to check the temperature leads of an electronic thermometer for high temperature, they should be .

Answer:

total current

Explanation:

In a parallel circuit, the current leaves the power source and then splits into multiple branches that have individual components. Each branch operates independently of the other branches, and the current flows through each component and then recombines and enters back into the power source.

Because each branch is independent, the current flowing in each branch can be different, depending on the resistance of the components in that branch. However, the total current entering the circuit is equal to the sum of the currents in each branch, as long as the circuit is properly wired.

So the total current that leaves and reenters the power source in a parallel circuit is equal to the sum of the currents in each branch, and this is known as the total or overall current of the circuit.

(a) The magnitude of the tangential acceleration of a point on the edge of the turntable is 0.290 m/s².

(b) The time it takes the turntable to come to rest is 48.3 s.

(c) The number of revolutions the turntable makes before stopping is 4.79 revolutions.

(d) The magnitude of the torque exerted on the turntable is 0.116 Nm.

(e) The magnitude of the frictional force is 0.116 N.

(f) The magnitude of the initial angular momentum of the turntable is 0.056 kg m²/s.

(a) The tangential acceleration can be calculated using the formula a = α × r, where α is the angular acceleration and r is the radius of the turntable. Given α = -0.580 rad/s² and r = 0.20 m (half of the diameter), we find a = 0.290 m/s².

(b) The time it takes for the turntable to come to rest can be determined using the equation vf = vi + at, where vf is the final velocity (zero in this case), vi is the initial velocity (28.0 rad/s), a is the acceleration (-0.580 rad/s²), and t is the time. Rearranging the equation, we have t = (vf - vi) / a = -28.0 rad/s / (-0.580 rad/s²) = 48.3 s.

(c) The number of revolutions the turntable makes before stopping can be found using the equation θ = ωi × t + 0.5 × α × t², where θ is the angle in radians, ωi is the initial angular velocity, t is the time, and α is the angular acceleration. Since ωi = 28.0 rad/s, α = -0.580 rad/s², and t = 48.3 s, we can calculate θ = 4.79 revolutions.

(d) The magnitude of the torque exerted on the turntable can be determined using the equation τ = I × α, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. The moment of inertia for a disk rotating about its axis is given by I = (1/2) × m × r², where m is the mass of the disk and r is its radius. Substituting the given values, we find τ = 0.116 Nm.

(e) The magnitude of the frictional force can be calculated using the equation f = m × a, where f is the force, m is the mass, and a is the acceleration. Substituting the given values, we find f = 0.116 N.

(f) The magnitude of the initial angular momentum of the turntable can be calculated using the equation L = I × ω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. Substituting the given values, we find L = 0.056 kg m²/s.

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The most important advantage of observing living things in a closed setting is to observe the causes and effects of a given variable.

The expression 'cause and effect in experimentation' refers to which are the dependent variables whose changes depend on the independent variable, which only can be tested in a closed system during experimental procedures.

In conclusion, the most important advantage of observing living things in a closed setting is to observe the causes and effects of a given variable.

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The value of the resistance of R₂, given that the total resistance in the circuit is 15 ohms is 5 ohms (option D)

The following data were obtained from the question:

Total resistance in series connection is given by the following formula:

R = R₁ + R₂ + R₃ + ... + Rₙ

Inputting the various parameters given, we can obtain the resistance of R₂ as follow:

R = R₁ + R₂

15 = 10 + R₂

Collect like terms

R₂ = 15 - 10

R₂ = 5 ohms

Thus, we can conclude from the above calculation that the resistance of R₂ is 5 ohms (option D)

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Complete question:

See attached photo

The time of flight is obtained as 25.9 seconds

The time of flight is the time taken to move the object that have been projected along the parabolic path. In this case, we have a cannonball that have been fired  from level ground with a velocity of 240.0 m/s at an angle of 32 degrees.

We know that from the question

T = 2usinθ/g

T = time of flight

u = initial velocity

θ = angle of projection

g = acceleration

T = 2 * 240.0 m/s * sin 32 degrees/9.8 m/s^2

T = 25.9 seconds

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The velocity of the ion released from rest and accelerated through a potential difference of 152V is 6.34 × 10^5m/s.

The electric potential difference is a scalar quantity that measures the energy required per unit of electric charge to transfer the charge from one point to another. The electric potential difference between two points in an electric circuit determines the direction and magnitude of the electric current that flows between those two points. A lithium-ion containing three protons and four neutrons has a mass of 1.16 × 10-26 kg. The ion is released from rest and accelerates as it moves through a potential difference of 152 V.

The change in electric potential energy of an object is equal to the product of the charge and the potential difference across two points. The formula to calculate the velocity of the ion released from rest and accelerated through a potential difference of 152V is:

v = √(2qV/m) where q is the charge of the ion, V is the potential difference, and m is the mass of the ion.

Substituting the values in the formula, we get:

v = √(2 × 1.6 × 10-19 C × 152 V/1.16 × 10-26 kg)v = 6.34 × 10^5m/s

Therefore, the velocity of the ion released from rest and accelerated through a potential difference of 152V is 6.34 × 10^5m/s.

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The maximum current delivered by the AC source connected across the 3.70 µF capacitor is approximately 0.0369 amperes.

To calculate the maximum current delivered by an AC source connected across a capacitor, you need to use the below formula.

I = C × δVmax × 2πf.

Where:

I = Maximum current (in amperes)

C = Capacitance (in farads)

δVmax = Maximum voltage (in volts)

f = Frequency (in hertz)

Given:

δVmax = 46.0 V

f = 80.0 Hz

C = 3.70 µF = 3.70 × 10⁻⁶ F

Plugging in the values into the formula:

I = (3.70 × 10⁻⁶ F) × (46.0 V) × (2π × 80.0 Hz)

Calculating:

I = (3.70 × 10⁻⁶ F) × (46.0 V) × (502.65)

I = 0.0369 A

Therefore, the maximum current delivered by the AC source connected across the 3.70 µF capacitor is approximately 0.0369 amperes.

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Answer:

wow

Explanation:

Answer:

I can't see the picture, can you right down the same thing that is the picture in the comments so I can help you.

Explanation:

Thx.

The answer is A.

The answer is A.

The answer is A.

Answer:

4.8 A

Explanation:

Voltage = current x resistance

(V = IR)

12 = I (0.5 +2)

I = 4.8 A

Answer:

Can't see the pic

Explanation:

Answer: straight

Explanation:

If an object moves at a constant speed, the line on a distance/time graph would be straight.

Since the object is moving at a speed which is constant, it'll have a straight line because the direction of the object is not changed.

1. The total change in dB between points X and Z is a net loss of 6 dB due to a boost of 10 dB at point Y and losses of 8 dB each between X and Y and between Y and Z.

2. The power of the signal at point Z, starting with 100 Watts at point X, is approximately 79.43 Watts after the net loss of 6 dB.

3. To ensure a minimum power of 50 Watts at point Z, the initial strength of the signal should be increased to approximately 398.11 Watts.

4. Shannon's theorem calculates a maximum data transfer rate of approximately 66,419 bps for a signal with a bandwidth of 10,000 Hz, signal power of 2 Watts, and noise power of 0.002 Watts.

5. To increase the maximum data transfer rate to 200 kbps, options include increasing the bandwidth or reducing the noise level.

Part 1: The total change in dB between points X and Z is 10 dB (boost at point Y) minus 16 dB (8 dB loss between X and Y, and another 8 dB loss between Y and Z), resulting in a net loss of 6 dB.

Part 2: The signal starts at point X with a power of 100 Watts and experiences a net loss of 6 dB, the power of the signal at point Z can be calculated using the logarithmic formula:

Power (Z) = Power (X) * 10^(dB/10)

Power (Z) = 100 Watts * 10^(-6/10) ≈ 79.43 Watts

The power of the signal at point Z will be approximately 79.43 Watts.

Part 3: To ensure that the signal arrives at point Z with a minimum power of 50 Watts, the initial strength of the signal should be increased. Using the same logarithmic formula:

Power (Z) = Power (X) * 10^(dB/10)

50 Watts = Power (X) * 10^(-6/10)

Power (X) = 50 Watts / 10^(-6/10) ≈ 398.11 Watts

The initial strength of the signal should be increased to approximately 398.11 Watts.

Part 1: The maximum data transfer rate (in bits per second) can be calculated using Shannon's theorem formula:

Maximum Data Transfer Rate = Bandwidth * log2(1 + (Signal Power / Noise Power))

Bandwidth = 10,000 Hz

Signal Power = 2 Watts

Noise Power = 0.002 Watts

Maximum Data Transfer Rate = 10,000 * log2(1 + (2 / 0.002)) ≈ 66,419 bps (bits per second)

The maximum data transfer rate is approximately 66,419 bps.

Part 2: To increase the maximum data transfer rate to 200 kbps (kilobits per second), two possible ways are:

1. Increase the bandwidth: Increasing the available bandwidth will allow for a higher data transfer rate. To achieve 200 kbps, the bandwidth needs to be increased accordingly.

2. Decrease the noise level: Reducing the noise level will improve the signal-to-noise ratio, allowing for a higher data transfer rate. By reducing the noise power, the maximum data transfer rate can be increased.

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Answer:

t = 0.553

Explanation:

The mass has nothing to do with an object falling.

Givens

d = 1.5 m

a = 9.8

vi = 0

t = ?

Formula

d = vi * t + 1/2 a t^2

Solution

1.5 = 0 + 1/2 9.81 * t^2

1.5 = 4.905 t^2

1.5/4.905 = t^2

0.3058 = t^2

sqrt(t^2) = sqrt(0.3058)

t = 0.553

About 1/2 a second which is really surprising wouldn't you think?

Answer:

Explanation:

Initial velocity is 0. In the equation v = v0+at where v0 is the initial velocity of 0, we only have to fill in -9.8 for a and 2 for t to get the velocity after 2 seconds -19.6 m/s; after 5 seconds, when it hits the ground, a = -9.8 and t = 5 to give a velocity of -49 m/s. Gravity pulls down everything at the same rate, it doesn't matter whether we drop a feather or an elephant from the window!

The sum of gravitational and elastic potential energies of the system increases as the mass is pulled down in a spring-mass system.

When the mass is pulled down in a spring-mass system, two potential energies are at play: gravitational potential energy and elastic potential energy. Gravitational potential energy depends on the height of the mass above a reference point, while elastic potential energy is related to the displacement of the spring from its equilibrium position.

As the mass is pulled down, the gravitational potential energy increases because the height decreases. Simultaneously, the elastic potential energy increases because the spring is stretched or compressed further from its equilibrium position. Therefore, the sum of gravitational and elastic potential energies of the system increases when the mass is pulled down in a spring-mass system.

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Answer:

2.3.1

Explanation:

Answer:

They store floodwater

Explanation:

Assuming the raindrop was stationary relative to the vertical distance to the ground at the start:

D=0.5at where d is distance, a is acceleration and t is time

D is 300 meters

a is 9.8 meter/sec squared

Solve for t in seconds

t = 61.2 seconds

v=at where v is velocity

a is 9.8 meters per second squared

t is 61.2 seconds

solve for v

v = 600 meters per second.

If it had an initial vertical velocity (v0) at the start :

d= 0.5at+v0t

and

v=at+v0

Faster-moving air above an airplane wing reduces pressure on the wing, creating lift due to the pressure difference between the upper and lower surfaces of the wing.

This phenomenon is explained by Bernoulli's principle, which states that as the velocity of a fluid (in this case, air) increases, its pressure decreases. When an airplane wing moves through the air, its shape causes the air to move faster over the top surface than the bottom surface. This leads to a decrease in pressure above the wing and an increase in pressure below the wing.

The pressure difference creates an upward force called lift, which counteracts the weight of the airplane, allowing it to stay airborne. The greater the speed of the airplane, the stronger the lift force generated.

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Answer: Exothermic

Explanation:

An exothermic reaction causes an increase in temperature. This is because the energy released when the bonds are broken among the reactants is greater than than the energy absorbed when new bonds are formed among the products. This leftover energy gets passed along to the surroundings, where it produces a measurable increase in heat.

The rotational kinetic of the two rotating solid cylinders is 44.2 (r₁²+r₂²). The result is obtained by inserting total inertia value in rotational kinetic equation.

The moment of inertia of a solid cylinder can be expressed as

\(I = \frac{1}{2} mr^{2}\)

Where

If there are two of more cylinders, the total moment of inertia would be

\(I = I_{1} + I_{2} + ...\)

The kinetic energy of a rotating body can be expressed as

\(E_{rot} = \frac{1}{2} I \omega^{2}\)
Where

If given

What is the rotational kinetic energy?

We first calculate the moment of inertia of the two cylinders.

\(I = \frac{1}{2} m_{1} r_{1}^{2} + \frac{1}{2} m_{2}r_{2}^{2}\)

\(I = \frac{1}{2} m r_{1}^{2} + \frac{1}{2} mr_{2}^{2}\)

\(I = \frac{1}{2} m (r_{1}^{2} + r_{2}^{2})\)

\(I = \frac{1}{2} \times 1,25 (r_{1}^{2} + r_{2}^{2})\)

\(I = 0,625 (r_{1}^{2} + r_{2}^{2})\)

Then, the rotational kinetic energy is

\(E_{rot} = \frac{1}{2} I \omega^{2}\)

\(E_{rot} = \frac{1}{2} \times 0,625 (r_{1}^{2} + r_{2}^{2}) \times \omega^{2}\)

\(E_{rot} = \frac{1}{2} \times 0,625 (r_{1}^{2} + r_{2}^{2}) \times 235^{2}\)

\(E_{rot} = 117.5 \times 0.625 (r_{1}^{2} + r_{2}^{2}) \times 235\)

\(E_{rot} = 44.2 (r_{1}^{2} + r_{2}^{2}) Joule\)

Hence, the rotational kinetic energy of the two rotating solid cylinders is 44.2 (r₁²+r₂²).

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Technician B is correct. A typical MacPherson strut suspension system uses only one control arm and one ball joint per side.

The MacPherson strut is a common type of suspension system used in many vehicles. It consists of a single control arm and a strut assembly that includes a shock absorber and a coil spring. The control arm is connected to the steering knuckle, which holds the wheel, through a single ball joint. This setup allows for vertical movement of the wheel and controls its alignment.

When the vehicle turns, the steering system applies rotational force to the control arm, which pivots around the ball joint. This movement allows the wheel to turn while maintaining suspension function. However, it's important to note that the entire strut assembly itself does not rotate during normal steering maneuvers.

Technician A's statement is incorrect because the rotation of the front wheels is primarily achieved through the steering system and the control arm's movement, not by rotating the entire strut assembly. Technician B's statement is accurate, as a typical MacPherson strut suspension system does indeed use only one control arm and one ball joint per side to support and control the front wheels.

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Answer:

An article written by a scientist .

Explanation:

Primary sources information are those that contain first  hand information probably from the place of the activities or by the original contact of the information (source ) .

So primary information is normally contained in artifact , journals , letter , dissertations manuscripts  , videos and audio recordings .

The fact that the article is written by the scientist means he is the one who has the original information hence he is the first person to access the information , meaning he is the original owner .

The position of the spaceship when the engine shuts off is \($\vec{r} = (600 \hat{i}-400 \hat{j}+200,882,200 \hat{k}) \times 10^3 km$\).

We can use the equations of motion to determine the final position of the spaceship when the engine shuts off. We know that the position of an object can be expressed as a function of time using the following equation:

\($\vec{r} = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2$\)

where \($\vec{r}$\) is the position of the object, \($\vec{r}_0$\) is the initial position of the object, \($\vec{v}_0$\) is the initial velocity of the object, \($\vec{a}$\) is the acceleration of the object, and \($t$\) is the time elapsed.

In this case, the initial position of the spaceship is \($\hat{r}=(600 \hat{i}-400 \hat{j}+200 \hat{k}) \times 10^3 km$\), the initial velocity is \($\vec{v}=9500 \hat{\imath} m / s$\), and the acceleration is \($\vec{a}=(40 \hat{\imath}-20 \hat{k}) m / s ^2$\). We need to first convert the time elapsed to seconds, since all other units are expressed in SI units. 35 minutes is equal to \($35\cdot60=2100$\) seconds.

Plugging in the values into the equation, we get:

\($\vec{r} = (600 \hat{i}-400 \hat{j}+200 \hat{k}) \times 10^3 km + 9500 \hat{\imath} \cdot 2100 s + \frac{1}{2} (40 \hat{\imath}-20 \hat{k}) \cdot (2100)^2 s^2$\)

Expanding the equation, we get:

\($\vec{r} = (600 \hat{i}-400 \hat{j}+200 \hat{k}) \times 10^3 km + 19,95,000 \hat{\imath} + 20 \cdot (2100)^2 \hat{k}$\)

\($\vec{r} = (600 \hat{i}-400 \hat{j}+200 \hat{k}) \times 10^3 km + 19,95,000 \hat{\imath} + (20 \cdot 4410000) \hat{k}$\)

\($\vec{r} = (600 \hat{i}-400 \hat{j}+200 \hat{k}) \times 10^3 km + 19,95,000 \hat{\imath} + 882,200,000 \hat{k}$\)

So, the position of the spaceship when the engine shuts off is \($\vec{r} = (600 \hat{i}-400 \hat{j}+200,882,200 \hat{k}) \times 10^3 km$.\)

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Complete question:

A spaceship maneuvering near Planet Zeta is located at \(\hat{r}=(600 \hat{i}-400 \hat{j}+200 \hat{k}) \times 10^3 km\), relative to the planet, and traveling at \(\vec{v}=9500 \hat{\imath} m / s\). It turns on its thruster engine and accelerates with \($\vec{a}=(40 \hat{\imath}-20 \hat{k}) m / s ^2$\) for 35 min. What is the spaceship's position when the engine shuts off? Give your answer as a position vector measured in km.

Answer:

(a) Both the charges are positive or negative.

(b) Teh value of each charge is 1.53 x 10^-5 C.

Explanation:

Spring constant, K = 340 N/m

Natural length, L = 0.4 m

stretch, y = 0.033 m

(a) Let the charge on each sphere is q and they repel each other so the nature of charge of either sphere may be both positive or both negative.  

(b) The electrostatic force is balanced by the spring force.

\(\frac{kq^2}{(L + y)^2}=Ky\\\\\\\frac{9\times 10^9 q^2}{(0.4 +0.033)^2} = 340\times0.033\\\\q= 1.53\times 10^{-5} C\)

Answer:

The sun is located in the Milky Way Galaxy.

The transducer took 0.077 seconds to detect the reflected waves from the metal fragment.

To calculate the time taken by the transducer to detect the reflected waves, we can use the formula: time = distance/speed. Here, the distance is twice the depth of the metal fragment in the muscle tissue, which is 10 cm or 0.1 m.

The speed of sound waves in muscle tissue is 1,300 m/s. So, time = 0.1/1300 = 0.000077 s or 0.077 ms.

This means it took the transducer 0.077 seconds to detect the reflected waves from the metal fragment after they were first emitted. This time delay is used by the ultrasound machine to determine the depth and location of structures within the body.

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Answer:

Pulse & Heart Rate. Your heart rate is the number of times each minute that your heart beats, which is normally between 60 and 100 times per minute for adults. Your pulse is a way you can feel each time your heart beats

Explanation: