A 120-V rms signal at 60.0 Hz is applied across a 30.0-mH inductor and a resistor. If the rms value of the current in this circuit is 0.600 A, what is the value of the resistor

Know vernier caliper to analysis this image.

What is vernier caliper?

A tool for measuring linear dimensions Utilizing the measuring jaws, it is also used to determine the diameter of a sphere.

What is contained in the vernier caliper?

A large vernier caliper used to measure interior building dimensions.

The smallest reading on the vernier caliper's main scale is equal to its least count, which is equal to 1 mm divided by 10 divisions, or 0.1 mm. Length = (VSR * LC) + MSR is used to calculate the length of the object. MSR: the principal scale reading.

Along the vernier calliper's body is a substantial scale. Depending on how it is being used, the main scale's reading may be in centimeters or millimeters. According to SI units, 1 mm is the smallest main scale division.

Therefore, 1 mm is the smallest main scale division.

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Weight of the bag of sugar on the moon = 3.34 N

Weight of the bag of sugar on Venus = 18.09 N

Mass of the bag of sugar on the Earth = 2.04 kg

Mass of the bag of sugar on the moon = 2.04 kg

Mass of the bag of sugar on venus = 2.04 kg

The weight of the object on the earth

The acceleration due to gravity on the earth:

The mass of the bag of sugar on the earth = 2.04 kg

Note that the mass of the sugar is the same everywhere

Acceleration due to gravity on the moon is one-sixth of acceleration due to gravity on the earth

The weight of the bag of sugar on the moon is:

The acceleration due to gravity on the Venus is 0.904 times that of the earth

The weight of the bag of sugar on the venus is:

Mass of the bag of sugar on the Earth = 2.04 kg

Mass of the bag of sugar on the moon = 2.04 kg

Mass of the bag of sugar on venus = 2.04 kg

Chester will need to exert a force of 120 Newtons to accelerate the cart at a rate of 1.2 m/s^2, neglecting the mass of the cart and assuming there is no friction.

To determine the force exerted by Chester to accelerate the cart, we can utilize Newton's second law of motion, which states that the force acting on an object is equal to the product of its mass and acceleration. In this scenario, the mass of the cart itself is neglected, so the total mass to consider includes the two 50 kg sacks, resulting in a total mass of 100 kg.

Newton's second law can be expressed as F = m * a, where F is the force, m is the mass, and a is the acceleration. Substituting the given values, we have:

F = (100 kg) * (1.2 m/s^2) = 120 N

Therefore, Chester will need to exert a force of 120 Newtons to accelerate the cart at a rate of 1.2 m/s^2, neglecting the mass of the cart and assuming there is no friction. This force will provide the necessary push to overcome the inertia of the combined mass and achieve the desired acceleration. However, it is important to note that in real-world scenarios, additional factors such as friction and air resistance would need to be considered, which may require greater force exertion by Chester to achieve the desired acceleration.

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A. The period of vibration is the time it takes for one complete oscillation to occur. It is the reciprocal of the frequency. The period of vibration can be calculated using the formula:

Period = 1/Frequency

Given that the frequency of vibration is 5.0Hz, the period of vibration is:

Period = 1/5.0 s

Period = 0.2 s

B. If the cart's mass is doubled while the spring constant remains unchanged, the frequency of vibration will not change. This is because frequency is dependent on the spring constant and the mass of the cart and not on the amplitude of the oscillation.

C. If the spring constant doubles while the cart's mass remains the same, the frequency of vibration will also double. This is because frequency is directly proportional to the square root of the spring constant. Therefore, doubling the spring constant would lead to an increase in the frequency of vibration by a factor of square root of 2.

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Oscillatory frequency is the number of complete oscillations, cycles or vibrations of a system that occur in one second. It is measured in hertz (Hz) and is the reciprocal of the period of oscillation.

The oscillatory frequency (f) of a vibrating stretched wire in SHM is given by:

f = 1/T

where T is the period of vibration.

We know that 1000 cycles of vibration take 2.00 seconds. Therefore, the period of vibration (T) is:

T = time taken for 1000 cycles / number of cycles

T = 2.00 s / 1000 = 0.002 s

Substituting this value of T in the formula for frequency, we get:

f = 1/T = 1/0.002 s = 500 Hz

Therefore, the oscillatory frequency of the vibrating stretched wire is 500 Hz and its period is 0.002 s.

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The number of full oscillations, cycles, or vibrations of the a system that take place in a second is known as the oscillatory frequency. It is the inverse of the oscillation's period and expressed in hertz (Hz).

According to SHM, the oscillation rate (f) of a stretched wire in motion is given by:

f = 1/T

where T is the period of vibration.

We know that 1000 cycles of vibration take 2.00 seconds. Therefore, the period of vibration (T) is

T = time taken for 1000 cycles / number of cycles

T = 2.00 s / 1000 = 0.002 s

Substituting this value of T in the formula for frequency, we get:

f = 1/T = 1/0.002 s = 500 Hz

As a result, the vibrating strained wire has an oscillation frequency of 500 Hz and a period of 0.002 s.

It is the amount of oscillations in a single time unit, such as one second. A pendulum which takes 0.5 second to complete one full oscillation does have a frequency of one oscillation every 0.5 second, or two oscillations per second.

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Answer:

When a branch of a tree is shaken, some of the fruits may fall down. Why? Solution : The fruits fall down due to inertia of rest.

Answer:

I HOPE IT WILL HELP YOU A LOT....

Answer:

i think its  c

Explanation:

Answer:10

Explanation: if you look it has a 170 but some people have 70 if you have 170 then it will be 10 if you have a 70 then it will be 110 if I'm right then Yw

Answer:

10

Explanation:

i took the quiz xoxo

27.44 J of the potential energy of the book relative to the ground, when a 1.60 m tall person lifts 1.40 kg of the book off the ground.

A 1.60 m m tall person lifts a 1.40 kg book off the ground so it is 2.00 m m above the ground.

The potential energy of the book relative to the ground would be:

PE = mgh

Where PE is the potential energy,

m is the mass,

g is the acceleration due to gravity, and

h is the height

The acceleration due to gravity is constant and is 9.8 m/s²

mass m = 1.40 kg

Height h = 2.00 m

mgh = 1.40 × 9.8 × 2.00 = 27.44 J

Ans: The potential energy of the book relative to the ground is 27.44 J.

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In physics and engineering, a free frame diagram (FBD; also referred to as a force diagram) is a graphical instance used to visualize the applied forces, moments, and ensuing reactions on a frame in a given situation.

A free-body diagram is a sketch of an item of interest with all the surrounding items stripped away and all of the forces acting on the frame shown. The drawing of an unfastened-body diagram is a critical step in the solving of mechanics issues because it helps to visualize all the forces appearing on an unmarried object.

A free-body diagram is a graphic, dematerialized, symbolic representation of the body (structure, element, or segment of an element) wherein all connecting "pieces" have been removed. An FBD is a convenient method to model the structure, structural detail, or phase that is under scrutiny.

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Answer:

I had the same question a while ago !

Explanation:

Answer:

"Faster" is most likely the answer, I'm not 100% sure though

Answer:

im pretty sure its d

Explanation:

Answer:

In general, A/Ps win, but some synaptic signal transduction is much faster.

Explanation:

The answer depends on the synapse type:

A high-speed action potential (A/P) moves along an axon at ≈ 1 mm every 8.3 μs (≈ 120 m/s). Arriving at a terminal, the A/P voltage spike causes voltage-gated calcium channels to open, flooding the axon bouton with Ca ++

 ions, triggering neurotransmitter release.

Three different types of synapse:

In chemical synapses, transport vesicles dump their payloads into the synaptic gap (via exocytosis), neurotransmitter molecules diffuse across a ≈ 20–40 nm gap, to be sensed by receptor proteins on the post-synaptic cell membrane. Together, this may result in significant synaptic delays (milliseconds).

In electrical synapses, gap junctions open, allowing charged ions (an electric current) to flow cross, with a very short delay from axon bouton polarization ⟶ ion flow ⟶ post-synaptic cell reception.

In retina horizontal cells — said to be using the fastest known synapse type — there is no synaptic delay.

The speed of the east wind is 0.75 km/h.

To solve this problem, we can use the concept of vector addition.

Let's assume :

1) The speed of the boat is B km/h

2) The speed of the east wind is E km/h.

Given Information:

1) Time sailed by the boat = 120 minutes = 2 hours

2) Distance between the buoy and the harbor = 3 km.

Calculation for the wind speed.

Using the Pythagorean theorem, we can relate the boat's speed, the wind's speed, and the resulting velocity(v):

\(V^2=B^2+E^2\)

The Resultant velocity (V) can be calculated by dividing the distance traveled (3 km) by the time taken (2 hours):

\(V=\frac{3 \ km}{2 \ h}\)

\(= 1.5 \ km/hr\)

Now, \(Eastward \ wind \ velocity = Resultant \ velocity \times cos(60^\circ)\)

\(E=V\times cos(60)\)

\(E=0.75 \ km/h\)

Therefore, the speed of the east wind is 0.75 km/h.

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The wave's magnetic field amplitude is \(1.2 x 10^{-5} T.\)

The relationship between the electric and magnetic fields in an electromagnetic wave is given by the wave impedance, which is approximately equal to 377 ohms in free space.

The maximum rate of change of the electric field is given by:

Where c is the speed of light in free space, Bmax is the maximum amplitude of the magnetic field, and Emax is the maximum amplitude of the electric field. Solving for Bmax, we get:

Plugging in the given values, we get:

However, this is the peak value of the magnetic field. The root mean square (rms) value of the magnetic field is given by:

Plugging in the value for Bmax, we get:

Therefore, the wave's magnetic field amplitude is \(1.06 x 10^{-3} T\), or \(1.2 x 10^{-5\)T in scientific notation.

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A passenger with a mass of 80 kg is on a roller coaster doing a loop-the-loop. the weight of the passenger at the top of the loop of radius r = 15 m, assuming the coaster is moving at 22 m/s is  -1333.3 N

To determine the weight of the passenger at the top of the loop-the-loop, we need to consider the forces acting on the passenger. At the top of the loop, the passenger experiences both gravitational force and centripetal force.

The gravitational force acting on the passenger is given by the formula:

F_gravity = m * g

Where:

M is the mass of the passenger (80 kg)

G is the acceleration due to gravity (approximately 9.8 m/s^2)

The centripetal force required to keep the passenger moving in a circular path at the top of the loop is given by:

F_centripetal = m * (v^2 / r)

Where:

V is the velocity of the roller coaster (22 m/s)

R is the radius of the loop (15 m)

At the top of the loop, the weight of the passenger is the net force acting on them. So, we can subtract the centripetal force from the gravitational force:

Weight = F_gravity – F_centripetal

Weight = m * g – m * (v^2 / r)

Substituting the given values:

Weight = (80 kg) * (9.8 m/s^2) – (80 kg) * ((22 m/s)^2 / 15 m)

Weight ≈ 784 N – 2117.3 N

Weight ≈ -1333.3 N

The negative value indicates that the weight of the passenger at the top of the loop is directed downward, which is in the opposite direction to the gravitational force. However, this negative sign signifies that the passenger experiences a “feeling of weightlessness” at the top of the loop. In reality, the passenger’s apparent weight is zero due to the balance between gravitational and centripetal forces.

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Answer:

0.03s

Explanation:

Period of wave and its frequency are closely related. Period is the inverse of frequency.

The period of a wave is the time taken for a number of waves to pass through a particular point.

The frequency of a wave is the number of waves that passes through a point at a particular time.

So;

          f  = \(\frac{1}{T}\)  

f is the frequency

T is the period

   Period of the wave  = \(\frac{1}{30}\)   = 0.03s

Answer:

a

Explanation:

Answer:

(b) is correct

object moving towards you - higher frequency heard

object moving away from you - lower frequency heard

Answer:

i guess if the a glass gets to cold it will bust

Explanation:

a) The expression to determine the velocity of cart A is \(v_{A} = 2\cdot \sqrt{g\cdot H}\).

b) The expression to determine the velocity of cart B is

\(v_{B} =\sqrt{2\cdot g\cdot H }\).

c) The final velocity of the carts after the collision is \(v = \left(\frac{1-\sqrt{2}}{3} \right)\cdot \sqrt{g\cdot H}\).

d) The collision is inelastic since a part of the energy of the entire system is lost when they stick together.

In this question we shall apply principle of energy conservation and principle of linear momentum conservation to model an inelastic collision between two carts.

a) The combination of cart and ramp can be considered a conservative system as there are no non-conservative forces (i.e. friction), the final velocity of cart A (\(v_{A}\)) is related to the change in gravitational potential energy:

\(\frac{1}{2}\cdot M\cdot v_{A}^{2} = M\cdot g \cdot 2\cdot H\) (1)

Now we clear \(v_{A}\) and simplify the resulting expression:

\(v_{A} = 2\cdot \sqrt{g\cdot H}\)

The expression to determine the velocity of cart A is \(v_{A} = 2\cdot \sqrt{g\cdot H}\). \(\blacksquare\)

b) We apply the same approach used in part b) to find the final velocity:

\(\frac{1}{2}\cdot (2\cdot M) \cdot v_{B}^{2} = (2\cdot M)\cdot g \cdot H\) (2)

Now we clear \(v_{B}\) and simplify the resulting expression:

\(v_{B} =\sqrt{2\cdot g\cdot H }\)

The expression to determine the velocity of cart B is

\(v_{B} =\sqrt{2\cdot g\cdot H }\). \(\blacksquare\)

c) The final velocity (\(v\)system is determined by principle of linear momentum conservation:

\(3\cdot M\cdot v = M\cdot 2\cdot \sqrt{g\cdot H}-2\cdot M\cdot \sqrt{2\cdot g\cdot H}\)

\(3\cdot v = 2\cdot (1-\sqrt{2})\cdot \sqrt{g\cdot H}\)

\(v = \left(\frac{1-\sqrt{2}}{3} \right)\cdot \sqrt{g\cdot H}\)

The final velocity of the carts after the collision is \(v = \left(\frac{1-\sqrt{2}}{3} \right)\cdot \sqrt{g\cdot H}\). \(\blacksquare\)

d) The collision is inelastic since a part of the energy of the entire system is lost when they stick together. \(\blacksquare\)

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when preasure decreases, air expands and it cools

i hope you will accept my answer and please put a picture next for your next question

Answer:

O ice

Explanation:

Albedo refers to the proportion of the incident light that is reflected by a surface to the amount of incoming light.

Here, only ice will reflect even some amount of light. Pavement and dirt do not reflect at all, meanwhile, where glass reflects slightly.

Answer:

5.59 m/s2

Explanation:

F = 1900 N

m = 340 kg

F = ma

Therefore, a = 1900/340 = 5.59

Answer:

Heat

Explanation:

Because chemical energy is stored, it is a form of potential energy. When a chemical reaction takes place, the stored chemical energy is released. Heat is often produced as a by-product of a chemical reaction – this is called an exothermic reaction.

Hope this helped.

Explanation:

Answer Should be B.

Its in Form of Heat.

Often they are not accurate because scientists may not have all the data.

A scientific model is a bodily and/or mathematical and/or conceptual representation of a system of ideas, occasions, or procedures. Scientists are seeking to perceive and recognize patterns in our world by drawing on their scientific understanding to offer motives that enable the styles to be anticipated.

Models are beneficial gear in studying science which may be used to enhance reasons, generate discussion, make predictions, provide visual representations of summary standards and generate intellectual models.

Benefits of modeling and simulation :

* Can be more secure and less expensive than the real international.

* Able to test a product or device that works earlier than building it.

* Can use it to discover unexpected troubles.

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(a) The speed of the block when it leaves the spring is 3.87 m/s.

(b) The speed of the block the moment it first become airborne is 2.28 m/s.

(c) The speed of the block when it hits the ground is 5.88 m/s.

(d) The largest mass of the block that will still allow it to reach the ground is 459.3 g.

The speed of the block when it leaves the spring is calculated by applying the principle of conservation of energy.

¹/₂mv² = ¹/₂kx²

mv² = kx²

v² = kx²/m

v = √(kx²/m)

where;

v = √(200 x 0.15²/0.3)

v = 3.87 m/s

The moment the block moves upwards, gravity will set in. The speed of the block the moment it first become airborne is calculated as follows;

v² = u² - 2gh

where;

v² = 3.87² - 2(9.8 x 0.5)

v² =  5.18

v = √5.18

v = 2.28 m/s

The speed of the block when it hits the ground is calculated as follows;

v² = u² + 2gh

where;

v² = 2.28² + 2(9.8)(1.5)

v² = 34.598

v = √34.598

v = 5.88 m/s

The maximum potential energy of the spring is calculated as;

U(max) = ¹/₂kx²

U(max) = ¹/₂ x 200 x 0.15²

U(max) = 2.25 J

The minimum speed required to cross the 1.5 m of the ramp from a height of 1 m.

v = √2gΔh

v = √(2 x 9.8 x 0.5)

v = 3.13 m/s

Apply the principle of conservation of energy and determine the largest mass required to cross the 1.5 m ramp from a height of 1 m.

¹/₂mv² = U(max)

m = 2U/v²

m = (2 x 2.25) / (3.13²)

m = 0.4593 kg

m = 459.3 g

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