the battery terminals 1 and 35 are connected directly with a wire, it would create a short circuit. In a short circuit, the resistance of the circuit becomes very low or almost zero. This results in a very high current flowing through the circuit. In this case, the internal resistance of the battery plays a crucial role. Since the internal resistance is not zero, there will be a significant voltage drop across the internal resistance, and the battery may heat up. Connecting the battery terminals directly with a wire can be dangerous as it may cause overheating and potentially damage the battery. It is important to use appropriate circuitry and components to regulate the current and protect the battery from excessive discharge or damage.
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Three source charges are used to create an electric field at a point P in space located at x=9.0 m, y=7.0 m. The first charge is -7.0uC and is located at the origin. The second charge is 1.0 uC and is located on the y-axis at y=4.0 m. The third charge is 4.0uC and is located on the x-axis at x=4.0m.
The x-component of the electric field at point P is?
The y-component of the electric field at point P is?
A new charge of 2.0uC is now placed at the point P. The x-component of the electric force on the new charge due to the original three charges is?
The y-component of the electric force on the new charge due to the original three charges is?
1. Calculating the x-component of the electric field at point P:
Using Coulomb's law, the electric field due to a point charge is given by:
E = k * q / r^2
where E is the electric field, k is the Coulomb's constant (approximately 8.99 * 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge to the point of interest.
For the first charge (q1 = -7.0 μC) at the origin (0, 0), the distance between q1 and P is:
r1 = sqrt((9.0 - 0)^2 + (7.0 - 0)^2) = sqrt(81 + 49) = sqrt(130)
The electric field due to q1 at P is:
E1 = k * q1 / r1^2 = (8.99 * 10^9 N m^2/C^2) * (-7.0 * 10^-6 C) / (sqrt(130))^2
Calculating the value of E1:
E1 ≈ -2.32 * 10^6 N/C
For the second charge (q2 = 1.0 μC) at y = 4.0 m, the distance between q2 and P is:
r2 = sqrt((0 - 0)^2 + (7.0 - 4.0)^2) = sqrt(9) = 3
The electric field due to q2 at P is:
E2 = k * q2 / r2^2 = (8.99 * 10^9 N m^2/C^2) * (1.0 * 10^-6 C) / (3)^2
Calculating the value of E2:
E2 ≈ 1.00 * 10^6 N/C
For the third charge (q3 = 4.0 μC) at x = 4.0 m, the distance between q3 and P is:
r3 = sqrt((9.0 - 4.0)^2 + (0 - 0)^2) = sqrt(25) = 5
The electric field due to q3 at P is:
E3 = k * q3 / r3^2 = (8.99 * 10^9 N m^2/C^2) * (4.0 * 10^-6 C) / (5)^2
Calculating the value of E3:
E3 ≈ 1.44 * 10^6 N/C
Now, let's calculate the x-component of the electric field at point P by summing the contributions from each charge:
Ex = E1 * cos(theta1) + E2 * cos(theta2) + E3 * cos(theta3)
where theta1 = atan(7.0/9.0), theta2 = 90 degrees, and theta3 = 0 degrees.
Substituting the values and calculating Ex:
Ex = (-2.32 * 10^6 N/C) * cos(atan(7.0/9.0)) + (1.00 * 10^6 N/C) * cos(90°) + (1.44 * 10^6 N/C) * cos(0°)
Calculating the value of Ex:
Ex ≈ -1.54 * 10^6 N/C
Therefore, the x-component of the electric field at point P is approximately -1.54 * 10^6 N/C.
2. Calculating the y-component of the electric field at point P:
Ey = E1 * sin(theta1) + E2 * sin(theta2) + E3 * sin(theta3)
where theta1 = atan(7.0/9.0), theta2 = 90 degrees, and theta3 = 0 degrees.
Substituting the values and calculating Ey:
Ey = (-2.32 * 10^6 N/C) * sin(atan(7.0/9.0)) + (1.00 * 10^6 N/C) * sin(90°) + (1.44 * 10^6 N/C) * sin(0°)
Calculating the value of Ey:
Ey ≈ 2.68 * 10^6 N/C
Therefore, the y-component of the electric field at point P is approximately 2.68 * 10^6 N/C.
3. Calculating the x-component of the electric force on the new charge (q4 = 2.0 μC) at point P:
Fx = q4 * Ex
Substituting the values:
Fx = (2.0 * 10^-6 C) * (-1.54 * 10^6 N/C)
Calculating the value of Fx:
Fx ≈ -3.08 N
Therefore, the x-component of the electric force on the new charge at point P is approximately -3.08 N.
4. Calculating the y-component of the electric force on the new charge (q4 = 2.0 μC) at point P:
Fy = q4 * Ey
Substituting the values:
Fy = (2.0 * 10^-6 C) * (2.68 * 10^6 N/C)
Calculating the value of Fy:
Fy ≈ 5.36 N
Therefore, the y-component of the electric force on the new charge at point P is approximately 5.36 N.
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A rock is lossed straight up with a velocity of +20 m/s Part A When it returns, it talls into a hole 10 moeep. What is the rock's volocily as it hits the bottom of the hole? Part B How long is the rock in the air, from the instant it is roleased until it hits the bottom of the hole?
Part A: Finding the velocity of the rock as it hits the bottom of the hole.
v = u + at
u = +20 m/s (upwards)
a = -9.8 m/s²
h = 10 m (distance fallen)
We need to find v when the rock hits the bottom of the hole, so the final position will be h = 0.
Using the equation for displacement in vertical motion:
h = ut + (1/2)at²
0 = (20)t + (1/2)(-9.8)t²
0 = 20t - 4.9t²
Since t cannot be zero (that would be the initial time), we take t = 20/4.9 ≈ 4.08 seconds.
v = 20 - 9.8 * 4.08
v ≈ -39.98 m/s (approximately -40 m/s)
So, the velocity of the rock as it hits the bottom of the hole is approximately -40 m/s, where the negative sign indicates it is moving downwards.
Part B: Finding the time the rock is in the air.
We have already found that the time it takes for the rock to hit the bottom of the hole is approximately 4.08 seconds.
Therefore, the rock is in the air for approximately 4.08 seconds from the instant it is released until it hits the bottom of the hole.
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A physics student stands on a cliff overlooking a lake and decides to throw a baseball to her friends in the water below. She throws the baseball with a velocity of 25.5 m/s25.5 m/s at an angle of 37.5∘37.5∘ above the horizontal. When the baseball leaves her hand, it is 12.5 m12.5 m above the water. How far does the baseball travel horizontally before it hits the water? Neglect any effects of air resistance when calculating the answer.
horizontal distance:
The horizontal distance traveled by the baseball before it hits the water is 133.3 meters.
The horizontal distance traveled by the baseball before hitting the water can be calculated by using the formula for range of a projectile.
Range of a projectile:
Range= 2v₀²sinθ/g
Where v₀ is the initial velocity,
θ is the angle of projection,
and g is the acceleration due to gravity.
Substituting the given values in the above formula, we get:
Range= 2(25.5 m/s)²sin(37.5°) /9.8 m/s²= 133.3 m
Therefore, the horizontal distance traveled by the baseball before it hits the water is 133.3 meters.
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A 44.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 142 m/s from the top of a cliff 138 m above level ground, where the ground is taken to be y = 0. (a) W
The work done against gravity by the projectile is approximately 59,729.6 Joules.
To solve this problem, we need to analyze the motion of the projectile and calculate its work done against gravity.
(a) Work Done Against Gravity:
The work done against gravity can be calculated using the formula:
W = m * g * Δh
Where:
W is the work done against gravity
m is the mass of the projectile
g is the acceleration due to gravity
Δh is the change in height
Given:
m = 44.0 kg
g ≈ 9.8 m/s²
Δh = 138 m (height above the ground)
Substituting these values into the formula, we have:
W = 44.0 kg * 9.8 m/s² * 138 m
Calculating this expression, we find:
W ≈ 59,729.6 J
Note: This calculation assumes no air resistance or other external forces acting on the projectile during its flight.
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A body emits most light at a wavelength of 430 nm. Which radiation temperature corresponds to the?
The wavelength of light at which a body emits the most light is determined by its temperature and follows a relationship known as Wien's displacement law. According to Wien's law, the wavelength of maximum emission (λ_max) is inversely proportional to the temperature (T) of the body.
The equation for Wien's displacement law is:
λ_max = b / T
where λ_max is the wavelength of maximum emission, T is the temperature of the body in Kelvin, and b is Wien's displacement constant, which is approximately equal to 2.898 × 10^-3 meters per Kelvin.
In this case, the given wavelength is 430 nm, which needs to be converted to meters (1 nm = 10^-9 m). So, 430 nm is equal to 430 × 10^-9 m.
We can now rearrange the equation to solve for T:
T = b / λ_max
Substituting the values, we have:
T = (2.898 × 10^-3) / (430 × 10^-9)
Simplifying the expression, T ≈ 6736 K.
Therefore, the radiation temperature corresponding to a body emitting most light at a wavelength of 430 nm is approximately 6736 Kelvin.
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A uniform electric freld of 2 N/C points in the +x direction. What is the change in electiv potertal energy U
b
−U
a
of a −2 C test charge as it is moved from point at (x,y)=(−1m+3m) to poirt b at (x,y)=(+2m,+3m) ?
The change in electric potential energy (Ub−Ua) of the -2 C test charge as it is moved from point A to point B is -12 J.
The electric potential energy of a test charge in an electric field is given by the formula:
U = qV,
where U is the potential energy, q is the charge, and V is the electric potential. To calculate the change in electric potential energy (Ub−Ua), need to find the potential energy at point A (Ua) and point B (Ub), and then subtract them.
Given that the electric field is uniform and points in the +x direction, the electric potential V at any point in the field can be calculated using the formula:
V = Ex * x,
where Ex is the magnitude of the electric field and x is the displacement in the x-direction.
At point A, the displacement in the x-direction is -1 m, and at point B, it is +2 m. Therefore, the potential energy at point A (Ua) is:
Ua = (-2 C) * (2 N/C) * (-1 m) = 4 J,
and at point B (Ub), it is:
Ub = (-2 C) * (2 N/C) * (+2 m) = -8 J.
For finding the change in electric potential energy (Ub−Ua), subtract Ua from Ub:
Ub−Ua = -8 J - 4 J = -12 J.
Therefore, the change in electric potential energy (Ub−Ua) of the -2 C test charge as it is moved from point A to point B is -12 J.
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How much energy is contained in a 200 -Calorie chocolate bar? 1. A) 4184 Joules B) 8368 Joules C) 840 Joules D) 84000 Joules E) 840000 Joules 34). Which is an example of a system utilizing chemic
The correct answer to the given question is option (D) 84000 Joules. The calorie is a unit of energy that is typically used in nutrition and chemistry. One calorie is the amount of heat energy that is required to raise the temperature of one gram of water by one degree Celsius.
Here are the steps to calculate the energy contained in a 200-Calorie chocolate bar:
One calorie = 4.184 Joules
So, 200 Calories = 200 x 4.184 Joules= 836.8 Joules = 8.368 x 10² Joules
(scientific notation
Now, 8.368 x 10² Joules = 84000 Joules
Therefore, the energy contained in a 200-Calorie chocolate bar is 84000 Joules
Examples of systems utilizing chemical energy:
Combustion engines, batteries, nuclear power plants.
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