Answer:
10.17 m/s in the direction the compact car was going (+)
Explanation:
Law of conservation of momentum states that total P before the collision must equal total momentum after the collision.
P = mv
solve for v, velocity of the two cars stuck together after the collision:
(187kg)(-38 m/s) + (702kg)(23 m/s) = (187 + 702 kg)v
momentum before = momentum after
-7106 kg·m/s + 16146 kg·m/s = 889 kg(v)
9040 kg·m/s = 889 kg(v)
v = 9040 kg·m/s / 889 kg = 10.17 m/s
A cheetah stalking her prey is moving at 1 m/s. She accelerates to a speed of
25 m/s to catch the zebra. Her acceleration is 10 m/s². Calculate the time
caken to reach her top speed.
The time taken to reach zebra's top speedis 2.4 second by the cheetah.
How to calculate time?using the suvat formulav=u+at 25=1+10t because of this, t=2.4 secondsWhat is the formula for Suvat?V = U + A T. It belongs to the family of equations known as the SUVAT. Their namesake motion in a straight line with constant acceleration involves the following variables: displacement (S), starting velocity (U), final velocity (V), acceleration (A), and time (T).The majority of the suvat equations are rather simple to construct once you realize that acceleration (, assumed constant) is the derivative of velocity () with respect to time and velocity is the derivative of position (), likewise with respect to time.Physics frequently mentions Issac Newton. For more information on suvat equation kindly visit to
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a front-loading washing machine is mounted on a thick rubber pad that acts like a spring; the weight (with m/s ) of the machine depresses the pad exactly 0.89 cm. when its rotor spins at radians per second, the rotor exerts a vertical force of newtons on the machine. neglecting friction, at what speed (in revolutions per minute) will resonance vibrations will occur? resonance occurs at rpms
The speed (in revolutions per minute) at which resonance vibrations will occur is 273.91 RPM.
The front-loading washing machine is mounted on a thick rubber pad that acts as a spring. When the rotor spins at radians per second, the rotor exerts a vertical force of newtons on the machine.
Resonance vibrations occur when the natural frequency of the machine matches the frequency of the external force applied to it. This can cause the machine to vibrate excessively and even damage it.
In this scenario, the natural frequency of the washing machine is related to its spring constant and the mass of the machine. Neglecting friction, resonance vibrations occur at RPMs.
Given,
The weight (with m/s) of the machine depresses the pad exactly 0.89 cm.
The rotor exerts a vertical force of N on the machine.
The natural frequency of a machine is given by:
f = 1/2π x √(k/m)
where f is the frequency of oscillation,
k is the spring constant of the spring,
m is the mass of the machine.
In this case, the weight of the machine depresses the spring by 0.89 cm.
So, the effective spring constant is given by,
k = F/x = (mg)/x
where F is the weight of the machine,
m is its mass,
g is the acceleration due to gravity,
and x is the amount by which the spring is compressed.
Substituting the given values,
k = (m x g) / x = (m x 9.8) / 0.0089 = 1101.12m N/m
Substituting the given values,
f = 1/2π x √(k/m) = 1/2π x √(1101.12m/m) = 4.9448 √m Hz
The frequency of oscillation is given in Hz.
To convert it to RPMs, we need to divide it by 60 (the number of seconds in a minute) and multiply it by 2π (the number of radians in a circle).
f = 4.9448 √m Hz = (4.9448 √m x 2π)/60 RPM
At resonance, the frequency of oscillation should match the frequency of the rotor.
So, the frequency of the rotor is given by,
f = ω/2π
where ω is the angular velocity of the rotor in radians per second.
Substituting the given values,
N = mg = F,
ω = N/(m x r) = 94.4912/r rads/s
where r is the radius of the rotor in meters.
Substituting these values in the equation for frequency,
f = ω/2π = (94.4912/r) / 2π = 15.0415 / r Hz
At resonance, the frequency of oscillation should match the frequency of the rotor.
Therefore,4.9448 √m = 15.0415 / r
Squaring both sides,19.4542 m = (15.0415/r)^2
Solving for r, r = 0.0275 m = 2.75 cm
Therefore, the radius of the rotor is 2.75 cm.
The speed (in revolutions per minute) at which resonance vibrations will occur is 273.91 RPM.
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a fidget spinner rotates at about 4000 rpm (revolutions per minute). calculate the speed of the edge of a fidget spinner that has a diameter of 6.0 cm.
The speed of the edge of a fidget spinner that has a diameter of 6.0 cm is 12.58 m/s.
To calculate the speed of the edge of the fidget spinner, we need to convert the rotational speed from revolutions per minute to radians per second, and then multiply by the radius of the spinner.
First, we convert 4000 rpm to radians per second:
4000 rpm = 4000 revolutions / 60 seconds = 66.67 revolutions per second
Each revolution is 2π radians, so:
66.67 revolutions per second = 66.67 * 2π radians per second = 419.47 radians per second
Now that we have the rotational speed in radians per second, we can calculate the speed of the edge of the fidget spinner:
speed = rotational speed x radius
The radius of the fidget spinner is half of its diameter, which is 6.0 cm / 2 = 3.0 cm = 0.03 m.
So, the speed of the edge of the fidget spinner is:
speed = 419.47 radians per second x 0.03 m = 12.58 m/s
Therefore, the speed of the edge of the fidget spinner is approximately 12.58 m/s.
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At which location is contact metamorphism least likely to occur?
Contact metamorphism is least likely to occur in low-temperature areas with sedimentary rocks, such as the Earth's oceanic crust.
Contact metamorphism occurs due to heating, with or without burial, of rocks that lie close to a magma intrusion. It is characterized by low P/T gradients, as strong thermal gradients between an intruding magma and adjacent country rock are best established at shallow crustal level. It is a type of metamorphism where rock minerals and texture are changed, mainly by heat, due to contact with magma. Regional metamorphism is a type of metamorphism where rock minerals and texture are changed by heat and pressure over a wide area or region.
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A 2. 72-nF parallel-plate capacitor is charged to an initial potential difference ΔVi = 100 V and is then isolated. The dielectric material between the plates is Pyrex glass, with a dielectric constant of 5. 6
The work done by the capacitor is obtained as 1.36 * 10^-5 J
What is the capacitor?A capacitor is an electrical component that stores electrical energy in an electric field. It consists of two conductive plates separated by a non-conductive material, known as the dielectric.
When a voltage is applied across the plates, one plate becomes positively charged and the other becomes negatively charged, creating an electric field between them. The dielectric material prevents the charges from flowing between the plates and instead stores the energy in the form of an electric field.
Work done by the Capacitor = 1/2CV^2
= 0.5 * 2.72 * 10^-9 * (100)^2
= 1.36 * 10^-5 J
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Missing parts;
A 2. 72-nF parallel-plate capacitor is charged to an initial potential difference ΔVi = 100 V and is then isolated. The dielectric material between the plates is Pyrex glass, with a dielectric constant of 5. 6. What is the work done by the capacitor?
Stopping Distance Activity
Background: This activity involves observing the effect of original car speed upon the
skid distance. The simulation involves a toy car rolling down a hill, hitting a box, and
skidding to a stop. The height from which the car is released can be modified. The
speed of the car at the bottom of the hill (prior to contact with the box) is reported. The
distance that the car (and box) skid before stopping can be measured. Before answering
the questions, it would be useful to first run the simulation for each of the provided
heights and to record the corresponding pre-collision car speed and the stopping
distance. The background grid on the simulation screen can be used to determine the
stopping distance. Each square on the grid is 10 cm in length along its edge.
Data:
Initial Height Pre-Collision Speed Skid Distance
(m/s)
(cm)
(cm)
10. 0
20. 0
30. 0
40. 0
50. 0
60. 0
70. 0
80. 0
By using the formula:- [tex]mgh= \frac{1}{2} mv^{2} , v = \sqrt{2gh}[/tex], as the release height of the car increase the speed of the car increases.
As the release height of the car increases, the skid distance of the car increaseincrease due to increased velocity.
The directional speed of an object in motion, as measured by a specific unit of time and witnessed from a specific point of reference, is what is referred to as velocity.
Velocity can be defined as the rate at which something travels in a specific direction. Like the speed of a vehicle travelling north on a highway or the speed of a rocket after launch, for instance.
Less momentum with increased height. Less height, more speed. The rate at which it is exchanging height for velocity at any given moment in the fall is therefore equal to the velocity divided by the gravitational constant. As a result, at the top when its velocity is low, it loses only a tiny amount of height with each increase in velocity.
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complete question:-
A 80 kg passenger is seated 12 m from the center of the loop of a roller coaster. What centripetal force does the passenger experience when the roller coaster reaches an angular speed of 3.14 rad/s? Fc = (m)(r)(ω2)
The centripetal force the passenger experience is 30167.52 N
Centripetal force calculation'The centripetal force experienced by the passenger can be calculated using the formula:
Fc = (m)(r)(ω^2)
where Fc is the centripetal force, m is the mass of the passenger, r is the radius of the loop, and ω is the angular speed of the roller coaster.
Given:
m = 80 kg
r = 12 m
ω = 3.14 rad/s
Substituting these values into the formula, we get:
Fc = (80 kg)(12 m)(3.14 rad/s)^2
Fc = 30167.52 N
Therefore, the passenger experiences a centripetal force of approximately 30167.52 N when the roller coaster reaches an angular speed of 3.14 rad/s.
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mike has a linear production possibilities frontier that shows maximum production of trucks or computers of 10 million each. debra also has a linear production possibilities curve that shows maximum production of 9 million trucks or 3 million computers.
Mike and Debra both have linear production possibilities frontiers (PPF), which show the maximum amount of two goods that can be produced given their resources and technology. However, their PPFs are different because they have different maximum production levels for trucks and computers.
Mike's PPF shows that he can produce a maximum of 10 million trucks or 10 million computers. This means that if he allocates all of his resources to producing trucks, he can produce 10 million of them, and if he allocates all of his resources to producing computers, he can produce 10 million of them. Debra's PPF, on the other hand, shows that she can produce a maximum of 9 million trucks or 3 million computers. This means that if she allocates all of her resources to producing trucks, she can produce 9 million of them, and if she allocates all of her resources to producing computers, she can produce 3 million of them.
Overall, Mike and Debra's PPFs show the trade-offs they face when deciding how to allocate their resources between producing trucks and computers. They must decide how much of each good to produce in order to maximize their production and meet their needs.
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The 0. 20 kg puck on the frictionless, horizontal table in the figure is connected by a string through a hole in the table to a hanging 1. 20 kg block. With what speed must the puck rotate in a circle of radius r = 0. 40 m if the block is to remain hanging at rest?
The puck must rotate with a speed of 3.44 m/s to keep the hanging block at rest.
To keep the hanging block at rest, the tension in the string must be equal to the weight of the block. We can start by finding the tension in the string using the weight of the hanging block:
T = mg = (1.20 kg)(9.81 m/s^2) = 11.77 N
The tension in the string provides the centripetal force that keeps the puck moving in a circle. The centripetal force is given by:
F = ma = m(v^2/r)
where m is the mass of the puck, v is its speed, r is the radius of the circle, and a is the centripetal acceleration.
Solving for v, we get:
v = sqrt(ra) = sqrt(r(T/m)) = sqrt((0.40 m) × (11.77 N)/(0.20 kg)) = 3.44 m/s
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a particle with charge 18 nc travels from an equipotential line with 46 volts to another equipotential with 10 volts as part of miss frizzle's science experiment of the day for her students. what is the change in potential energy of the system during this time interval? answer in nj (nano-joules) to the nearest nj.
The change in potential energy of the system during this time interval is -648 nJ.
The change in potential energy of a particle with charge q that moves from a point with potential V1 to a point with potential V2 is given by:
ΔPE = q x (V2 - V1)
where q is the charge of the particle, and
V2 and V1 are the potentials at the final and initial positions.
Therefore, the charge of the particle is 18 nC, and it moves from a point with potential V1 = 46 V to a point with potential V2 = 10 V.
Therefore, the change in potential energy is:
ΔPE = q x (V2 - V1) = 18 nC x (10 V - 46 V) = -648 nJ
The negative sign indicates that the potential energy of the particle has decreased, which means that the particle has lost potential energy as it moved from the point with higher potential to the point with lower potential.
Therefore, the change in potential energy of the system during this time interval is -648 nJ
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describe the single touch method of making magnet. Also describe the process of making electromagnet.
Answer:
The single touch method of making a magnet is a simple way to magnetize a ferromagnetic material, such as a piece of iron. The process involves taking a magnet and bringing it into contact with the material to be magnetized, while moving it along the length of the material in a single direction. This process aligns the magnetic domains within the material, causing it to become magnetized. The strength of the resulting magnet depends on the strength of the magnet used and the length of time it is in contact with the material.
The process of making an electromagnet involves using an electric current to create a magnetic field. To make an electromagnet, a wire is wound into a coil around a magnetic core, such as an iron rod. When an electric current flows through the wire, it creates a magnetic field around the coil, which in turn magnetizes the magnetic core. The strength of the resulting magnet depends on the strength of the current flowing through the wire, the number of coils in the wire, and the type of magnetic core used. Electromagnets are commonly used in a variety of applications, including electric motors, speakers, and MRI machines.
Explanation:
what is $12.7\ge s-5.3$ .
The solution of expression $12.7\ge s-5.3$ is $s \leq 18$.
The expression $12.7\ge s-5.3$ is an inequality, which means that it expresses a range of values for the variable $s$. The symbol $\ge$ means "greater than or equal to," and in this context it indicates that $12.7$ is greater than or equal to $s-5.3$.
To solve for $s$, we can start by adding $5.3$ to both sides of the inequality:
\begin{align*}
12.7 &\ge s-5.3 \
12.7+5.3 &\ge s-5.3+5.3 \
18 &\ge s
\end{align*}
So the solution is $s \leq 18$. This means that any value of $s$ that is less than or equal to $18$ will satisfy the inequality. For example, $s=15$ would satisfy the inequality, because $12.7 \ge 15-5.3$. However, $s=20$ would not satisfy the inequality, because $12.7 \ngeq 20-5.3$.
What are the variables?
A variable is a symbol or letter that represents a value that can vary or change in a given context or problem. Variables are used to write mathematical expressions, equations, and functions.
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Complete question is: $12.7\ge s-5.3$ is $s \leq 18$.
A horizontal pipe of diameter 0.945 m has a smooth constriction to a section of diameter 0.567 m. the density of oil flowing in the pipe is 821 kg/m° if the pressure in the pipe is 8940 n/m° and in the constricted section is 6705 n/m?, what is the rate at which oil is flowing?
If the pressure in the pipe is 8940 n/m° and in the constricted section is 6705 n/m, the rate at which oil is flowing through the pipe is 2.74 m^3/s.
The volume flow rate (Q) can be expressed as:
Q = Av
We can use the continuity equation to relate the velocities at the two different cross-sectional areas:
A1v1 = A2v2
Where:A1 and A2 are the cross-sectional areas at the two different points in the pipe
v1 and v2 are the fluid velocities at the two different points in the pipe
We can rearrange this equation to solve for v1 in terms of v2:
v1 = (A2/A1) * v2
Now we can use Bernoulli's equation to relate the pressure and velocity at the two different points in the pipe:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
We can rearrange this equation to solve for v2:
v2 = sqrt(2(P1 - P2) / ρ + v1^2)
Now we can plug in the given values to find v2:
v1 = ?
A1 = (π/4)(0.945 m)^2 = 0.664 m^2
A2 = (π/4)(0.567 m)^2 = 0.100 m^2
ρ = 821 kg/m^3
P1 = 8940 N/m^2
P2 = 6705 N/m^2
v2 = sqrt(2(8940 - 6705) / 821 + ((0.100/0.664) * v1)^2)
We can simplify this equation by assuming that the velocity at the wider section of the pipe (v1) is much smaller than the velocity at the constriction (v2), due to the principle of continuity. Therefore, we can neglect the second term in the equation, giving:
v2 = sqrt(2(8940 - 6705) / 821) = 11.98 m/s
Now we can use the continuity equation to find the volume flow rate:
Q = A1v1 = A2v2 = (π/4)(0.945 m)^2 * 11.98 m/s =2.74 m^3/s
Therefore, the rate at which oil is flowing through the pipe is 2.74 m^3/s.
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Exercise 2: Julie, the clown, carries two balloons that rub against a circus
elephant, causing the balloons to separate. Each balloon acquires 2. 0 x107 C of
in charge. How large is the electric force between them when they are separated by
a distance of 0. 50 m?
When the spheres are separated by 0.5 m, the electric force between them is 1.44 x [tex]10^{20}[/tex] N.
To determine the electric force between two balloons, we can use Coulomb's law:
F = k * (q₁ * q₂) / r²
where F is the electric force, k is Coulomb's constant (9 × [tex]10^{9}[/tex] N·m²/C²), q1 and q2 are the charges on the balloons (2.0 x [tex]10^{7}[/tex] C), and r is the distance between the balloons (0.5 m).
Plugging in these values, we get:
F = 9 × [tex]10^{9}[/tex] * (2.0 x [tex]10^{7}[/tex])² / (0.5)²
F = 1.44 x [tex]10^{20}[/tex] N
Therefore, the electric force between the balloons is 1.44 x [tex]10^{20}[/tex] N when they are separated by a distance of 0.5 m.
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A hungry 165 kg lion running northward at 76.4 km/hr attacks and holds onto a 35.9kg Thomson’s gazelle running eastward at 64.6 km/hr. Find the final speed of the lion-gazelle system immediately after the attack.
Final speed: _____ m/s
as a van travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the van by the air and the road. if the power developed by the engine is 2.01 hp, calculate the total friction force acting on the van (in n) when it is moving at a speed of 28 m/s. one horsepower equals 746 w.
When travelling at a speed of 28 m/s, the van experiences a total amount of friction force of 748 N.
Calculate the van's overall friction force (measured in N) at a speed of 28 m/s with an engine that produces 2.01 horsepower.
Engine power is calculated as
P = 2.01 hp, or 2.01 x 746, or 1502.46 W.
Van's top speed is 28 m/s.
The van's total work is equal to its change in kinetic energy, according to the work-energy theorem.
Hence, total work completed = Change in kinetic energy (K) -Change in kinetic energy (K)
= [tex](1/2) m (vf_2 - vi_2)[/tex] = [tex](1/2) m v^2[/tex]
where m is the van's mass
([tex]v_f[/tex]) = The van's final speed
([tex]v_i[/tex] )= The van's initial speed
As a result, the van is subject to a 748 N friction force (approx.). So, at a speed of 28 m/s, the van experiences a total friction force of 748 N.
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You plunge a basketball beneath the surface of a swimming pool until half the volume of the basketball is submerged. If the basketball has a radius of 12cm what is that buoyancy force on the ball due to the water
If the basketball has a radius of 12cm,the buoyant force on the basketball due to the water is 35.5 N.
When an object is submerged in a fluid, it experiences an upward force called the buoyant force. The magnitude of the buoyant force is equal to the weight of the fluid displaced by the object.
In this case, the basketball is submerged in water until half its volume is submerged. The volume of a sphere of radius r is given by:
V = (4/3) * pi * r^3
So, the volume of the basketball is:
V_basketball = (4/3) * pi * (0.12 m)^3 = 7.24 x 10^-3 m^3
Half of the volume of the basketball that is submerged is:
V_submerged = (1/2) * V_basketball = 3.62 x 10^-3 m^3
The weight of the water displaced by the basketball is equal to the weight of the volume of water that is submerged. The density of water is 1000 kg/m^3, so the mass of the displaced water is:
m_water = rho * V_submerged = 1000 kg/m^3 * 3.62 x 10^-3 m^3 = 3.62 kg
The buoyant force on the basketball due to the water is equal to the weight of the displaced water, which is:
F_buoyant = m_water * g
= 3.62 kg * 9.81 m/s^2
= 35.5 N
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Why is it difficult to walk on wet marble floor?
Answer:
On a wet marble floor, the friction is very less because marble floor has irregularities that get filled with water. We can walk only when there is more friction. Due to the less friction, we may slip. So, it is very difficult to walk on a wet marble floor.
Answer:
because marbles are always sliding and water makes them more slimy
Starting from rest, a car accelerates at 2.2 m/s2 up a hill that is inclined 5.3 ∘
above the horizontal.
Part A: How far horizontally has the car traveled in 15 s?
Part B: How far vertically has the car traveled in 15 ?
Express your answer using two significant figures.
The car travels approximately 245 m horizontally and 14.1 m vertically in 15 seconds up an inclined hill.
Section A: To find how far evenly the vehicle has gone in 15 s, we really want to utilize the equation:distance = starting speed × time + 1/2 × speed increase × time^2.Since the vehicle is beginning from rest, the underlying speed is 0. We are searching for the even distance, which is the distance along the ground. Hence, we can disregard the grade of the slope. Hence, we can utilize the level part of the speed increase, which is given by:a_horizontal = a × cosθ.where θ is the point of grade and an is the speed increase. Subbing the qualities given, we get:a_horizontal = 2.2 m/s^2 × cos(5.3°) ≈ 2.18 m/s^2.Utilizing this worth, we can find the even distance went in 15 s:distance = 0 + 1/2 × 2.18 m/s^2 × (15 s)^2 ≈ 245 m.Hence, the vehicle has voyaged around 245 m evenly in 15 s.
Part B: To find how far in an upward direction the vehicle has gone in 15 s, we want to utilize the upward part of the speed increase, which is given by:a_vertical = a × sinθ
Subbing the qualities given, we get:
a_vertical = 2.2 m/s^2 × sin(5.3°) ≈ 0.197 m/s^2
Utilizing this worth, we can find the upward distance went in 15 s:
distance = 0 + 1/2 × 0.197 m/s^2 × (15 s)^2 ≈ 14.1 m
Hence, the vehicle has voyaged roughly 14.1 m in an upward direction in 15 s.
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Compare resistance of electrical flow through wires to the resistance of water flow in a pipe. How are they similar?
Answer:
Resistance to the flow of electricity in wires and resistance to the flow of water in pipes are similar in several ways:
Both are affected by the size of the conductor: In both cases, the resistance to flow is affected by the size of the conductor. A larger diameter wire or pipe will offer less resistance to the flow of electricity or water than a smaller diameter one.
Both are affected by the length of the conductor: The resistance to flow also increases with the length of the conductor. A longer wire or pipe will offer more resistance to the flow of electricity or water than a shorter one.
Both are affected by the material of the conductor: Different materials have different resistances to flow. In electrical wires, metals are commonly used as they have low resistance. In pipes, materials such as copper, plastic, or steel may be used depending on the specific application.
Both can be reduced by minimizing obstructions: Resistance to flow can be reduced in both cases by minimizing obstructions in the path of the flow. In pipes, for example, the use of smooth inner walls and proper pipe bends can help reduce resistance. In electrical wires, minimizing the number of connections and using high-quality connections can help reduce resistance.
Overall, resistance to flow is a fundamental concept that applies to the flow of electricity and water, and the factors affecting resistance are similar in both cases.
Pls help! Could you guys verify if the correct answer is B because of the v^2? See the attached image below :)
Answer:
It could either be b, or d, Your choice! (hint: I know the answer but pick one)
What is the dewpoint when the air temperature is 28°C and the relative humidity is 47%?
The dewpoint when the air temperature is 28°C and the relative humidity is 47% is 15.9°C.
What is temperature?Temperature is a physical property of matter that quantitatively expresses hot and cold. It is measured with a thermometer, which may work through the bulk behavior of a thermometric material, detection of thermal radiation, or particle kinetic energy. Temperature is one of the principal parameters of thermodynamics. Temperature is also important in all fields of natural science, including physics, geology, chemistry, atmospheric sciences, medicine and biology.
The dewpoint is the temperature at which the air has reached its saturation point. At this point, water vapor will begin to condense into liquid droplets, forming dew.
To calculate the dewpoint when the air temperature is 28°C and the relative humidity is 47%, we use the formula:
Dewpoint (°C) = (Air Temperature (°C) * Relative Humidity) / 100 + (112 - Relative Humidity) / (x + 0.444)
Where x = (17.27 * Air Temperature (°C)) / ( Air Temperature (°C) + 237.3 ).
In this example, x = (17.27 * 28) / (28 + 237.3) = 0.39.
Plugging the values into the formula gives us:
Dewpoint (°C) = (28 * 47) / 100 + (112 - 47) / (0.39 + 0.444) = 15.9°C.
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An electron is accelerated through a distance of 15mm the work done on the electron is 1.2x10^-13J calculate the force on the electron
help pleaseee
A pipe, 30 cm long, is open at both ends. Which harmonic mode of the pipe resonates with a 1. 1 kHz source? Speed of sound in air = 330 m/s
The harmonic mode of the pipe that resonates with a 1. 1 kHz source is second harmonic.
The length of the pipe is given as 30 cm = 30/100 m = 0.3 m
Speed of light is known to be 330 m/s
Frequency of the source is given as 1.1 kHz = 1100 Hz
We know the relation between frequency, harmonic mode and length as,
vn = n v / 2 l
where,
n is number of harmonic mode
v is speed of light
l is length
vn is frequency of nth harmonic in an open pipe
Putting in the values, by making n as subject we have,
n = vn × 2 l/v = 1100 × 2 × 0.3/330 = 2
Thus, the harmonic mode of the pipe is calculated to be 2nd harmonic mode.
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An operating lamp draws a current of 0. 50 ampere. The amount of charge passing through the lamp in 10. Seconds is
please help asap
An operating lamp draws a current of 0. 50 ampere. the amount of charge passing through the lamp in 10 seconds is 5.00 Coulombs.
The amount of charge (Q) passing through a conductor is given by the product of current (I) and time (t):
Q = I * t
In this problem, we are given the following information:
The current passing through the lamp is I = 0.50 A
The time for which the current flows is t = 10 s
Using the above formula, we can calculate the amount of charge passing through the lamp as:
Q = I * t
Q = 0.50 A * 10 s
Q = 5.00 C
Thus, the amount of charge is 5.00 Coulombs.
The ampere, often shortened to amp, is the unit of electric current in the International System of Units.Ampere meter, commonly known as Ammeter is an electrical instrument used to measure electrical current in Amperes.
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A metal cube with a negative charge -Q_1−Q
1
minus, Q, start subscript, 1, end subscript is brought into contact with a cylinder of a negative charge -Q_2−Q
2
minus, Q, start subscript, 2, end subscript, where |Q_1| > |Q_2|∣Q
1
∣>∣Q
2
∣vertical bar, Q, start subscript, 1, end subscript, vertical bar, is greater than, vertical bar, Q, start subscript, 2, end subscript, vertical bar
The metal cube will transfer some of its charge to the cylinder, causing both objects to have a net positive charge. This is because when two objects with opposite charges are brought into contact, they will exchange charges until they have the same magnitude and the same sign.
In this case, the cube with a greater charge will transfer charge until both objects have a positive charge.When an object has more electrons than protons is called a negative charge. A positive charge occurs when an atom has more protons than electrons. Electrons have a negative charge and protons have a positive charge. The unit of charge is the coulomb(C). Like charges repel each other.
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Question:
A metal cube with a negative charge -Q_1−Q 1 minus, Q, start subscript, 1, end subscript is brought into contact with a cylinder of a negative charge -Q_2−Q 2 minus, Q, start subscript, 2, end subscript, where
|Q_1| > |Q_2|Q1∣>∣Q 2∣vertical bar, Q, start subscript, 1, end subscript, vertical bar, is greater than, vertical bar, Q, start subscript, 2, end subscript, vertical bar.
a luminous object moves along the optical axis of a concave spherical mirror. if the object approaches the mirror and its focal point, where does the image of the object move?
When a luminous object moves along the optical axis of a concave spherical mirror, if the object approaches the mirror and its focal point, the image of the object moves to infinity.
A concave mirror is a mirror with a reflecting surface that curves inward. A concave mirror, often known as a converging mirror, causes light to concentrate at a point known as the focal point when light rays hit it. The distance between the mirror's center and its focal point is known as the focal length. A concave mirror, for example, is used in searchlights to direct the beam of light, in telescopes to gather more light from faraway objects, and in headlights to concentrate the beam of light to illuminate the road ahead.
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In a physics lab experiment, a 6kg box is pushed across a flat table by a horizontal force F . What the magnitude of force F if the box is speeding up with a constant acceleration of 0.18ms^-2?
Answer:
The F's accelerated
where m is the box's mass, an is its acceleration, and F net is the net force.
The net force in this situation is equal to the applied force less any potential frictional forces. The net force is just the force applied to the box, assuming the table is smooth and there is no friction:
F net = F
Thus, the equation can be rewritten as follows:
F = m * a
Inputting the values provided yields:
F = 6 kg times 0.18 m/s2 to get 1.08 N
As a result, 1.08 N of force F is needed to accelerate a 6 kilogramme box at a constant rate of 0.18 m/s2.
Determine the power consumed by
The power consumed by 18 watts.
What is power ?
Work is done at a certain rate according to power. P=Wt. The watt (W), which is equal to 1 joule per second (1J/s), is the SI unit of power. Power is the rate at which energy is used up since power is the transfer of energy. For instance, a 60-W light bulb uses about 60 J of energy every second.
What is current ?
A flow of electrical charge carriers, most frequently electrons or atoms lacking in electrons, is referred to as current. I in capital letters is a typical symbol for current. Ampere, which is denoted by the letter A, is the common unit.I, which is derived from the French term intensité du courant, is the commonly used symbol for current (current intensity).
Therefore, power consumed by 18 watts.
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a football is kicked from the ground with a speed of 28.65 m/s at an angle of 37.18 degrees. what is the horizontal displacement of the football when it lands back on the ground?
49.87 m. is the horizontal displacement of the football when it lands back on the ground.
The horizontal displacement of the football can be determined using the formula
x = vxt,
where vx is the initial x-component of the velocity, and t is the time the football is in the air.
The initial x-component of the velocity (vx) can be calculated using the formula
vx = vcos(θ),
where v is the initial velocity, and θ is the angle of projection.
Therefore, vx = 28.65cos(37.18) = 25.36 m/s
The time the football is in the air (t) can be calculated using the formula t = 2vsin(θ)/g, where g is the acceleration due to gravity (9.81 m/s2).
Therefore, t = 2(28.65sin(37.18))/9.81 = 1.97 seconds
Finally, the horizontal displacement (x) can be calculated using the formula x = vxt.
Therefore, x = 25.36*1.97 = 49.87 m.
Therefore, the horizontal displacement of the football when it lands back on the ground is 49.87 m.
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