There is a bell at the top of a tower that is 45 m high. The bell has a mass of 40kg. The bell has__ energy. Calculate it.

To answer this question, we need to remember what distance really is. Distance is far something moves in a scalar motion, which is not relative to where she starts out originally.

Since Diane travels around the track 10 times and the track is 400 meters long, she ultimately ends up traveling 4000 meters.

Answer:

Explanation:

Given data

mass m= 45g

volume v= 9mL

we know that density=m/v

substituting our given data we have

\(density=\frac{45}{9} =5g/mL\)

What is Density?

The Density of a body can be defined as the ratio of mass to volume,

or

Density, mass of a unit volume of a material substance. The formula for density is \(Density= \frac{mass}{volume}\),

where d is density,

M is mass, and

V is volume.

Density is commonly expressed in units of grams per cubic centimetre.

Answer:

4

Explanation:

Impact Force = mv/2t

Impact force is inversely proportional to t

If m and v are constant and the only thing that changes is t, then:

0.005 s increases by a factor of 4 to 0.020 s

Therefore, impact force reduces by a factor of 4 (1/4 less than when the time of impact was 0.005 s)

The size of the region responsible for powering an Active Galactic Nucleus (AGN) is typically c. Solar System size. This region, which includes the supermassive black hole and the surrounding accretion disk, has dimensions comparable to those of our Solar System.

First, it's important to understand what an AGN is. An AGN (Active Galactic Nucleus) is a compact region at the center of a galaxy that emits a tremendous amount of energy across the electromagnetic spectrum, from radio waves to gamma rays. The energy output of an AGN is believed to be powered by the accretion of matter onto a supermassive black hole at the center of the galaxy. As matter falls toward the black hole, it becomes heated and emits radiation before eventually crossing the event horizon and being swallowed by the black hole.


In summary, the size of the region responsible for powering an AGN is not a simple answer, but rather a complex question that depends on the specific AGN being observed and the method used to measure its size. While estimates can vary widely, the emission region of an AGN is typically much larger than the black hole itself but still relatively compact compared to the overall size of the galaxy, making "d. galaxy size" the most appropriate answer to this question.

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Answer:

(C) Radiation

Explanation:

The weak nuclear forces causes radiation to form.

The period of a rotating object is the time it takes for one complete revolution. In this case, the wheel takes 2.75 seconds to complete one revolution. its period is 2.75 seconds. The angular speed of the wheel is 235.62 radians per second.

The angular speed of the wheel is the rate at which it rotates in radians per second. To find the angular speed, we need to first convert the diameter to radius. The radius of the wheel is half of its diameter, so the radius is 107 cm. We then use the formula for angular speed, Angular speed = (2 x π x radius) / time where π is the mathematical constant pi (approximately 3.14). Substituting the values, we have: Angular speed = (2 x 3.14 x 107) / 2.75 Angular speed = 235.62 radians per second Therefore, the angular speed of the wheel is 235.62 radians per second.

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In the given question, we can infer from the table that the average velocity for the car upto one decimal place is 7.3 m/s.

Velocity is defined as the ratio of Displacement of a body to the time it has take for the displacement. We can find the average velocity but adding up all the velocities and then divide it by number of occurrences. Using the data, after adding up all the average velocities, we get the value as 87.497 m/s. As the number of velocities are 12, we divide the total velocity by 12 and fing the mean velocity. The mean or average velocity is : 87.497 / 12 = 7.291416. Upto one decimal place, we get the value as 7.3 m/s.

Hence the average or mean velocity from the given data is 7.3 m/s

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Answer:

h = 10000 m

Explanation:

The pressure applied at a depth of the liquid is given by:

P =ρgh

where,

P = Maximum Pressure to Survive = (1000)(Atmospheric Pressure)

P = (1000)(101325 Pa) = 1.01 x 10⁸ Pa

ρ = Density of sea water = 1025 kg/m³

g = 9.8 m/s²

h = maximum depth to survive = ?

Therefore,

1.01 x 10⁸ Pa = (1025 kg/m³)(9.8 m/s²)h

h = (1.01 x 10⁸ Pa)/(1025 kg/m³)(9.8 m/s²)

h = 10000 m

Answer:

But if you place a clear container filled with hydrogen gas between the flashlight and the prism, gaps appear in the smooth rainbow of colors, places where the light literally goes missing. The dark absorption lines of a star at rest (left) get shifted towards red if the star is moving away from Earth (right)

Explanation:

The flow of electricity in a certain path is the circuit.

The detection of quasars from such great distances is due to their extreme luminosity and the fact that they emit massive amounts of energy.

Quasars are the most energetic objects in the universe, and their bright emissions make them visible even at very large distances. On the other hand, even the biggest and most luminous galaxies cannot be seen from such distances because their emissions are not as powerful as those of quasars.

Quasars are actually supermassive black holes at the center of galaxies that are actively feeding on surrounding matter. As they consume matter, they emit large amounts of energy in the form of light, X-rays, and other types of radiation. This energy is what makes them visible from great distances, even beyond the limits of what other galaxies can achieve.

The fact that quasars can be detected from distances beyond the reach of other galaxies is a testament to their extreme power and luminosity. It also helps astronomers to study the universe at a deeper level and gain insight into the evolution of galaxies and supermassive black holes.

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To calculate the IV (intravenous) flow rate for an LVP (large volume parenteral) infusion, a technician must know the total volume of the LVP and either the infusion time or the desired infusion rate.

1. Infusion Time: If the technician knows the total volume of the LVP and the desired infusion time, they can use the formula:

Flow Rate (mL/h) = Total Volume (mL) / Infusion Time (hours)

This formula will give the flow rate required to infuse the entire volume of the LVP over the specified infusion time.

2. Infusion Rate: If the technician knows the total volume of the LVP and the desired infusion rate, they can use the formula:

Flow Rate (mL/h) = Infusion Rate (mL/h)

In this case, the infusion rate is already provided, and it represents the desired flow rate at which the LVP should be administered.

Both of these formulas assume a constant flow rate throughout the infusion process. It's important to note that the actual flow rate might vary slightly due to factors such as line resistance, gravity, or specific equipment used. Therefore, close monitoring of the infusion process is necessary to ensure accurate delivery of the LVP.

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Answer:

B. 15 m/s

Explanation:

p = m × v

480 = 32 × v

32v = 480

v = 480/32

v = 15 m/s

Answer:

B. 15 m/s

Explanation:

P=MV

480=32×v

v=15m/s

The tension in the rope between the 20 kg mass and the pulley is 176.47 N, and the 20 kg mass falls 0.6125 m in the first 0.5 s.

1. Calculate the net torque acting on the pulley: τ = Iα, where α is the angular acceleration.

2. Use the 20 kg mass to find the torque: τ = rF, where r is the radius (0.125 m) and F is the force (20 kg * 9.81 m/s²).

3. Solve for α: α = τ/I = (0.125 * 20 * 9.81)/0.05.

4. Calculate the linear acceleration of the 20 kg mass: a = rα.

5. Find the tension in the rope: T = m(a + g), where m is the 20 kg mass and g is the acceleration due to gravity (9.81 m/s²).

6. Determine the distance the 20 kg mass falls in the first 0.5 s using the equation: d = 0.5 * a * t², where t is the time (0.5 s).

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The uterus lining is rebuilt by the end of the menstrual period. The rebuilding of the uterus lining starts after menstruation and the lining is typically completely rebuilt by day 14 of the menstrual cycle, which is when ovulation occurs and the uterus is preparing to potentially receive a fertilized egg.

The endometrium is the inner lining of the uterus, and it thickens every month to prepare for pregnancy. After menstruation, the endometrium grows and thickens to prepare for the implantation of a fertilized egg. The cells in the lining multiply and enlarge, and the glands in the lining begin to secrete mucus and other substances that help support the fertilized egg and promote its growth.

The rebuilding of the endometrium usually takes about two weeks after menstruation. This process is closely regulated by hormones such as estrogen and progesterone, which are produced by the ovaries and other parts of the body. These hormones help control the growth and development of the endometrium and other reproductive tissues.

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Answer:

The expansion and contraction of materials can also cause problems. For example, bridges expand in the summer heat and need special joints to stop them bending out of shape.A bimetallic strip has two metal strips glued together. One of these metals expands more for each degree temperature rise than that other.

When a motor is spinning, most of the power is utilized to overcome friction and add kinetic energy to the wheel.

However, when preventing the wheel from spinning, the energy is dissipated as heat and sound due to the work done against friction and other resistive forces.

When the wheel is prevented from spinning, the power that would have been used to overcome friction and add kinetic energy is no longer converted into useful work. Instead, it is dissipated as heat and sound. The work done against friction generates heat as the surfaces interact and create frictional forces.

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Answer:

Only (B) is correct:

PE + KE = constant

When Potential Energy is a maximum., the KE is zero

Also, PE depends on the height of the ball

Answer:

Resultant horizontal force  :  16 N

Resultant vertical force :   4 N

Explanation: Force acting on left : 3N

                      Force acting on left : 1 N

                      Forces acting on left : 3N + 1N = 4N

                       Force acting on right : 12 N

                       Resultant horizontal force  : 12 N - 4 N = 16 N

                      Force acting upwards : 4 N

                      Force acting upwards : 2 N

                      Forces acting upwards : 4 N + 2 N = 6 N

                       Force acting downwards : 10 N

                        Resultant vertical force : 10 N - 6N = 4 N

Answer: There are many problems associated with digitized DNA storage.

Explanation:

The cost of encoding the DNA data is high and lot of working efficiency is required for the same. The encoding process requires a considerable amount of time and encoding is slow. It takes about 400 bytes per second. This is done in a silicon memory chip. The process is millions of times slower.

The two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

In the given case, the figure shows an exaggerated view of a rifle that has been sighted in for a 91.4-meter target. Let the muzzle speed of the bullet be v0 = 427 m/s.

Now, we are required to find the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target.

It is known that the horizontal displacement of the bullet from the gun can be given by the equation: x = v0 t cosθ ..........(i)and the vertical displacement of the bullet from the gun can be given by the equation: y = v0 t sinθ - (1/2) g t^2..........(ii).

Here, t is the time of flight of the bullet and g is the acceleration due to gravity.

As the bullet hits the target, its final vertical displacement from the gun is equal to the height of the target, i.e.,y = 91.4m.Now, we can substitute equations (i) and (ii) in place of t and y in equation (ii) to get:x tanθ - (g/2v0^2) x^2 sec^2θ = 91.4 ..........(iii)This is a quadratic equation in tanθ.

On solving this equation using the quadratic formula, we get:tanθ = [-b ± √(b^2 - 4ac)]/2aWhere,a = -gx^2/(2v0^2) = -4.9x^2/v0^2, b = x, and c = -91.4.

Rearranging the terms, we get:2a tanθ^2 + b tanθ - 91.4 = 0On substituting the given values, we get:2(-4.9x^2/v0^2) tanθ^2 + x tanθ - 91.4 = 0θ1 and θ2 are the two possible angles which can be found by solving the above quadratic equation.

Using the trigonometric identity given in the hint, we can write: sin 2θ = 2 sinθ cos θ = 2 tanθ/ (1 + tan^2θ)Now, we can substitute tanθ = (-b ± √(b^2 - 4ac))/2a in the above equation to get: sin 2θ = (-4bx ± 2x√(b^2 - 4ac))/(b^2 + 4a^2)Now, we can substitute the given values to get: sin 2θ1 = -0.999sin 2θ2 = 0.998.

Thus, we get two values of sin 2θ, one is close to -1 and the other is close to 1. As sin 2θ = -1 when 2θ = -π/2 + nπ and sin 2θ = 1 when 2θ = π/2 + nπ, where n is an integer, we get two possible values of θ for each of these two cases.

Hence, the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

As one of these angles is so large that it is never used in target shooting, we only need to consider the other angle.

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The two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

In the given case, the figure shows an exaggerated view of a rifle that has been sighted in for a 91.4-meter target. Let the muzzle speed of the bullet be v0 = 427 m/s.

Now, we are required to find the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target.

It is known that the horizontal displacement of the bullet from the gun can be given by the equation: x = v0 t cosθ ..........(i)and the vertical displacement of the bullet from the gun can be given by the equation: y = v0 t sinθ - (1/2) g t^2..........(ii).

Here, t is the time of flight of the bullet and g is the acceleration due to gravity.

As the bullet hits the target, its final vertical displacement from the gun is equal to the height of the target, i.e.,y = 91.4m.Now, we can substitute equations (i) and (ii) in place of t and y in equation (ii) to get:x tanθ - (g/2v0^2) x^2 sec^2θ = 91.4 ..........(iii)This is a quadratic equation in tanθ.

On solving this equation using the quadratic formula, we get:tanθ = [-b ± √(b^2 - 4ac)]/2aWhere,a = -gx^2/(2v0^2) = -4.9x^2/v0^2, b = x, and c = -91.4.

Rearranging the terms, we get:2a tanθ^2 + b tanθ - 91.4 = 0On substituting the given values, we get:2(-4.9x^2/v0^2) tanθ^2 + x tanθ - 91.4 = 0θ1 and θ2 are the two possible angles which can be found by solving the above quadratic equation.

Using the trigonometric identity given in the hint, we can write: sin 2θ = 2 sinθ cos θ = 2 tanθ/ (1 + tan^2θ)Now, we can substitute tanθ = (-b ± √(b^2 - 4ac))/2a in the above equation to get: sin 2θ = (-4bx ± 2x√(b^2 - 4ac))/(b^2 + 4a^2)Now, we can substitute the given values to get: sin 2θ1 = -0.999sin 2θ2 = 0.998.

Thus, we get two values of sin 2θ, one is close to -1 and the other is close to 1. As sin 2θ = -1 when 2θ = -π/2 + nπ and sin 2θ = 1 when 2θ = π/2 + nπ, where n is an integer, we get two possible values of θ for each of these two cases.

Hence, the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

As one of these angles is so large that it is never used in target shooting, we only need to consider the other angle.

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The 100 Ω and 50 Ω have been joined in series and their combination is joined in parallel with 75 Ω, in the given picture. The circuit diagram shown below.

A circuit diagram uses electrical symbols to show the various components of an electrical circuit graphically or visually. There is a lot of resistance in the voltmeter.

For the design (circuit design), fabrication (like PCB layout), and maintenance of electrical and electronic equipment, circuit diagrams are utilized. When illustrating Boolean algebraic expressions, circuit diagrams are helpful in the field of computer science.

We may create parallel and series circuits, which are two different sorts of circuits. Wires connect the various parts of a circuit. If there are no branches, the circuit is in series. If there are branches, the circuit is parallel.

Consider the 100 Ω and 50 Ω have been joined in series and their combination is joined in parallel with 75 Ω.

The circuit diagram drawing is as follows:

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The complete question is as follows:

Draw a circuit diagram for the circuit of Figure P23.1.

Answer:

4

Explanation:

Because that is when they're mind learns how to read

Answer:

B.she cannot use it because the credit card is not collareral

Explanation:

sorry if am wrong

The charge stored in capacitor is 22.8 mC.

We need to know about capacitors to solve this problem. Capacitor is a device that stores electric charge. The charge stored in capacitor can be determined as

Q = C . V

where Q is charge, C is capacitance and V is voltage.

From the question above, we know that

C = 152 µF = 152 x 10¯⁶ F

V = 150 V

By substituting the following parameters, we can calculate the charge

Q = C . V

Q = 152 x 10¯⁶ . 150

Q = 0.0228 C

Convert to millicoulombs

Q = 0.0228 C

Q = 0.0228 x 10³ mC

Q = 22.8 mC

Hence, the charge stored in capacitor is 22.8 mC.

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Answer:

m = 7 kg

Explanation:

F = m × a

49 = m × 7

7m = 49

m = 49/7

m = 7 kg

The distance of the fourth leg and the angle is Ф = 29.94 degree and  d = 6.73 km

Vector addition is the process of combining two or more vectors to obtain a resultant vector. The resultant vector is determined by adding the corresponding components of the vectors.

To add vectors, we add their horizontal components together and their vertical components together separately.

The horizontal component of the resultant vector is the sum of the horizontal components of the individual vectors.

The vertical component of the resultant vector is the sum of the vertical components of the individual vectors.

By adding the horizontal and vertical components, we can find the resultant vector in terms of its magnitude and direction.

GIven :A = 3.20km

B = 5.10km

C = 4.80 km

D = ?

adding components all along the x-axis

3.60 cos40 - 5.10 cos 35 - 4.80 cos 23 + d (cosФ) = 0

d (cosФ) = -5.8383

similarly for y components

d (sinФ) = -3.363

so tanФ = 0.576

Ф = 29.94 degree

d = 6.73 km

Therefore, the distance of the fourth leg and the angle is Ф = 29.94 degrees and  d = 6.73 km

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The velocity of the golf ball is 65.2 m/s

The principle of conservation of momentum is that the momentum before collision is equal to the momentum after collision.

Given that;

(0.2kg *  55 m/s) + (0.046kg * 0 m/s) = (0.2kg * 40 m/s) + (0.046kg  * v)

11 = 8 + 0.046v

v = 11 - 8/0.046

v = 65.2 m/s

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Answer:

Some modifications that can be done to a galvanometer to increase its sensitivity:

It is important to note that these modifications can only be done up to a certain point. For example, if the number of turns in the coil is too high, the coil will become too heavy and will not be able to deflect as easily. Similarly, if the magnetic field is too strong, the coil will be damaged. Therefore, it is important to find a balance between these factors in order to achieve the desired sensitivity.

Answer and Explanation:

To increase the sensitivity of a galvanometer, which is an instrument used to detect and measure small electric currents, several modifications can be made. Here are some possible modifications and their explanations:

1. Decrease the resistance: By reducing the resistance in the galvanometer, the current flowing through it will increase, resulting in higher sensitivity. This can be achieved by using a lower resistance coil or by adding a shunt resistor in parallel with the galvanometer.

2. Increase the number of turns in the coil: Increasing the number of turns in the galvanometer's coil will amplify the effect of the current passing through it, making it more sensitive to small currents. This can be achieved by winding the coil with a greater number of turns of wire.

3. Use a more sensitive suspension system: The suspension system of a galvanometer plays a crucial role in its sensitivity. By using a more delicate and sensitive suspension system, such as a fine fiber or a torsion wire, the deflection caused by small currents can be magnified, enhancing the sensitivity of the galvanometer.

4. Decrease the moment of inertia: The moment of inertia of the galvanometer's moving parts affects its responsiveness to current changes. By reducing the mass or size of the moving parts, the moment of inertia decreases, allowing the galvanometer to respond more quickly and accurately to small current variations.

5. Increase the strength of the magnetic field: The sensitivity of a galvanometer is directly proportional to the strength of the magnetic field in which it operates. Increasing the magnetic field strength can be achieved by using a stronger permanent magnet or by increasing the current flowing through the coil, if it is an electromagnet.

It's important to note that these modifications may have limitations and trade-offs. For example, reducing the resistance may increase the power dissipation and affect the galvanometer's accuracy. Therefore, careful consideration and calibration are necessary to optimize the sensitivity while maintaining the desired performance of the galvanometer.

Earth's inner movements resemble dynamos and produce the magnetosphere. The magnetosphere of Earth is a dynamic, networked system that reacts to solar, planetary, and interstellar circumstances.

The area surrounding a planet where the magnetic field of the planet is dominant is called a magnetosphere. All of the rocky planets in our solar system have magnetospheres, but Earth's is the strongest. The enormous, comet-shaped bubble that makes up Earth's magnetosphere has been essential to the planet's capacity to support life. This magnetic environment has protected life on Earth from its inception and continues to do so. The magnetosphere protects our planet from solar and cosmic ray radiation as well as the solar wind's steady stream of charged particles that stream off the sun, which may erode the atmosphere.

Due to the convective motion of charged,

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