An automobile of mass 1500 kg has an initial velocity of 7.20 m/s toward the north and a final velocity of 6.50 m/s toward the west. The magnitude and direction of the change in momentum of the car are

Answers

Answer 1

Answer:

Explanation:

Momentum is the product of mass and velocity.

Momentum = mass * velocity

Given

Mass = 1500kg

resultant velocity

v² = 6.5²+7.20²

v² = 42.25+51.84

v² = 94.09

v = √94.09

v= 9.7m/s

Momentum = 1500*9.7

Momentum = 14550kgm/s


Related Questions

A pail of water is being swung in a vertical circle at the end of a 0.55-m string. How slowly can the pail go through its top positionwithout having the string go slack?

Answers

Given data:

* The length of the string (or the radius of the vertical circle) is,

\(r=0.55\text{ m}\)

Solution:

Let the tension of the string is T.

The string will go slack after zero tension.

Thus,

\(T=0\text{ N}\)

At the top, the vertical forces acting are,

The weight of the pail of water ,

\(W=mg\)

where m is the mass, g is the acceleration due to gravity,

The centripetal force acting on the apil of water moving in vertical circular motion is

\(F=\frac{mv^2}{r}\)

As the weight of pail of water and tension of the string provides the centripetal force for the circular motion,

Thus,

\(\begin{gathered} T+W=F \\ 0+mg=\frac{mv^2}{r} \\ mg=\frac{mv^2}{r} \\ g=\frac{v^2}{r} \\ v^2=gr \\ v=\sqrt[]{gr} \end{gathered}\)

Substituting the known values,

\(\begin{gathered} v=\sqrt[]{9.8\times0.55} \\ v=2.32ms^{-1} \end{gathered}\)

Thus, the velocity of the pail at the top position is 2.32 meter per second.

5. If the momentum of a body is increased by 100% then the percentage increase in kinetic
energy is
150%
b. 200%
225%
d. 300%
a.
С.​

Answers

The answer is the letter ( d )

what is the similaries between parallel and series circuits

Answers

Answer:

same power source

can have more than 1 light bulb in a circuit

requires a source of energy

Explanation:

A small sphere of mass 10 kg
is released from rest at a height of
15.0 m above the ground level.
The sphere experiences a constant
resistive force (due to air
resistance) of magnitude R = 10.0
N.
a) Calculate the speed of the
sphere after it has fallen
through a distance of 5.00 m

bCalculate the speed of the ball just before a it hits the gound.

Answers

Answer:

Approximately \(9.39 \; {\rm m\cdot s^{-1}}\) after the sphere has travelled a distance of \(5\; {\rm m}\).

Approximately \(16.3\; {\rm m\cdot s^{-1}}\) right before touching the ground (a distance of \(15\; {\rm m}\).)

Assumption: \(g = 9.81\; {\rm N\cdot kg^{-1}}\).

Explanation:

Weight of the sphere: \(m\, g = 9.81\; {\rm N \cdot kg^{-1}} \times 10\; {\rm kg} = 98.1\; {\rm N}\), downwards.

Drag on the sphere: \(10.0\; {\rm N}\) upwards.

Net force on the sphere: \(98.1\; {\rm N} - 10\; {\rm N} = 88.1\; {\rm N}\) downwards.

Acceleration of the sphere: \(a = F_\text{net} / m = 88.1\; {\rm N} / (10\; {\rm kg}) = 8.81\; {\rm m\cdot s^{-2}}\).

Apply the SUVAT equation \(v^{2} - u^{2} = 2\, a\, x\), where \(v\) is the final velocity, \(u\) is the initial velocity (\(0\) in this case, as the sphere was released from rest,) and \(x\) is the distance (displacement) that the sphere has travelled so far.

Rearrange this equation to obtain an expression for \(v\):

\(\displaystyle v = \sqrt{2\, a\, x + u^{2}}\).

For example, after the ball travelled a distance of \(5.00\; {\rm m}\), \(x = 5.00 \; {\rm m}\):

\(\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 5.0\; {\rm m} + 0} \\ &\approx 9.39\; {\rm m\cdot s^{-1}}\end{aligned}\).

Similarly, \(x = 15.0\; {\rm m}\) right before landing, such that:

\(\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 15.0\; {\rm m} + 0} \\ &\approx 16.3\; {\rm m\cdot s^{-1}}\end{aligned}\).

If the current in each wire is the same, which wire produces the strongest magnetic field?
-a wire that is 1 mm thick and not coiled
-a wire that is 2 mm thick and not coiled
-a 1-mm-thick coiled wire with ten loops
-a 2-mm-thick coiled wire with two loops

Answers

Answer:

C

Explanation:

I did this before

Answer:

it is c

Explanation:

cuz i know

Pls help me 10 points

Pls help me 10 points

Answers

Answer:

d is the right answer

it's was helpful to you

What are the major energy transformations that occur between when energy starts from a primary energy source, becomes a piece of wood, which is then burned in a fire?​

Answers

Sunlight to chemistry energy conversion, energy transfer as from sun to plants, kinetic energy is created by converting thermal energy. when wood is burned, mechanical energy is converted to thermal energy.

What are the primary six energy forms?

Chemical, electrical, infrared, mechanical, thermal, et nuclear energy are the six main types of energy. Different research may focus on other forms, such as electric, acoustic, electromagnetic, and some others.

What is an indication of an energy revolution?

Energy transition describes the change in the global energy industry away from fossil-based methods of hydropower, such as oil, natural gas, and coal, and toward rechargeable batteries with renewable sources of electricity like wind and solar.

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A container with a height of 7.7 inches with an open top has a 5.6 inch diameter and is open to the atmosphere. The container is filled with water. The bottom of the container has a 0.87 inch diameter hole. Calculate ρgh at the top of the container if the datum is set at the bottom of the container.

Answers

This is the pressure at the top of the container due to the height of the water. The pressure at the bottom of the container due to the height of the water is 0 Pa since the datum was set at the bottom.

What is pressure?

Pressure is a physical quantity used to measure the force applied by an object to another object or by an object to a surface. It can be expressed as the force per unit area and is typically measured in units such as pounds per square inch (psi) or pascals (Pa). Pressure can be applied to gases, liquids, and solids, and is generated by the weight of the atmosphere, the force of gravity, and by the movement of air or liquids. Pressure can also be created by mechanical devices such as pumps and compressors, as well as by chemical reactions.

ρ = density of water = 1000 kg/m3

g = acceleration due to gravity = 9.81 m/s2

h = height of the container = 7.7 in = 0.198 m

Using the equation ρgh = density x gravity x height, we can calculate the pressure at the top of the container to be:

ρgh = (1000 kg/m3)(9.81 m/s2)(0.198 m) = 1960.38 Pa

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The 20-cm diameter disk in (Figure 1) can rotate on an axle through its center. What is the net torque about the axle?

The 20-cm diameter disk in (Figure 1) can rotate on an axle through its center. What is the net torque

Answers

This question involves the concepts of torque, applied force, and force arm.

The net torque about the axle will be "0.94 N.m clockwise".

The net torque can simply be found by adding the torques produced by every force about the axle, that is at some perpendicular distance from the axle and whose line of action does not touch with the center point. We will take the clockwise direction as negative here, and the counter-clockwise direction as positive. The torque due to a force can simply be given as the product of the force and the perpendicular distance between center point and force, also known as force arm.

\(Net\ Torque = T = -(30\ N)(0.1\ m)+(20\ N)(0.05\ m)+(30\ N)(Sin\ 45^o)(0.05\ m)\)

T = -3 N.m + 1 N.m + 1.06 N.m

T = -0.94 N.m (negative sign shows clockwise direction)

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The Sun radiates energy at a rate of about 4×1026W. At what rate is the mass decreasing?

Answers

4.44×\(10^{9}\) kg/s is the rate at which the sun mass is decreasing.

The Sun radiates energy through a process called nuclear fusion, where hydrogen atoms combine to form helium, releasing a tremendous amount of energy in the process. According to Einstein's mass-energy equivalence principle (E=mc²), this energy release corresponds to a decrease in mass.

To calculate the rate at which the Sun's mass is decreasing, we can use the formula ΔE = Δmc², where ΔE is the change in energy, Δm is the change in mass, and c is the speed of light.

Given that the Sun radiates energy at a rate of 4×10^26 W, we can substitute this value into the equation as ΔE and solve for Δm.

ΔE = 4×10^26 W

c = 3×10^8 m/s (speed of light)

Using the equation ΔE = Δmc² and rearranging it, we get Δm = ΔE / c².

Substituting the values, we have:

Δm = (4×10^26 W) / (3×10^8 m/s)²

Evaluating this expression, we find that the rate at which the Sun's mass is decreasing is approximately 4.44×10^9 kg/s.

This calculation demonstrates that the Sun's mass is gradually decreasing as it continuously radiates energy into space, primarily through the process of nuclear fusion in its core.

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7.0 J of work is done to draw a bowstring back. The bow launches an arrow with a mass of 0.09 kg straight upward.
(a) What is the arrow's kinetic energy as it leaves the bow? (Round your answer to one decimal place.)
J
(b) What is the arrow's speed? (Round your answer to one decimal place.)
m/s
(c) What maximum height does the arrow reach? (Round your answer to one decimal place.)
m

Answers

The final answer is

(a) The KE = 3.88 * 10

(b) The speed of the arrow is 2.93 * 10

(c) Height which the arrow attains is 4.38 * 10

When the arrow is drawn from a bowstring back then there will be some work done. The pulling of the bowstring will have a distance moved and the energy used to do this.

Given,

Work done = 7 J

Mass of the arrow = 0.09 kg

The computations for a and b are the following:

To calculate the Kinetic energy,

                  Work done = force x distance

                  7.0 J = force x 0.09

                   force = 7/0.09=77.77 = 7.77 x 10

                   force = m. a = 0.09 x a

                    77.77 / 0.09 = a = 864.1 = 8.64*10²

KE = 0.5 x m x v²

     = 0.5 x 0.09 x 864.1

     = 3.88 * 10

The speed of the arrow can be calculated as

       speed 38.8/(0.5 x 0.09) = v² = 862 sq-root

       Hence, v = 2.93*10¹ m/s is the answer

Height that the arrow reaches

       29.3² = 2gh

      = 29.3²/(2 x 9.8) = 858.49 / (2 x 9.8) = 43.8 m

h = 43.8 m or 4.38 * 10¹ is the answer

Therefore, we can conclude that the kinetic energy of the arrow that leaves bowstring is 3.88*10 with the speed of 29.3 m/s and it reaches the height of 43.8 m.

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Bill drives and sees a red light. He slows down to a stop. A graph of his velocity over time is shown below.
What is his average acceleration from 0 to 10 seconds?

Bill drives and sees a red light. He slows down to a stop. A graph of his velocity over time is shown

Answers

Answer: its actually 0

Explanation: i read the khan academy hints and the answer is 0

yes the answer is 0 and if its not and have this graph its -2.0

To test the performance of its tires, a car
travels along a perfectly flat (no banking) circular track of radius 96.6 m. The car increases
its speed at uniform rate of
at ≡((d |v|)/dt) = 4.87 m/s^2
until the tires start to skid.
If the tires start to skid when the car reaches
a speed of 21.1 m/s, what is the coefficient of
static friction between the tires and the road?
The acceleration of gravity is 9.8 m/s^2

Answers

The coefficient of  static friction between the tires and the road is 1.987.

What is Static friction?

Radius of the track, r =  516 m, Tangential Acceleration =  3.89 m/s^2 and Speed,v =  32.8 m/s

The radial Acceleration is given by, Now the total acceleration is The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of  static friction is, From (1) and (2), Substituting the values, we get friction is 1.987.

Therefore, The coefficient of  static friction between the tires and the road is 1.987.

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Two blocks, 1 and 2, are connected by a massless string that passes over a massless pulley. 1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. 2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105

. The figure illustrates the configuration.

A system of two blocks connected by a rope passing over a pulley. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle. Box M subscript 1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M sub 1 and the surface is mu subscript 1. Box M subscript 2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle. The coefficient of friction between box M sub 2 and the surface is mu subscript 2.

Answers

The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.

The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.

The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.

The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.

The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).

The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).

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Yanni just turned one. He loves to play with his cars and trucks. He can also name the animals he sees in books.

In which stage of development is Yanni?

childhood
infancy
adolescence
adulthood

Answers

infancy
i think hope this helps


PLEASE HELP AND SHOW WORK,THANK YOU!!
4) Suppose that two identical
mass planets are sitting
million miles apart. At that
distance the planets have a
gravitational force of 1,000,000 N.
If the planets are moved
to two million miles apart, what
is the new gravitational force
between them?

Answers

The new gravitational force between the two planets, when they are moved to two million miles apart, is 250,000 N

The gravitational force between two objects can be calculated using Newton's Law of Universal Gravitation, which states that the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Given:

Initial distance between the planets = 1 million miles

Initial gravitational force = 1,000,000 N

Final distance between the planets = 2 million miles

To determine the new gravitational force, we need to compare the ratios of the distances and apply the inverse square law.

Let's denote the initial distance as d1, the initial gravitational force as F1, the final distance as d2, and the unknown final gravitational force as F2.

According to the inverse square law, the ratio of the gravitational forces is the square of the ratio of the distances:

(F2/F1) = (d1/d2)²

Substituting the given values:

(F2/1,000,000 N) = (1 million miles / 2 million miles)²

Simplifying:

(F2/1,000,000 N) = (1/2)²

(F2/1,000,000 N) = 1/4

F2 = (1/4) * 1,000,000 N

F2 = 250,000 N

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A squirrel runs across a branch on an oak tree and knocks an acorn and a leaf off the tree.
Which item will hit the ground first? Explain.

Answers

Answer: Acorn


Explanation: Acorns have more mass

If an object is placed between a convex lens and its focal point, which type of image will be produced?

Answers

Blury image or some type of incorrect postural image

is a pot plant a closed system?​

Answers

Answer:

No because a pot plant system must interact with its surroundings (air, water, etc) and these are not closed systems.

Only if the pot plant were completely enclosed, say in a large glass jar would the system be closed.

As part of astronaut training, a prospective astronaut is spun around in a human centrifuge such that the candidate experiences a centripetal acceleration that is 2.8 times the acceleration due to gravity on the surface of the earth. If the candidate is 11.05 m from the center, determine the candidate's speed in meters per second.

Answers

The candidate's speed (m/s), given that the candidate experiences a centripetal acceleration that is 2.8 times the acceleration due to gravity is 17.4 m/s

How do I determine the candidate's speed?

We understood that the centripetal acceleration is related to speed and radius according to the following formula:

a = v² / r

Cross multiply

v² = ar

Take the square root of both sides

v = √ar

Where

v is the speeda is the centripetal accelerationr is the radius

Withe the above formula, we can determin the speed of the candidate. Details below:

Acceleration due to gravity (g) = 9.8 m/s²Centripetal acceleration = 2.8 × g = 2.8 × 9.8 = 27.44 m/s²Radius (r) = 11.05 mSpeed of candidate (v) =?

v = √ar

v = √(27.44 × 11.05)

v = 17.4 m/s

Thus, the speed of the candidate is 17.4 m/s

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1) Which of the following is not true about vectors.a) Must have magnitude and direction.b) When doing math with vectors, we can usually treat the x and y components independently.c) If several vectors are added together, their order does not matter.d) The magnitude of a vector is the sum of the magnitude of its x and y components.2) Explain briefly your argument or reasoning.

Answers

a) A vector must have a magnitude and direction. A physical quantity with only the magnitude is called a scalar.

b) When doing the math we can treat the x and y components independently. We can use only x-component or only y-component for the necessary calculations. For example, the projectile motion. In projectile motion, we use the components of velocities independently.

c) We can add or subtract more than two vectors in any order. For example, If there are 3 vectors, A, B, and C then, from the associative law of vector addition,

\(\vec{A}+(\vec{B}+\vec{C})=(\vec{A}+\vec{B})+\vec{C}\)

d) The magnitude of the vectors is not the sum of their x and y components. Let vector A be represented as

\(\vec{A}=x\hat{i}+y\hat{j}\)

Where x and y are the x and y components of vector A respectively. And î and j cap are the unit vectors along the x and y direction respectively.

Then, the magnitude of vector A is given by,

\(A=\sqrt[]{x^2+y^2}\)

Thus the correct answer is option d. That is the statement in option d is not true.

12.
A hiker walks for 5km on a bearing of 053" true (North 53° East). She then turns and
walks for another 3km on a bearing of 107° true (East 17° South).
(a)
Find the distance that the hiker travels North/South and the distance that she travels
East/West on the first part of her hike.

Answers

The hiker travelled 4.02 km North/South and 4.74 km East/West during her hike.

This question involves vector addition, the resolution of vectors, the use of bearings, and trigonometry in the calculation of the hiker's movement.

This may appear to be a difficult problem, but with some visual aid and the proper use of mathematical formulas, the issue can be addressed correctly.

Resolution of VectorThe resolution of a vector is the process of dividing it into two or more components.

The angle between the resultant and the given vector is equal to the inverse tangent of the two rectangular components.

Angles will always be expressed in degrees in the solution.

The sine, cosine, and tangent functions in trigonometry are denoted by sin, cos, and tan.

The tangent function can be calculated using the sine and cosine functions as tan x = sin x/cos x. Also, in right-angled triangles, Pythagoras’ theorem is used to find the hypotenuse or one of the legs.

Distance Travelled North/SouthThe hiker traveled North for the first part of the hike and South for the second.

The angles that the hiker traveled in the first part and second parts are 53 degrees and 17 degrees, respectively.

The angle between the two is (180 - 53 - 17) = 110 degrees.

The angle between the resultant and the Northern direction is 110 - 53 = 57 degrees.

Using sine and cosine, we can calculate the north/south distance traveled to be 5 sin 57 = 4.02 km, and the east/west distance to be 5 cos 57 = 2.93 km.

Distance Travelled East/WestThe hiker walked East for the second part of the hike.

To calculate the distance travelled East/West, we must first calculate the component of the first part that was East/West.

The angle between the vector and the Eastern direction is 90 - 53 = 37 degrees.

Using sine and cosine, we can calculate that the distance travelled East/West for the first part of the hike is 5 cos 37 = 3.88 km.

To determine the net distance travelled East/West, we must combine this component with the distance travelled East/West in the second part of the hike.

The angle between the second vector and the Eastern direction is 17 degrees.

Using sine and cosine, we can calculate the distance traveled East/West to be 3 sin 17 = 0.86 km.

The net distance traveled East/West is 3.88 + 0.86 = 4.74 km.

Therefore, the hiker travelled 4.02 km North/South and 4.74 km East/West during her hike.

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6. a block of ice of 4 kg is at 0 C. what is the amount of heat to change this ice to water at
50 C

Answers

Hence, 1344 10 3 J of heat are needed to melt the ice. From 0°C to 4°C, heating water causes it to continuously contract rather than expand.

From where does energy come?

In accordance with the U.S. Energy Statistics Administration, coal, nuclear power, and natural gas made up the majority of the country's electrical generation in 2020. Renewable energy sources like wind, hydropower, solar, biomass, and geothermal are also used to generate electricity.

How does energy become made?

Using coal and oil, nuclear energy, biomass, geothermal, & solar thermal energy, steam turbines are used to produce the majority of the world's power. Gas turbines, hydroelectric turbines, wind turbines, & solar photovoltaics are some additional significant electricity generation systems.

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Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.30 cm. If the potential difference across the plates was 24.5 kV, find the magnitude of the electric field (in V/m) in the region between the plates.

Answers

Modern LCD and LED screens have taken the role of the Cathode Ray Tubes (CRTs) that were once used in televisions. The magnitude of the electric field (in V/m) in the region between the plates is 1.88 x 10^6 V/m.

The CRT is made up of an electron gun that produces a beam of electrons that are then directed by an electric field towards a fluorescent screen at the end of the tube. The electric field between the plates of a CRT (Cathode Ray Tube) has to be found. The plates are 1.30 cm apart and there is a potential difference of 24.5 kV between the plates. Here’s how to find the electric field: The formula to calculate the electric field between the plates is: `E = V/d`, where E is the electric field.

Substitute the given values in the formula: E = V/d = 24.5 kV/0.013 m= 1.88 x 10^6 V/m. Therefore, the magnitude of the electric field (in V/m) in the region between the plates is 1.88 x 10^6 V/m.

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A 45.0-kg girl stands on a 13.0-kg wagon holding two 18.0-kg weights. She throws the weights horizontally off the back of the wagon at a speed of 6.5 m/s relative to herself . Assuming that the wagon was at rest initially, what is the speed of the girl relative to the ground after she throws both weights at the same time

Answers

Answer:

v = 4.0 m/s

Explanation:

Assuming no external forces acting during the instant that the girl throws the weights, total momentum must be conserved.Since all the masses at rest initially, the initial momentum must be zero.So, due to momentum must keep constant, final momentum must be zero too, as follows:

       \(p_{f} = m_{w} * v_{w} + m_{g+w} *v_{g+w} = 0 (1)\)

Assuming the direction towards the back of the wagon as positive, and replacing the masses in (1), we can solve for vg, as follows:

       \(v_{g+w} =- \frac{m_{w} *v_{w}}{m_{g+w} } = - \frac{36.0kg *6.5m/s}{58.0kg } = -4.0m/s (2)\)

This means that the girl (along with the wagon on she is standing) will move at a speed of 4.0 m/s in an opposite direction to the one she threw the weights.

The half-life of a radioactive isotope is 210 d. How many days would it take for the decay rate of a sample of this isotope to fall to 0.58 of its initial rate?

Answers

It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.

2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.

3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.

4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.

5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.

6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".

7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).

8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).

9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.

10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.

11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).

12. Evaluating the expression, we find: t ≈ 546.

13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

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Sergio has made the hypothesis that "the more time that passes, the farther away a person riding a bike will be." Do the data in the table below support his hypothesis? A. Yes, the data support the hypothesis. B. No, the data support the opposite of the hypothesis. C. The data show no relationship between the time passed and the distance.

Answers

Answer:

Option A

Explanation:

Given that

Distance = Speed / Time

So, they are in inverse relation.

Such that when the time passes, the distance from the reacing point will become less and vice versa.

So, Yes! The more time that passes, the farther away a person riding a bike will be.

A can of soup has a mass of 0.35 kg. The can is moved from a shelf that is 1.2 m off the ground to a shelf that is 0.40
m off the ground. How does the gravitational potential energy of the can change?

Answers

Answer:

2.744 difference

Explanation:

Use Pe=mgh

So when the soup is at a height of 1.2m, its Pe is (.35kg)(9.8m/\(s^{2}\))(1.2m)=4.116

when the soup is at a height of .40m, its Pe is (.35kg)(9.8m/\(s^{2}\))(.40m)=1.372

So youre looking at a 2.744 difference in pe


The potential at point P is the work required to bring a one-coulomb test charge from far
away to the point P?

True or false ?

Answers

true so true Its so true

Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 1.61THz. Calculate the wavelength of the infrared radiation. Round your answer to 3 significant digits.

Answers

Answer:

λ = 1.86 x 10⁻⁴ m = 186 μm

Explanation:

The relationship between the wavelength and the frequency of a wave is given by the following equation:

\(c = f\lambda\\\\\lambda = \frac{c}{f}\)

where,

λ = wavelength of infrared radiation = ?

c = speed of infrared radiation = speed of light = 3 x 10⁸ m/s

f = frequency of infrared radiation = 1.61 THz = 1.61 x 10¹² Hz

Therefore,

\(\lambda = \frac{3\ x\ 10^8\ m/s}{1.61\ x\ 10^{12}\ Hz}\)

λ = 1.86 x 10⁻⁴ m = 186 μm

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