Answer:
Explanation:
Formula
W = I * E
Givens
W = 150
E = 120
I = ?
Solution
150 = I * 120 Divide by 120
150/120 = I
5/4 = I
I = 1.25
Note: This is an edited note. You have to assume that 120 is the RMS voltage in order to go any further. That means that the peak voltage is √2 times the size of 120. The current has the same note applied to it. If the voltage is its rms value, then the current must (assuming the properties of the bulb do not change)
On the other hand, if the voltage is the peak value at 120 then 1.25 will be correct.
However I would go with the other answerer's post and multiply both values by √2
This question is about as sneaky as they ever get.
First let's do the easy part:
Power = (voltage) x (current)
150 watts = (120 volts) x (current)
current = (150 watts) / (120 volts)
current = 1.25 Amperes but this is NOT the answer to the question.
The voltage at the outlet is a "sinusoidal" wave ... it wiggles up and down 60 times every second. The number of "120 volts" is NOT the "peak" of the wave. In fact , the highest it ever gets is √2 greater than 120 volts. And all of this applies to the current too.
The RMS current through the lamp is (150/120) = 1.25 Amperes .
The peak current through the lamp is 1.25·√2 = about 1.77 Amperes .
The moment of inertia for a 5500 kg solid disc is 12100 kg-m^2. Find the radius of the disc? (a) 2.111 m (b) 2.579 m (c) 1.679 m (d) 2.574 m (e) 2.098 m (f) 2.457 m
Answer:
The radius of the disc is 2.098 m.
(e) is correct option.
Explanation:
Given that,
Moment of inertia I = 12100 kg-m²
Mass of disc m = 5500 kg
Moment of inertia :
The moment of inertia is equal to the product of the mass and square of the radius.
The moment of inertia of the disc is given by
[tex]I=\dfrac{mr^2}{2}[/tex]
Where, m = mass of disc
r = radius of the disc
Put the value into the formula
[tex]12100=\dfrac{5500\times r^2}{2}[/tex]
[tex]r=\sqrt{\dfrac{12100\times2}{5500}}[/tex]
[tex]r= 2.098\ m[/tex]
Hence, The radius of the disc is 2.098 m.
A 0.144 kg baseball moving 28.0 m/s strikes a stationary 5.25 kg brick resting on small rollers so it moves without significant friction. After hitting the brick, the baseball bounces back, and the brick moves forward at 1.1 m/s (a) For determining baseball's speed after the collision, do you need to use momentum conservation, energy conservation or both? Give reasons for your answer (Marks will only be awarded if reasoning is qiven). (b) What is the baseball's speed after the collision? (c) Find the total mechanical energy before and after the collision. (d) Show that the collision is inelastic. (e) Is the kinetic energy of the system conserved? Give a reason. (f Is the total energy of the system conserved? Give a reason
(a) Only momentum conservation
In order to find the speed of the ball, momentum conservation is enough. In fact, we have the following equation:
[tex]m u + M U = m v + M V[/tex]
where
m is the mass of the baseball
u is the initial velocity of the baseball
M is the mass of the brick
U is the initial velocity of the brick
v is the final velocity of the baseball
V is the final velocity of the brick
In this problem we already know the value of: m, u, M, U, and V. Therefore, there is only one unknown value, v: so this equation is enough to find its value.
(b) 12.1 m/s
Using the equation of conservation of momentum written in the previous part:
[tex]m u + M U = m v + M V[/tex]
where we have:
m = 0.144 kg
u = +28.0 m/s (forward)
M = 5.25 kg
U = 0
V = +1.1 m/s (forward)
We can solve the equation to find v, the velocity of the ball:
[tex]v=\frac{mu-MV}{m}=\frac{(0.144 kg)(+28.0 m/s)-(5.25 kg)(+1.1 m/s)}{0.144 kg}=-12.1 m/s[/tex]
And the sign indicates that the ball bounces backward, so the speed is 12.1 m/s.
(c) 56.4 J, 13.7 J
The total mechanical energy before the collision is just equal to the kinetic energy of the baseball, since the brick is at rest; so:
[tex]E_i = \frac{1}{2}mu^2 = \frac{1}{2}(0.144 kg)(28.0 m/s)^2=56.4 J[/tex]
The total mechanical energy after the collision instead is equal to the sum of the kinetic energies of the ball and the brick after the collision:
[tex]E_f = \frac{1}{2}mv^2 + \frac{1}{2}MV^2 = \frac{1}{2}(0.144 kg)(12.1 m/s)^2 + \frac{1}{2}(5.25 kg)(1.1 m/s)^2=13.7 J[/tex]
(d)
A collision is:
- elastic when the total mechanical energy is conserved before and after the collision
- inelastic when the total mechanical energy is NOT conserved before and after the collision
In this problem we have:
- Energy before the collision: 56.4 J
- Energy after the collision: 13.7 J
Since energy is not conserved, this is an inelastic collision.
(e) No
As shown in part (c) and (d), the kinetic energy of the system is not conserved. This is due to the fact that in inelastic collisions (such as this one), there are some internal/frictional forces that act on the system, and that cause the dissipation of part of the initial energy of the system. This energy is not destroyed (since energy cannot be created or destroyed), but it is simply converted into other forms of energy (mainly heat and sound).
The height of the upper falls at Yellowstone Falls is 33 m. When the water reaches the bottom of the falls, its speed is 26 m/s. Neglecting air resistance, what is the speed of the water at the top of the falls?
Answer:
Speed of water at the top of fall = 5.40 m/s
Explanation:
We have equation of motion
[tex]v^2=u^2+2as[/tex]
Here final velocity, v = 26 m/s
a = acceleration due to gravity
[tex]a=9.8m/s^2 \\ [/tex]
displacement, s = 33 m
Substituting
[tex]26^2=u^2+2\times 9.8 \times 33\\\\u^2=29.2\\\\u=5.40m/s \\ [/tex]
Speed of water at the top of fall = 5.40 m/s
Car goes 60min/hr. What is speed in m/s? 1 mile = 1.6 km
Answer:
26.7 m/s
Explanation:
The speed of the car is
v = 60 mi/h
We know that
1 mile = 1.6 km = 1600 m
1 h = 60 min = 3600 s
So we can convert the speed from mi/h into m/s by multiplying by the following factor:
[tex]v = 60 \frac{mi}{h} \cdot \frac{1600 m/mi}{3600 s/h}=26.7 m/s[/tex]
In the amusement park ride known as Magic Mountain Superman, powerful magnets accelerate a car and its riders from rest to 43.4 m/s in a time of 8.59 s. The mass of the car and riders is 3.00 × 10^3 kg. Find the average net force exerted on the car and riders by the magnets.
Answer:
Average net force, F = 15157.15 N
Explanation:
It is given that,
The mass of the car and riders is, [tex]m=3\times 10^3\ kg[/tex]
Initial speed of the car, u = 0
Final speed of the car, v = 43.4 m/s
Time, t = 8.59 seconds
We need to find the average net force exerted on the car and riders by the magnets. It can be calculated using second law of motion as :
F = m a
[tex]F=m(\dfrac{v-u}{t})[/tex]
[tex]F=3\times 10^3\ kg\times (\dfrac{43.4\ m/s-0}{8.59\ s})[/tex]
F = 15157.15 N
So, the average net force exerted on the car and riders by the magnets. Hence, this is the required solution.
Final answer:
The average net force comes out to be 15,150 N.
Explanation:
The student has provided information about the Magic Mountain Superman ride, where riders are accelerated by magnets. To find the average net force exerted on the car and riders by the magnets, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = m * a).
First, we determine the acceleration using the formula a = (v - u) / t, where 'v' is the final velocity, 'u' is the initial velocity (which is 0 since the car starts from rest), and 't' is the time taken to reach the final velocity.
Using the given data, the acceleration a = (43.4 m/s - 0 m/s) / 8.59 s = 5.05 [tex]m/s^2[/tex]. Now, we can use this acceleration to calculate the force with F = m * a. Substituting the values, we get [tex]F = 3.00 times 103 kg * 5.05 m/s^2 = 1.515 times 104[/tex]N. Hence, the average net force exerted by the magnets on the car and riders is 15,150 N.
A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. If the disk starts with an angular velocity of 7.0 rad/s and is subject to a constant angular acceleration of 3.0 rad/s2, find the angular displacement of a point on the rim of the disk as it rotates under these conditions for 15 s. (Assume the positive direction is in the initial direction of the rotation of the disk. Indicate the direction with the sign of your answer.)
Answer:
442.5 rad
Explanation:
w₀ = initial angular velocity of the disk = 7.0 rad/s
α = Constant angular acceleration = 3.0 rad/s²
t = time period of rotation of the disk = 15 s
θ = angular displacement of the point on the rim
Angular displacement of the point on the rim is given as
θ = w₀ t + (0.5) α t²
inserting the values
θ = (7.0) (15) + (0.5) (3.0) (15)²
θ = 442.5 rad
what is the radius of the circular path ofa charged particle in a magnetic field?
Answer:
The radius of a circular path of a charged particle orbit depends on the charge and velocity of the particle.
Explanation:
How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest? a.3.41 x 10^5 J b.2.73 x 10^5 J c.4.09 x 10^5 J d.4.77 x 10^5 J
Answer:
Work done by the frictional force is [tex]3.41\times 10^5\ J[/tex]
Explanation:
It is given that,
Mass of the car, m = 1000 kg
Initial velocity of car, u = 26.1 m/s
Finally, it comes to rest, v = 0
We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.
[tex]W=k_f-k_i[/tex]
[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]
[tex]W=\dfrac{1}{2}\times 1000\ kg(0^2-(26.1\ m/s)^2)[/tex]
W = −340605 J
or
[tex]W=3.41\times 10^5\ J[/tex]
Hence, the correct option is (a).
The work done by frictional forces to slow a 1000 kg car from 26.1 m/s to rest is 3.41 x 10^5 J (a). This value is derived from the car's initial kinetic energy, which is lost due to friction.
Explanation:The question is asking how much work the frictional forces need to do to bring a 1000 kg car to rest, from an initial speed of 26.1 m/s. In such a scenario, these forces would be working against the car's kinetic energy.
The car's initial kinetic energy can be calculated using the formula 1/2 m v^2 (m = mass, v = speed). So the kinetic energy = 1/2 * 1000 kg * (26.1 m/s)^2 = 3.41 x 10^5 J. Since work done by friction is equal to the change in kinetic energy, and the car is brought to rest, the total work done by frictional forces is 3.41 x 10^5 J. Therefore, the correct answer is (a) 3.41 x 10^5 J.
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A small bulb is rated at 7.5 W when operated at 125 V. Its resistance (in ohms) is : (a) 17 (b) 7.5 (c) 940 (d) 2100 (e) 0.45
Answer:
Resistance of the bulb is 2100 watts.
Explanation:
Given that,
Power of the bulb, P = 7.5 watts
Voltage, V = 125 volts
We have to find the resistance of the bulb. The power of an electrical appliance is given by the following formula as :
[tex]P=\dfrac{V^2}{R}[/tex]
[tex]R=\dfrac{V^2}{P}[/tex]
[tex]R=\dfrac{(125\ V)^2}{7.5\ W}[/tex]
R = 2083.34 ohms
or
R = 2100 ohms
Hence, the correct option is (d) "2100 ohms"
It takes 3 s for a rock to hit the ground when it is thrown straight up from a cliff with an initial velocity of 8.63 m/s. How long a time would it take to reach the ground if it is thrown straight down with the same speed?
Answer:
Explanation:
h = height of the cliff
Consider upward direction as positive and downward direction as negative
Consider the motion of rock thrown straight up :
Y = vertical displacement = - h
v₀ = initial velocity = 8.63 m/s
a = acceleration = - 9.8 m/s²
t = time taken to hit the ground = 3 s
Using the equation
Y = v₀ t + (0.5) a t²
- h = (8.63) (3) + (0.5) (- 9.8) (3)²
h = 18.21 m
Consider the motion of rock thrown down :
Y' = vertical displacement = - 18.21
v'₀ = initial velocity = - 8.63 m/s
a' = acceleration = - 9.8 m/s²
t' = time taken to hit the ground = ?
Using the equation
Y' = v'₀ t' + (0.5) a' t'²
- 18.21 = (- 8.63) t' + (0.5) (- 9.8) t'²
t' = 1.2 s
Which would have the highest frequency of vibration? (Prove mathematically.) Pendulum A with a 200 g mass on a 1.0 m string Pendulum B with a 400 g mass on a 0.5 string
Answer:
Pendulum B
Explanation:
The time period of a pendulum is given by :
[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]
Case 1.
Mass, m = 200 g = 0.2 kg
Length of string, l = 1 m
Time, [tex]T_1=2\pi\sqrt{\dfrac{1\ m}{9.8\ m/s^2}}[/tex]
T₁ = 2.007 Seconds
Since, [tex]f=\dfrac{1}{T_1}[/tex]
[tex]f_1=\dfrac{1}{2.007}[/tex]
f₁ = 0.49 Hz
Case 2.
Mass, m = 400 g = 0.4 kg
Length of string, l = 0.5 m
Time, [tex]T_2=2\pi\sqrt{\dfrac{0.5\ m}{9.8\ m/s^2}}[/tex]
T₂ = 1.41 seconds
[tex]f₂=\dfrac{1}{T_2}[/tex]
[tex]f₂=\dfrac{1}{1.41}[/tex]
f₂ = 0.709 seconds
Hence, pendulum B have highest frequency of vibration.
What is the final velocity of a 28 kg object that has an initial velocity of 5 m/s and has a force of 232 N exerted on in for 5.2 seconds? Do not leave a space between the units and the answer. No decimals-round to nearest whole number.
Answer:
48 m/s
Explanation:
m = mass of the object = 28 kg
F = magnitude of net force acting on the object = 232 N
acceleration of the object is given as
a = F/m
a = 232/28
a = 8.3 m/s²
v₀ = initial velocity of the object = 5 m/s
v = final velocity of the object
t = time interval = 5.2 s
using the kinematics equation
v = v₀ + a t
v = 5 + (8.3) (5.2)
v = 48 m/s
Calculate the power output of a 1.5 g fly as it walks straight up a windowpane at 2.4 cm/s .
Express your answer using two significant figures. (Watts)
The power output of a 1.5 g fly moving upwards at a speed of 2.4 cm/s can be calculated by applying the physics formula for power. The resultant power output is ~0.036 Watts.
Explanation:The power output of any moving body can be calculated using the formula P = mgh/t where 'P' is power, 'm' is mass, 'g' is acceleration due to gravity, 'h' is height and 't' is time. In order to calculate the power output of the fly, we need to convert the variables to the appropriate units.
So first, convert the mass of the fly to kg, which gives 0.0015 kg. Then, convert the speed from cm/s to m/s, giving us 0.024 m/s. The acceleration due to gravity is approximately 9.8 m/s2. We want to find the power as the fly walks 1 meter.
Substituting the values into the power equation, we get P = (0.0015 kg * 9.8 m/s2 * 1 m) / (1 m / 0.024 m/s) = 0.036 W, which when rounded off to two significant figures is 0.036 Watts.
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To calculate the power output of a fly walking up a windowpane, we need to use the equation Power = force x velocity. By converting the mass of the fly into kilograms, calculating the force exerted by the fly using the equation Force = mass x gravity, and multiplying the force by the velocity, we find that the power output of the fly is approximately 0.00004 Watts or 4 x 10^-5 Watts.
Explanation:To calculate the power output of the fly, we can use the equation: Power = force x velocity.
First, let's convert the mass of the fly from grams to kilograms. 1.5 g = 0.0015 kg. The force exerted by the fly is equal to its weight, which is given by the equation: Force = mass x gravity.
Assuming the acceleration due to gravity is 9.8 m/s^2, the force exerted by the fly is: Force = 0.0015 kg x 9.8 m/s^2. Finally, we can calculate the power output of the fly by multiplying the force by the velocity: Power = (0.0015 kg x 9.8 m/s^2) x 0.024 m/s. This gives us a power output of approximately 0.00004 Watts or 4 x 10^-5 Watts.
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Two children of mass 27 kg and 51 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 4 m, at what distance from the pivot point is the small child sitting in order to maintain the balance?
Answer:
2.62 m
Explanation:
Let the small child sit at a distance x from the pivot.
The distance of big child from the pivot is 4 - x .
By using the concept of moments.
Clockwise moments = anticlockwise moments
27 x = 51 ( 4 - x )
27 x = 204 - 51 x
78 x = 204
x = 2.62 m
g What is the specific heat of silver? The molar heat capacity of silver is 25.35 J/mol⋅∘C. How much energy would it take to raise the temperature of 9.00 g of silver by 18.3 ∘C? Express your answer with the appropriate units.
The specific heat of silver is 0.235 J/g⋅∘C. It would take 39.1 J of energy to raise the temperature of 9.00 g of silver by 18.3 ∘C.
Explanation:The specific heat of silver can be calculated using the information given. The molar heat capacity of silver is 25.35 J/mol⋅∘C. To find the specific heat, we need to convert the molar heat capacity to g⋅∘C. The molar mass of silver is 107.87 g/mol. Therefore, the specific heat of silver is 0.235 J/g⋅∘C.
To calculate the amount of energy required to raise the temperature of 9.00 g of silver by 18.3 ∘C, we can use the formula Q = mcΔT, where Q is the energy, m is the mass, c is the specific heat, and ΔT is the change in temperature. Plugging in the values, we get Q = (9.00 g)(0.235 J/g⋅∘C)(18.3 ∘C) = 39.1 J.
Therefore, it would take 39.1 J of energy to raise the temperature of 9.00 g of silver by 18.3 ∘C.
Consider a car that is initially traveling along a straight stretch of highway at 15 m/s. At t=0 the car begins to accelerate at 2.0 m/s2 in order to pass a truck. The final velocity is vx=25m/s. Calculate how far the car travels during its 5.0 s of acceleration.
Answer:
The car travels 100 meter during its 5.0 s of acceleration
Explanation:
We have the equation of motion v²=u²+2as, where v is the final velocity, u is the initial velocity, a is the acceleration ans s is the displacement.
v = 25 m/s
a = 2 m/s²
u = 15 m/s
Substituting
25²=15²+2 x 2 x s
s = 100 m
The car travels 100 meter during its 5.0 s of acceleration
Suppose that the centripetal force acting on an object in circular motion were doubled to a new value, and the object remained in a circular path with the same radius. How would the motion be affected?
Explanation:
Centripetal force is mass times centripetal acceleration:
F = m v² / r
If force is doubled while mass and radius are held constant, then velocity will increase.
2F = m u² / r
2 m v² / r = m u² / r
2 v² = u²
u = v√2
So the velocity increases by a factor of √2.
Doubling the centripetal force acting on an object in circular motion, while maintaining the same radius, necessitates an increase in the object's velocity, leading to faster circular motion.
The question explores how doubling the centripetal force acting on an object in uniform circular motion, while maintaining the same radius, affects its motion. According to Newton's second law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The formula for centripetal force is given as Fc = m(v2/r), where m is mass, v is velocity, and r is the radius of the circular path. Doubling the centripetal force while keeping the radius constant means that for the force to remain balanced and the object to stay in circular motion, the velocity of the object must increase. This is because the square of the velocity (v2) is directly proportional to the force applied. Therefore, the object's speed around the circular path will increase, resulting in a faster circular motion.
The actual size of a gold nucleus is approximately 7 fm. Knowing this, calculate the kinetic energy (KE, or KEa, however you may choose to label the measurement) that an alpha particle would need to just touch the outside of the nucleus. Does this seem like a reasonable number?
Hint - the LHC is currently the most powerful particle accelerator on Earth, and i?t operates at 13 TeV (TeV = 1012 eV).
Answer:
(bruh moment)
Explanation:
In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A rider is towed at a constant speed by a rope that is at an angle of 15 ∘ from horizontal. The tension in the rope is 1900 N. The force of the sail on the rider is 30∘ from horizontal. What is the weight of the rider? Express your answer with the appropriate units.
Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
The weight of the rider is 566.89 N.
The given parameters;
Tension in the rope, T = 1900 Nangle of inclination of Tension, = 15⁰The force of sail on the rider, = Fangle of inclination of Force, = 30 ⁰Let the weight of the rider = W
Apply Newton's second law of motion, to determine the net force in horizontal and vertical direction.
F = ma
Sum of the forces in horizontal direction is calculated as follows;
[tex]\Sigma F_x = 0\\\\Fcos(30) - Tcos(15) = 0\\\\0.866 F - 0.965T = 0\\\\0.866F = 0.965T\\\\F = \frac{0.965T}{0.866} , \ \ T = 1900 \ N\\\\F = \frac{0.965 \times 1900}{0.866} \\\\F = 2,177.21 \ N[/tex]
Sum of the forces in the vertical direction is calculated as follows;
[tex]\Sigma F_y = 0\\\\Fsin(30) - W -Tsin(15) = 0\\\\W = Fsin(30)- Tsin(15)\\\\W = (2117.21 \times 0.5) - (1900\times 0.2588)\\\\W = 566.89 \ N[/tex]
Thus, the weight of the rider is 566.89 N.
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wo ropes are attached to a 40kg object. The rope with a 50N force points along a direction 49 degrees from the positive x-axis, and the other robe with a 55N force points along a direction 220 degrees. What is the acceleration of the object?
Answer:
Acceleration is 0.24 m/s² at 104.44° to positive x-axis.
Explanation:
Refer the figure given.
Let us take component i along positive X axis and component j along positive Y axis.
50 N force can be resolved in to 50 cos 49 i + 50 sin 49 j
F1 = 32.80 i + 37.74 j
55 N force can be resolved in to 55 cos 220 i + 55 sin 220 j
F2 = -42.13 i -35.35 j
Total force
F = F1 + F2 = 32.80 i + 37.74 j -42.13 i -35.35 j = -9.33 i + 2.39 j
We have
F = ma = 40a =-9.33 i + 2.39 j
Acceleration
a =-0.233 i + 0.060 j
[tex]\texttt{Magnitude}=\sqrt{(-0.233)^2+(0.060)^2}=0.24m/s^2[/tex]
[tex]\texttt{Direction},\theta =tan^{-1}\left ( \frac{0.060}{-0.233} \right )=104.44^0[/tex]
Acceleration is 0.24 m/s² at 104.44° to positive x-axis.
A uniform conducting rod of length 22 cm has a potential difference across its ends equal to 41 mV (millivolts). What is the magnitude of the electric field inside the conductor in units of N/C?
Answer:
0.186 N/C
Explanation:
The relationship between electric field strength and potential difference is:
[tex]E=\frac{\Delta V}{d}[/tex]
where
E is the electric field strength
[tex]\Delta V[/tex] is the potential difference
d is the distance
Here we have
[tex]\Delta V=41 mV=0.041 V[/tex]
d = 22 cm = 0.22 m
So the electric field magnitude is
[tex]E=\frac{0.041 V}{0.22 m}=0.186 N/C[/tex]
Final answer:
To find the electric field inside a conductor, use the formula E = V/d. With a potential difference of 41 mV and length of 22 cm, the electric field is 0.186 N/C.
Explanation:
The magnitude of the electric field inside a conductor is calculated using the formula E = V/d, where V is the potential difference and d is the length across which the potential difference is applied.
In this case, the conducing rod has a length of 22 cm (0.22 meters) and a potential difference across its ends of 41 mV (0.041 volts). Therefore, the magnitude of the electric field E in the conductor can be calculated as:
E = V/d = 0.041 V / 0.22 m = 0.18636... V/m
Since 1 V/m is equivalent to 1 N/C, the magnitude of the electric field in the conductor is approximately 0.186 N/C.
At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6,9) and constant acceleration 2i−4j−2k. Find an equation for the position vector of the particle.
The particle has constant acceleration according to
[tex]\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k[/tex]
Its velocity at time [tex]t[/tex] is
[tex]\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du[/tex]
[tex]\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t[/tex]
[tex]\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k[/tex]
Then the particle has position at time [tex]t[/tex] according to
[tex]\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du[/tex]
[tex]\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k[/tex]
At at the point (3, 6, 9), i.e. when [tex]t=0[/tex], it has speed 8, so that
[tex]\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64[/tex]
We know that at some time [tex]t=T[/tex], the particle is at the point (5, 2, 7), which tells us
[tex]\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}[/tex]
and in particular we see that
[tex]v_{0y}=-2v_{0x}[/tex]
and
[tex]v_{0z}=-v_{0x}[/tex]
Then
[tex]{v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3[/tex]
[tex]\implies v_{0y}=\mp\dfrac{8\sqrt6}3[/tex]
[tex]\implies v_{0z}=\mp\dfrac{4\sqrt6}3[/tex]
That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is
[tex]\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k[/tex]
To find the equation for the position vector of the particle, use the formula r(t) = r(0) + v(0) * t + (1/2) * a * t^2.
Explanation:To find the equation for the position vector of the particle, use the following steps:
Write down the initial position vector, r(0), which is (3,6,9).Use the formula for the position vector of a particle with constant acceleration, r(t) = r(0) + v(0) * t + (1/2) * a * t^2, where r(t) is the position vector at time t, v(0) is the initial velocity vector, and a is the constant acceleration vector.Substitute the values into the formula and simplify to get the equation for the position vector.For the given problem, the equation for the position vector of the particle is: r(t) = (3 + 4t)i + (6 - 4t + 2t^2)j + (9 - 2t + t^2)k.
g You are standing on a skateboard, initially at rest. A friend throws a very heavy ball towards you. You can either catch the object or deflect the object back towards your friend (such that it moves away from you with the same speed as it was originally thrown). What should you do in order to MINIMIZE your speed on the skateboard?
Answer:
catch the ball
Explanation:
if you push the ball back, that equal amount of force will applied to you. however if you catch it, you absorb less of the directional energy
A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the bullet hits the ground? (b) How far does the bullet travel horizontally?
Explanation:
This situation is a good example of the projectile motion or parabolic motion, in which the travel of the bullet has two components: x-component and y-component. Being their main equations as follows:
x-component:
[tex]x=V_{o}cos\theta t[/tex] (1)
Where:
[tex]V_{o}=200m/s[/tex] is the bullet's initial speed
[tex]\theta=0[/tex] because we are told the bullet is shot horizontally
[tex]t[/tex] is the time since the bullet is shot until it hits the ground
y-component:
[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (2)
Where:
[tex]y_{o}=1.5m[/tex] is the initial height of the bullet
[tex]y=0[/tex] is the final height of the bullet (when it finally hits the ground)
[tex]g=9.8m/s^{2}[/tex] is the acceleration due gravity
Part (a):Now, for the first part of this problem, the time the bullet elapsed traveling, we will use equation (2) with the conditions given above:
[tex]0=1.5m+200m/s.sin(0) t-\frac{9.8m/s^{2}.t^{2}}{2}[/tex] (3)
[tex]0=1.5m-\frac{9.8m/s^{2}.t^{2}}{2}[/tex] (4)
Finding [tex]t[/tex]:
[tex]t=\sqrt{\frac{1.5m(2)}{9.8m/s^{2}}}[/tex] (5)
Then we have the time elapsed before the bullet hits the ground:
[tex]t=0.553s[/tex] (6)
Part (b):For the second part of this problem, we are asked to find how far does the bullet traveled horizontally. This means we have to use the equation (1) related to the x-component:
[tex]x=V_{o}cos\theta t[/tex] (1)
Substituting the knonw values and the value of [tex]t[/tex] found in (6):
[tex]x=200m/s.cos(0)(0.553s)[/tex] (7)
[tex]x=200m/s(0.553s)[/tex] (8)
Finally:
[tex]x=110.656m[/tex]
The bullet will hit the ground after approximately 0.553 seconds, and in that time, it will travel horizontally around 110.6 meters.
Explanation:Calculating Projectile Motion for a Horizontally Fired BulletTo determine how much time elapses before the bullet hits the ground (1.5 m drop) when fired horizontally at 200 m/s, we use the equation for free fall motion h = (1/2)gt², where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time in seconds. Solving for time t, we find that the bullet will hit the ground after approximately 0.553 seconds.
For part (b), to find how far the bullet travels horizontally, we simply multiply the time in the air by the bullet's initial horizontal velocity. This gives a horizontal distance of 200 m/s * 0.553 s = 110.6 meters. Therefore, the bullet travels roughly 110.6 meters before hitting the ground.
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A baseball approaches home plate at a speed of 40.0 m/s, moving horizontally just before being hit by a bat. The batter hits a pop-up such that after hitting the bat, the baseball is moving at 57.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 2.10 ms. What is the average vector force the ball exerts on the bat during their interaction? (Let the +x-direction be in the initial direction of motion, and the +y-direction be up.) F
Answer:
[tex]F = (3.9\times 10^3)\hat j - 2.8 \times 10^3\hat i[/tex]
Explanation:
Initial momentum of the ball is given as
[tex]P_i = mv_i[/tex]
[tex]P_i = 0.145 (40) = 5.8 kg m/s \hat i[/tex]
now final momentum of the ball is given as
[tex]P_f = 0.145(57) = 8.3 kg m/s \hat j[/tex]
now by the formula of force we have
[tex]F = \frac{P_f - P_i}{\Delta t}[/tex]
now we have
[tex]F = \frac{8.3 \hat j - 5.8 \hat i}{2.10 \times 10^{-3}}[/tex]
[tex]F = (3.9\times 10^3)\hat j - 2.8 \times 10^3\hat i[/tex]
The average vector force the ball exerts on the bat during their interaction is
[tex]\rm F = 3.9\times10^3\;\hat{j}-2.8\times10^3\;\hat{i}[/tex]
Given :
Initial Speed = 40 m/sec
Final Speed = 57 m/sec
Mass = 0.145 Kg
Time = 0.0021 sec
Solution :
Initial Momentum is,
[tex]\rm P_i = mv_i[/tex]
[tex]\rm P_i = 0.145\times 40 = 5.8\;Kg .m/sec\;\hat{i}[/tex]
Final momentum is,
[tex]\rm P_f = mv_f = 0.145\times57=8.3\; Kg.m/sec\; \hat{j}[/tex]
Now,
[tex]\rm F=\dfrac{P_f-P_i}{\Delta t}= \dfrac{8.3\hat{j}-5.8\hat{i}}{2.10\times10^-^3}[/tex]
[tex]\rm F = 3.9\times10^3\;\hat{j}-2.8\times10^3\;\hat{i}[/tex]
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Red light has a wavelength of 500 nm. What is the frequency of red light?
Answer:
[tex]0.6 \times 10 {}^{15} [/tex]
Explanation:
By using relation,
Speed of light = frequency × wavelength (in m)
A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field. What is the radius of the proton's resulting orbit? (mproton = 1.67 × 10-27 kg, e = 1.60 × 10-19 C)
Answer:
Radius = 0.11 m
Explanation:
To find the speed of the proton we know that
[tex]KE = PE[/tex]
here we have
[tex]\frac{1}{2}mv^2 = qV[/tex]
now we have
[tex]v = \sqrt{\frac{2qV}{m}}[/tex]
now we have
[tex]v = \sqrt{\frac{2(1.60 \times 10^{-19})(1000)}{(1.67\times 10^{-27})}}[/tex]
[tex]v = 4.38 \times 10^5 m/s[/tex]
Now for the radius of the circular motion of charge we know
[tex]\frac{mv^2}{R} = qvB[/tex]
[tex]R = \frac{mv}{qB}[/tex]
[tex]R = \frac{(1.67\times 10^{-27})(4.38 \times 10^5)}{(1.60\times 10^{-19})(0.040)}[/tex]
[tex]R = 0.11 m[/tex]
The radius of the proton orbit is 0.114m
kinetic energy:When the proton travels through the potential V it gains kinetic energy given below:
[tex]\frac{1}{2}mv^2=qV\\\\v=\sqrt[]{\frac{2qV}{m} }\\\\v=\sqrt{\frac{2\times (1.6\times10^{-19})\times10^3}{1.67\times10^{-27}}[/tex]
[tex]v=4.37\times10^5m/s[/tex]
magnetic force:Now, a moving charge under magnetic field B undergoes circular motion due to the magnetic force being perpendicular to the velocity of the charge, given by:
[tex]\frac{mv^2}{r}=qvB[/tex]
[tex]r=\frac{mv}{qB} \\\\r=\frac{1.67\times10^{-27}\times4.37\times10^5}{1.6\times10^{-19}\times0.040} m\\\\r=0.114m[/tex]
the radius of the orbit is 0.114m
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A 1.7cm diameter pipe widens to 4.7cm. Liquid flows through the first segment at a speed of 4.7m/s. What is the speed of the liquid in the second segment?
Answer:
Speed of the liquid in the second segment = 0.61 m/s
Explanation:
This discharge is constant.
That is
Q₁ = Q₂
A₁v₁ = A₂v₂
[tex]\frac{\pi d_1^2}{4}\times v_1=\frac{\pi d_2^2}{4}\times v_2\\\\\frac{\pi \times 1.7^2}{4}\times 4.7=\frac{\pi \times 4.7^2}{4}\times v_2\\\\v_2=0.61m/s[/tex]
Speed of the liquid in the second segment = 0.61 m/s
The star nearest to our sun is Proxima Centauri, at a distance of 4.3 light-years from the sun. How far away, in km, is Proxima Centauri from the sun?
Answer: [tex]4.068(10)^{13} km[/tex]
Explanation:
A light year is a unit of length and is defined as "the distance a photon would travel in vacuum during a Julian year at the speed of light at an infinite distance from any gravitational field or magnetic field. "
In other words: It is the distance that the light travels in a year.
This unit is equivalent to [tex]9.461(10)^{12}km[/tex], which mathematically is expressed as:
[tex]1Ly=9.461(10)^{12}km[/tex]
Doing the conversion:
[tex]4.3Ly.\frac{9.461(10)^{12}km}{1Ly}=4.068(10)^{13}km[/tex]
The Answer: The star closest to our sun is 4.2 light years away from the sun.
Explanation:
You throw a baseball directly upward at time t = 0 at an initial speed of 12.3 m/s. What is the maximum height the ball reaches above where it leaves your hand? At what times does the ball pass through half the maximum height? Ignore air resistance and take g = 9.80 m/s2.
Explanation:
At the maximum height, the ball's velocity is 0.
v² = v₀² + 2a(x - x₀)
(0 m/s)² = (12.3 m/s)² + 2(-9.80 m/s²)(x - 0 m)
x = 7.72 m
The ball reaches a maximum height of 7.72 m.
The times where the ball passes through half that height is:
x = x₀ + v₀ t + ½ at²
(7.72 m / 2) = (0 m) + (12.3 m/s) t + ½ (-9.8 m/s²) t²
3.86 = 12.3 t - 4.9 t²
4.9 t² - 12.3 t + 3.86 = 0
Using quadratic formula:
t = [ -b ± √(b² - 4ac) ] / 2a
t = [ 12.3 ± √(12.3² - 4(4.9)(3.86)) ] / 9.8
t = 0.368, 2.14
The ball reaches half the maximum height after 0.368 seconds and after 2.14 seconds.
The maximum height the ball reaches above where it leaves your hand is 7.72 m
The time taken for the ball to pass half of its maximum height moving upwards is 2.14s and 0.37 s when moving downwards.
The given parameters;
initial velocity, u = 12.3 m/s
acceleration due to gravity, g = 9.8 m/s²
The maximum height the ball reaches above where it leaves your hand is calculated as;
[tex]v_f^2 = v_0^2 - 2gh\\\\at \ maximum \ height \ final \ velocity \ v_f = 0\\\\2gh = v_0 ^2\\\\h = \frac{v_0^2}{2g} \\\\h = \frac{(12.3)^2}{2(9.8)} \\\\h = 7.72 \ m[/tex]
The time for the ball to reach half of the maximum height is calculated as;
[tex]half \ of \ the \ maximum \ height = \frac{7.72}{2} = 3.86 \ m[/tex]
[tex]h = v_0t - \frac{1}{2} gt^2\\\\3.86 = 12.3t - 0.5\times 9.8t^2\\\\3.86 = 12.3t - 4.9t^2\\\\4.9t^2- 12.3t + 3.86=0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 4.9, \ \ b = -12.3, \ \ c = 3.86\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-b \ \ +/- \ \ \sqrt{(-12.3)^2 - 4(4.9\times 3.86)} }{2(4.9)} \\\\t = 2.14 \ s \ \ or \ \ 0.37 \ s[/tex]
The time taken for the ball to pass half of its maximum height moving upwards is 2.14s and 0.37 s when moving downwards.
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