For how long must a current of 13 A flow to transfer:

a) 32.5 C of charge
b) 62.4 C of charge
c) 312 C of charge
d) 780 C of charge
e) 1612 C of charge
f) 812.5 C of charge
g) 468 kC of charge
h) 187.2 kC of charge
i) 163.8 kC of charge​

Answer:

The first one is: His weight on the Earth before take-off and the weight after take-off back on Earth once he gets back should be recorded as his Independent variable and his dependent variable.

The second one is: If he gained the weight back that he had lost while on the trip then you should disregard them unless that was the weight he was when he weighed himself after he got back.

The Third one is: The mass of an object is the amount of matter it contains, regardless of its volume or any forces acting on it. … Gravity is a force that attracts objects toward the Earth. The weight of the object is defined as the force caused by gravity on a mass.

Explanation:

I took  the quiz earlier. Hope this Helps you.

Sunspots at the equator take 26.9 days to move once around the sun. Therefore, sunspots A and B, which are close to the poles, would take longer to complete a revolution than sunspot C, which is located on the equator.

Sunspot C, which is at the equator, has a faster rotation speed than sunspots A and B, which are close to the poles. Sunspots are the regions on the sun's surface that appear dark and have a lower temperature than the surrounding areas. They are caused by magnetic activity in the sun's interior.

Sunspots move with the sun's rotation, but they do not move across the sky, from east to west, as the sun and moon do. Instead, sunspots appear at the eastern edge of the sun's disk, move across the sun, and then disappear from the western edge of the sun's disk, due to the sun's rotation. The sun's equator rotates faster than its poles, resulting in a shorter rotation period for sunspots at the equator.

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The 3 Ways to cooping with work related stress is to adopt healthy habits, seek social support, and engage in activities that promote relaxation.

The following are three ways to cope with work-related stress:

Exercise- Exercise is a simple yet effective way to reduce stress. When you exercise, your body releases endorphins that are natural mood boosters. Exercise helps to reduce the level of cortisol, which is a stress hormone. The best exercises to do when stressed include yoga, aerobics, walking, jogging, cycling, or dancing.

Engage in relaxation activities-  Engaging in relaxation activities such as meditation, deep breathing, or progressive muscle relaxation helps to relax your mind and body. Deep breathing helps to reduce muscle tension, lower blood pressure and reduce the level of cortisol in the body. Progressive muscle relaxation involves tensing and relaxing muscle groups in the body, one at a time. This technique helps to reduce muscle tension and improve relaxation.

Social support- Social support from family, friends, or colleagues can be a great way to cope with work-related stress. Talking to someone about your problems can help you to gain a different perspective on your situation and feel less isolated. Talking to a colleague can also help to create a supportive work environment. It is essential to identify a trusted confidant who can listen and provide support when you are feeling overwhelmed.

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Answer:What is the acceleration of a car that moves from rest to 15.0 m/s in 10.0 s? Vi=0, vf= 15.0 m/s,t=10.0s, a=? a= vf =vi/tA=15.0m/s-0m/s/10.0s = 15.0s/10.0s m/s*1/s =1.50 m/s^2 11.

Explanation:

Answer:

M g - T = NB         vertical force on box B

T - m g = NA         vertical force on box A

T must be uniform (same) throughout length of string

(M - m) g = NB + NA = (M + m) a        since F accelerates M and m

Note: one can enclose the entire system with no external forces

No net vertical force or horizontal force is present - the net vertical force is A + B + C - masses of these objects * g

One would write a = (M - m) * g / (M + m)

The masses and tensions influence the accelerations of the boxes while considering friction and normal forces on block C.

When box A and B are released, their accelerations can be calculated using Newton's second law, (F = ma), where (F) is the net force and (a) is the acceleration.

The net force is the difference between the gravitational force (mg) and the tension in the rope (T).

For box A:

Net force: \(\rm \(F_{\text{net,A}} = T - mg\)\)

Using \(\rm \(T = m \cdot a_A\)\), where \(\rm \(a_A\)\) is the acceleration of box A:

\(\rm \(ma_A = T - mg\)\\\rm \(a_A = \frac{T}{m} - g\)\)

For box B:

Net force: \(\rm \(F_{\text{net,B}} = mg - T\)\)

Using \(\rm \(T = M \cdot a_B\)\), where \(\rm \(a_B\)\) is the acceleration of box B:

\(\rm \(Ma_B = mg - T\)\\\rm \(a_B = g - \frac{T}{M}\)\)

Now, considering friction and the normal force between block C and the ground:

The friction force \(\rm (\(F_{\text{friction}}\))\) can be calculated using the equation \(\rm \(F_{\text{friction}} = \mu N\)\), where \(\rm \(\mu\)\) is the coefficient of friction and (N) is the normal force.

The normal force (N) is equal in magnitude to the weight of block C \((mg_{C})\) since it's not accelerating vertically:

\(\rm \(N = mg_C\)\)

To summarize:

Acceleration of box A: \(\rm \(a_A = \frac{T}{m} - g\)\)

Acceleration of box B: \(\rm \(a_B = g - \frac{T}{M}\)\)

Friction force on block C: \(\rm \(F_{\text{friction}} = \mu mg_C\)\)

Normal force on block C: \(\rm \(N = mg_C\)\)

The masses and tensions influence the accelerations of the boxes while considering friction and normal forces on block C.

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Answer:

d

Explanation:

because it also depends upon accleration acting upon it

Answer:

a spiral

Explanation:

It's the largest member of the Local Group of galaxies

Answer:

Barred spiral galaxy

Based on the information provided, we can determine that the object's acceleration is constant and equal to zero.

This is because the object is traveling the same distance in each second, indicating that its speed is constant. Acceleration is defined as the rate at which an object changes its velocity, and since the velocity of the object is not changing (it's constant), its acceleration is zero.

It's important to note that even though the object's acceleration is zero, it is still moving. This is because acceleration is only one aspect of an object's motion, and velocity and displacement are also important factors to consider. In this case, the object's displacement (total distance traveled) is 24 meters, and its velocity is constant.

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Answer:

C. a rolling bowling ball

I just answered this question on my quiz.  

Colle's fracture is an injury that occurs when a person falls and lands on an outstretched hand. The fracture affects the radius, one of the bones that makes up the forearm. The radial styloid process is the end of the radius bone closest to the thumb. It helps to keep the wrist stable when the hand moves.

This fracture is also known as a distal radius fracture or wrist fracture.A skateboarder, who hits the concrete with outstretched arms to stop their fall can break the radial styloid process resulting in Colle's fracture. Therefore, the given statement is true. Here is a detailed explanation for the same:Colles' fracture is a type of wrist fracture. It occurs when the bone that connects the wrist to the thumb (the radius) breaks and moves toward the back of the wrist. Colles' fracture usually occurs when a person falls and lands on an outstretched hand. It is also common in older adults with osteoporosis, a condition that weakens the bones. In the case of a skateboarder, who hits the concrete with outstretched arms to stop their fall can break the radial styloid process. This will result in a Colle's fracture. In addition, Colles' fracture may be classified as:Extra-articular, which means the fracture does not extend into the wrist joint. This type of fracture is more common.Intra-articular, which means the fracture extends into the wrist joint. This type of fracture is less common but is more severe.

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Answer:

The dilation of time.

The falling of objects.

The changing of paths of light.

Explanation:

I have explained in the image attached below.

From the explanation, the correct ones are;

The dilation of time.

The falling of objects.

The changing of paths of light.

Answer: 268.80 W/m^2

Explanation:

Intensity = Power/Area

first convert radius to meters

1.5mm= 1.5x10^-3m

area= pi (1.5x10^-3)^2

= 7.069m^2

convert power to W

1.9mW= 1.9x10^-3W

plug into intensity formula

(1.9x10^-3)/(7.069) = 268.80 W/m^2

Answer:

\(104\; {\rm g}\), assuming that the meter stick is uniform with the center of mass precisely at the \(50\; {\rm cm}\) mark.

Explanation:

Refer to the diagram attached. The meter stick could be considered as a lever. The string at the \(30\; {\rm cm}\) mark would then act as the fulcrum of this lever.

The \(m_{A} = 20\; {\rm g}\) mass at the \(80\; {\rm cm}\) mark is at a distance of \(r_{A} = 50\; {\rm cm}\) to the right of the fulcrum at \(30\; {\rm cm}\).

The weight of the \(80\; {\rm g}\) meter stick acts like a weight of \(m_{B} = 80\; {\rm g}\) attached to the center of mass of this meter stick. Under the assumptions, this center of mass of this meter stick would be at the \(50\; {\rm cm}\) mark, which is \(r_{B} = 20\; {\rm cm}\) to the right of the fulcrum at \(30\; {\rm cm}\).

Let \(m_{C}\) be the mass attached to the meter stick at the \(5\; {\rm cm}\) mark. This mass would be at a distance of \(r_{C} = 25\; {\rm cm}\) to the left of the fulcrum at \(30\; {\rm cm}\).

At equilibrium:

\(\begin{aligned} & m_{C}\, r_{C} && (\text{mass on the left of fulcrum})\\ &= m_{A}\, r_{A} + m_{B} \, r_{B} && (\text{mass on the right of fulcrum})\end{aligned}\).

Solve for \(m_{C}\), the unknown mass attached to the meter stick at the \(5\; {\rm cm}\) mark:

\(\begin{aligned}m_{C} &= \frac{m_{A}\, r_{A} + m_{B}\, r_{B}}{r_{C}} \\ &= \frac{20\; {\rm g} \times 50\; {\rm cm} + 80\; {\rm g} \times 20\; {\rm cm}}{25\; {\rm cm}} \\ &= 104\; {\rm g}\end{aligned}\).

3 is false 2 is true and the rest true

Answer;

True, False, False, True, True

Answer:

false

Explanation:

A resistor only reduces or regularizes the flow of current, not stop it totally.

The force exerted on the electron has the magnitude of  \(8\times10^{-14} N\). The force on an electric charge q in an electric field E is given by the formula F = qE.

Here, the charge on the electron is \(q = 1.60\times 10^{-19} C\), and the electric field is \(E = 5\times10^5 N/C\).

the electron's Magnitude of force is given by :\(F = (1.60\times10^{-19} C)(5\times10^5 N/C) = 8\times10^{-14} N\)

This force is directed opposite to the direction of the electric field because the electron has a negative charge. Since all other forces acting on the electron are negligible in comparison to the electric field force, we can assume that the electron moves with a constant acceleration given by F = ma. The acceleration of the electron can be determined by using the formula a = F/m.

Substituting the values, we get

\(a = \frac{8\times10^{-14} N)}{(9.11\times10^{-31} kg)} = 8.78\times10^{16} m/s^2.\)

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The slope of a velocity vs time scatterplot is the most appropriate method for estimating the value of the constant acceleration for an object that is constantly accelerating.

To estimate the value of the constant acceleration for an object that is constantly accelerating, we can use the slope of a velocity vs time scatterplot. This is because acceleration is defined as the rate of change of velocity over time, and the slope of a velocity vs time graph represents this rate of change.

By calculating the slope of the line of best fit for the velocity vs time data, we can determine the value of the constant acceleration. The other options listed, such as using the slope of a position vs time scatterplot or the peak of a histogram of accelerations or velocities, would not give us an accurate estimate of the constant acceleration. This is because acceleration is a rate of change, and is therefore best determined by examining changes in velocity over time.

To estimate the value of constant acceleration for an object with given velocity and time data, you can use the slope of a velocity vs. time scatterplot. This is because the relationship between acceleration, velocity, and time in a constantly accelerating object follows the equation a = (v - u) / t, where a is acceleration, v is final velocity, u is initial velocity, and t is time. The slope of the velocity vs. time scatterplot represents the rate of change of velocity over time, which is the acceleration.

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The time takes for the green ball to reach the ground is equal to 0.78 s. As both balls are thrown from the same height they will take the same time to reach the ground irrespective of their velocity.

The equations of motion can be described as the equations which establish the relationship between the time, velocity, acceleration, and displacement of a moving object.

The equations of motions as the mathematical expressions:

\(v = u +at\\S = ut +(1/2)at^2\\v^2-u^2= 2aS\)

Given, the height from which the ball is thrown, h = 3

From the 2nd equation of motion, calculate the time taken by the ball reach to the ground:

H = ut + (1/2)gt²

3 = 0 + (1/2)× 9.8×t²

t = 0.78 s

As the time is independent of the velocity of a ball so both balls take the same time to reach the ground as they are thrown from the same height.

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Dark energy has been hypothesized to exist in order to explain observations suggesting that the expansion of the Universe is accelerating.

This conclusion is based on observations of distant supernovae, which indicate that the rate of expansion of the Universe is increasing over time. This acceleration cannot be explained by the known laws of physics, and thus, the concept of dark energy was proposed as a possible explanation.

While dark energy may also have an effect on the high orbital speeds of stars far from the center of our galaxy, this phenomenon is more commonly explained by the presence of dark matter, which is thought to make up a large portion of the matter in the universe.

Dark matter is hypothesized to exist because the observed gravitational effects in the universe suggest that there is more matter present than can be accounted for by observable matter such as stars and galaxies.

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The kinetic energy of an object that has a mass of 50.0 kg and a velocity of 18m/s is 8,100J.

Kinetic energy refers to the energy of a body due to its motion. The kinetic energy can be calculated as follows:

K.E = ½ m × v²

Where;

According to this question, an object that has a mass of 50.0 kg and a velocity of 18m/s. The kinetic energy can be calculated as follows:

K.E = ½ × 50 × 18²

K.E = ½ × 16,200

K.E = 8100J.

Therefore, the kinetic energy of an object that has a mass of 50.0 kg and a velocity of 18m/s is 8,100J.

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Answer: \((a)10.812\ m/s\ (b)\ 0.75\ m/s\ \text{left}\)

Explanation:

Given

Mass of hockey puck \(m=0.160\ kg\)

Initial velocity of hockey puck is \(u=3\ m/s\)

First a horizontal force of \(25\ N\) is applied to the right for \(0.05\ s\)

acceleration associated with it is

\(\Rightarrow a=\dfrac{25}{0.160}\\\\\Rightarrow a=156.5\ m/s^2\)

Using equation of motion i.e.

\(\Rightarrow v=u+at\\\Rightarrow v=3+156.25\times 0.05\\\Rightarrow v=3+7.812\\\Rightarrow v=10.812\ m/s\)

(b) When a force of \(12\ N\) is applied for 0.05s

Using equation of motion i.e.

\(\Rightarrow v=3-\dfrac{12}{0.160}\times 0.05\\\\\Rightarrow v=3-75\times 0.05\\\Rightarrow v=3-3.75\\\Rightarrow v=-0.75\ m/s\\\Rightarrow v=0.75\ \text{towards left}\)

An appropriate value for the resistance in parallel with the smoothing capacitor would be 1.59 kΩ.

To ensure good tracking of the AM envelope, the resistance in parallel with the smoothing capacitor should be low enough to discharge the capacitor quickly during the troughs of the modulated signal, but high enough to avoid discharging it too quickly during the peaks of the signal.

The time constant (τ) of the RC circuit formed by the smoothing capacitor and the parallel resistance is given by the formula:

τ = RC

where R is the resistance and C is the capacitance.

To determine an appropriate value for the resistance, we need to calculate the time constant and compare it to the period of the modulated signal.

The period of a 500 kHz signal is T = 1/f = 2 μs. The modulating signal has a bandwidth of 5 kHz, which means its period is 200 μs.

Assuming a small signal approximation, we can use the formula for the time constant to calculate an appropriate value for the resistance:

τ = 20 nF × R = T/2π = 31.8 ns

Solving for R, we get:

R = τ/C = 31.8 ns / 20 nF = 1.59 kΩ

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Complete question is:

The input signal into an envelope detector is an am signal of carrier frequency 500 khz. the envelope detector employs a smoothing capacitor of 20 nf. the modulating signal has a bandwidth of 5 khz. specify an appropriate value for the resistance in parallel with the smoothing capacitor for a good tracking of the am envelope.

On the microscope slide, a specimen will appear upside-down and facing left while it is actually right-side up and facing right.

The orientation of the specimen image is reversed when viewed via a microscope compared to the orientation of the actual specimen. As a result, when viewed under a microscope, the specimen will appear to be upside-down and backwards.

Coarse Adjustment Knob: To concentrate the specimen, the coarse adjustment knob, which is placed on the microscope's arm, raises and lowers the stage. With just a partial turn of the adjustment knob, the gearing system creates a significant vertical movement of the stage.

Condenser-Objective lens-Body tube-Eyepiece is the proper order for the flow of light in a compound microscope.

Use two hands at all times when moving your microscope. Lift the scope while encircling the arm with one hand.

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4.88×10⁻⁵rad is the angular separation θ1 at which two point sources of wavelength 600 nanometers are just resolved when viewed through a circular aperture of diameter 1.5 centimeters.

The angle between two sightlines or between two point objects as seen by an observer is known as angular distance (also called as angular separation, apparent distance, and apparent separation).

All of the natural sciences and mathematics make use of angular distance, particularly geometry and trigonometry.  It appears with angular velocity, angular acceleration and torque in the classical mechanics of spinning objects.

The angular separation is calculated as given below.

θ   =1.220 ×λ/d

  = 1.220 ×600×10⁻⁹/1.5×10⁻²

 =4.88×10⁻⁵rad

Therefore, the angular separation is 4.88×10⁻⁵rad.

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Answer:

d. N*m

Explanation:

Workdone which is measured in Joules is calculated using the formula;

Workdone = force * distance

Where;

Force is measured in Newton

Distance is measured in meters.

Therefore, workdone also has the measurement of Newton-meters.

The distance the wagons travel until the spring is fully compressed is 2 × xmax = 3.42 m. To solve this problem, we need to apply the principle of conservation of momentum and conservation of energy.

Initially, the momentum of the system is given by:

p1 = (15 kg + 5 kg) × 3.75 m/s = 75 kg m/s

After the collision, the two wagons stick together and move as one unit, with a combined mass of 30 kg. Let v be the common velocity of the two wagons after the collision. Then, the momentum of the system is given by:

p2 = 30 kg × v

According to the principle of conservation of momentum, p1 = p2. Therefore, we have:

75 kg m/s = 30 kg × v

which gives:

v = 2.5 m/s

Now, let's consider the energy of the system. Initially, the kinetic energy of the system is given by:

K1 = (1/2) × (15 kg + 5 kg) × (3.75 m/s)² = 421.875 J

After the collision, the energy of the system is stored in the compressed spring. The potential energy of a spring is given by:

U = (1/2) × k × x²

where k is the spring constant and x is the displacement of the spring from its equilibrium position. At maximum compression, the displacement is xmax, and the potential energy is:

Umax = (1/2) × 250 N/m × xmax²

According to the principle of conservation of energy, the initial kinetic energy of the system is converted to potential energy of the spring at maximum compression. Therefore, we have:

K1 = Umax

which gives:

xmax = sqrt(2 × K1 / k) = 1.71 m

Therefore, the distance the wagons travel until the spring is fully compressed is 2 × xmax = 3.42 m.

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According to the Skygazer's Almanac for 2022 at 40 degrees, Venus and Saturn will be in opposite parts of the sky on December 17, 2022.

The Skygazer's Almanac provides astronomical information for a specific location and year. In this case, at a latitude of 40 degrees, the almanac indicates that Venus and Saturn will be in opposite parts of the sky on December 17, 2022. This means that Venus and Saturn will appear at opposite sides of the celestial sphere as observed from Earth. However, it's important to note that the almanac's predictions are approximate and can be influenced by various factors, including atmospheric conditions and the observer's specific location. To obtain the most accurate and up-to-date information, it is recommended to consult more recent astronomical sources or use specialized software that can provide precise positions and dates for celestial events.

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The speed of sound in air at the temperature of 136°F (331 K) = 365.71 m/s

The speed of sound in air depends on temperature, so the higher the temperature, the greater the value of the speed of sound.

The equation for determining the speed of sound with a change in temperature is:

v = v₀ + 0.6(∆T)

where:

v = the final sound velocity (m/s)

v₀ = the initial sound velocity (m/s)

∆T = temperature change (°C)

Note: The speed of sound in air at 0°C is 331 m/s.

We have ∆T = 136°F ⇒ 57.85°C

So,

v = 331 + 0.6 (57.85)

= 365.71 m/s

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One   worth of distance is covered by the object. The radius times 2 pi is the circumference. 1 metre is the radius.

The distance from a circle's centre to just about any place on its periphery is known as the radius. The radius of a circle is the distance measured from the centre to any point within the circle, according to another definition.

The area of a circle is the measurement of the area contained within the circle. Radius: The radius of the a circle is indeed the farthest from to a spot on the edge. The letter "r" or "R" stands in for it.

A closed this double figure is a circle.

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