A 0.140-kg baseball is thrown with a velocity of 27.1 m/s. It is struck by the bat with an average force of 5000 N, which results in a velocity of 37.0 m/s in the opposite direction from the original velocity. How long were the bat and ball in contact?

Answer:

Acceleration of the asteroid relative to the spacecraft = 2ε\(E^{2}\)A/m

Explanation:

The maximum electric field in the beam = 2E

cross-sectional area of beam = A

The intensity of an electromagnetic wave with electric field is

I = cε\(E_{0} ^{2}\)/2

for \(E_{0}\) = 2E

I = 2cε\(E^{2}\)    ....equ 1

where

I is the intensity

c is the speed of light

ε is the permeability of free space

\(E_{0}\)  is electric field

Radiation pressure of an electromagnetic wave on an absorbing surface is given as

P = I/c

substituting for I from above equ 1. we have

P = 2cε\(E^{2}\)/c = 2ε\(E^{2}\)    ....equ 2

Also, pressure P = F/A

therefore,

F = PA    ....equ 3

where

F is the force

P is pressure

A is cross-sectional area

substitute equ 2 into equ 3, we have

F = 2ε\(E^{2}\)A

force on a body = mass x acceleration.

that is

F = ma

therefore,

a = F/m

acceleration of the asteroid will then be

a = 2ε\(E^{2}\)A/m

where m is the mass of the asteroid.

The image will be formed at a distance of 30 cm in its front. Hence, this is the required solution.

Given that,

Object distance, u = -15 cm

The radius of curvature of the concave mirror, R = 20 cm

Focal length, f = R/2 = -10 cm (negative for concave mirror)

Let v is the distance between mirror and the formed image. Using mirror's formula to find it as :

\($\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$\)

\($\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$\)

\($\frac{1}{v}=\frac{1}{(-10)}-\frac{1}{(-15)}$\)

\($v=-30 \mathrm{~cm}$\)

So, the image will be formed at a distance of 30 cm in its front. Hence, this is the required solution.

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Answer: a

Explanation:

At a divergent boundary, heat and pressure can cause metamorphism, magma can solidify to form igneous rocks, and sediments can accumulate and lithify to form sedimentary rocks. These three processes contribute to the formation of various types of rocks.

At a divergent boundary, three main processes occur that can lead to the formation of metamorphic, igneous, and sedimentary rocks:

1. Metamorphic rocks: Heat and pressure from the divergent boundary can cause existing rocks to be metamorphosed, or transformed, into new rocks with different textures and mineral compositions. This process is called metamorphism.

2. Igneous rocks: Magma, which is molten rock, can rise to the surface at a divergent boundary and cool and solidify to form igneous rocks. This process is called solidification or crystallization.

3. Sedimentary rocks: Sediments, such as sand and mud, can accumulate in the low-lying areas near a divergent boundary, such as in rift valleys or on the continental shelf. Over time, these sediments can be compacted and cemented together to form sedimentary rocks. This process is called lithification.

Hence, Heat and pressure at a divergent boundary can produce metamorphism, solidify magma to create igneous rocks, and accumulate and lithify sediment to create sedimentary rocks. Different sorts of rocks are formed as a result of these three processes.

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Answer:

0.238 m/min

Explanation:

The volume of water in the trough V =Ah' where A = area of cross-section = area of isosceles trapezoid = 1/2(a + b)h where a = length of bottom of isosceles trapezoid = 40 cm = 0.4 m, b =  length of top of isosceles trapezoid = 100 cm = 1 m and h = height of isosceles trapezoid = 60 cm = 0.6 m. So,

A = 1/2(a + b)h = 1/2(0.4 m + 1 m)0.6 m = (1.4 m)0.3 m = 0.42 m² and h' = height of water level in trough = H - h" where H = length of trough = 10 m and h" = depth of water level in trough = 10 cm = 0.1 m

So, V = Ah'

V = A(H - h") = A(10 - h")

Now, the rate of change of volume of the trough with respect to time dV/dt = d[A(10 - h")]/dt

dV/dt = -Adh"/dt

dh"/dt = -dV/dt/A

Since dV/dt = 0.1 m³/min, substituting the other variables into the equation, we have

dh"/dt = -dV/dt/A

dh"/dt = -0.1 m³/min/0.42 m²

dh"/dt = -0.238 m/min

This is the rate at which the depth is decreasing

Since the height h' = 10 - h"

dh'/dt = d(10 - h")/dt

= -dh"/dt

= -(-0.238 m/min)

= 0.238 m/min

So the water level is increasing at a rate of 0.238 m/min

ANSWER AND EXPLAINATION:
To calculate the impulse imparted on the baseball, we can use the impulse-momentum principle, which states that the impulse experienced by an object is equal to the change in momentum of the object. Mathematically, it can be expressed as:

Impulse = Change in momentum

The momentum of an object is given by the product of its mass and velocity:

Momentum = mass × velocity

In this case, the baseball has an initial mass of 0.14 kg and an initial velocity of 32 m/s. After being struck by the bat, it travels in the opposite direction at a velocity of 49 m/s.

Therefore, the change in momentum is given by:

Change in momentum = (mass × final velocity) - (mass × initial velocity)

Change in momentum = mass × (final velocity - initial velocity)

Change in momentum = 0.14 kg × (49 m/s - (-32 m/s))

Change in momentum = 0.14 kg × (49 m/s + 32 m/s)

Change in momentum = 0.14 kg × 81 m/s

Change in momentum = 11.34 kg·m/s

So, the impulse imparted on the baseball is 11.34 kg·m/s.

Answer:

About 6 feet or higher is recommended.  

Explanation

The reaction of load L is 0 (no horizontal force), and the reaction of load R is 10 kN (vertical upward force).

To determine the reactions of the loads L and R, consider the equilibrium of the forces acting on the structure.

First, analyze the vertical equilibrium. The sum of the vertical forces must be zero:

ΣFy = R − 7 kN − 3 kN

ΣFy = 0

This gives:

R = 10kN

Next, analyze the horizontal equilibrium. The sum of the horizontal forces must be zero:

ΣFx = L

ΣFx = 0

This indicates that there is no horizontal force acting on the structure.

Therefore, the reaction of load L is zero (no horizontal force), and the reaction of load R is 10 kN (vertical upward force).

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The two strings are taut. When the system is released from rest, it accelerates at 2 m/s2 then, Mass of A = 8m/5 kg.

Let the mass of the body A be ‘m’.

The two strings are taut so they exert a tension ‘T’ on body A.

Let ‘a’ be the acceleration produced in the system.

The free body diagram of body A is given below: mA + 2T = mA + ma = mA + m(2)mA + 10T = mA + ma = mA + m(2)

As the two strings are taut, we can say that tension in both strings is equal.

Therefore 2T = 10T or T = 5T As the body A is resting on a smooth horizontal table, there is no friction force acting on the body A.

The net force acting on body A is the force due to tension in the strings. ma = 2T – mg …(1)

As per the given problem, the system is released from rest.

Hence the initial velocity is zero.

Also, we are given that the system accelerates at 2 m/s2.

Therefore a = 2 m/s2 …(2)

From the equations (1) and (2), we get, m(2) = 2T – mg …(3)⇒ m(2) = 2×5m – mg⇒ 2m = 10m – g⇒ g = 8m/5

Thus, the mass of A is 8m/5 kg.

Answer: Mass of A = 8m/5 kg.

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Answer:

Explanation:

f =ma

F 2500 kg (20m/s-0m/s/15s)

F=2500kg(1.33m/s²)

F =3,325 N

f =ma, F 2500 kg (20m/s-0m/s/15s) and F=2500kg(1.33m/s²), F =3,325 N.

A push or pull that an object experiences as a result of interacting with another item is known as a force. Every time two items interact, a force is exerted on each of the objects.

The force is no longer felt by the two objects when the interaction ends. Only when there is interaction do forces actually exist.

Action-at-a-distance forces are those that happen even when the two interacting objects are not physically touching each other but are nevertheless able to push or pull against each other despite their physical separation. Gravitational forces are an example of action-at-a-distance forces.

Therefore, f =ma, F 2500 kg (20m/s-0m/s/15s) and F=2500kg(1.33m/s²), F =3,325 N.

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1) The photoelectric effect is the phenomena by which, the metals release electrons when they are exposed to electromagnetic radiation with the suitable frequency. The photoelectrons are the electrons that are released in this process.

So, the statement is true.

2) In a particle model, energy transfer can be done through many ways such as:

The energy jumps between the particles in the form of photons.

Electrons can absorb or emit energy during energy transfer.

The energy is transferred between different objects in the form of photons.

So, all of them are true.

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The mass of the ball is approximately 10.24 kg, found using the conservation of energy principle.

What is the mass of a ball thrown upward with a speed of 10 m/s from a height of 20 m, using the conservation of energy principle?

To find the answer, we can use the conservation of energy principle, which states that the total energy of a system remains constant if there is no net work done on the system.

Initially, the ball has some potential energy due to its height above the ground and kinetic energy due to its motion. As it travels upwards, its potential energy increases while its kinetic energy decreases until it reaches its highest point, at which its kinetic energy is zero and its potential energy is at its maximum. Then, as it falls back down, its potential energy decreases while its kinetic energy increases until it reaches the ground, at which its potential energy is zero and its kinetic energy is at its maximum.

Let's use the following variables:

m = mass of the ball

v_i = initial velocity of the ball (10 m/s)

v_f = final velocity of the ball (0 m/s at the highest point)

h = height of the ball above the ground (20 m)

g =  (9.81 m/s^2)

The total energy of the ball initially is:

E_i = 1/2 * m * v_i^2 + m * g * h

At the highest point, the total energy of the ball is:

E_f = 1/2 * m * v_f^2 + m * g * 2h

Since the ball is at rest at the highest point, its final velocity is zero, so:

E_f = m * g * 2h

By the conservation of energy principle, the total energy of the ball is constant, so:

E_i = E_f

Substituting the expressions for E_i and E_f and solving for the mass of the ball:

1/2 * m * v_i^2 + m * g * h = m * g * 2h

1/2 * v_i^2 + g * h = 2g * h

m = (1/2 * v_i^2 + g * h) / g

m = (1/2 * 10^2 + 9.81 * 20) / 9.81

m = 10.24 kg

Therefore, the mass of the ball is approximately 10.24 kg.

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Thee tilt of earth's axis causes the northern hemisphere to be closer to the sun than the southern hemisphere in summer, and vice versa in winter.

In this case, option A is correct.

Describe the seasons: Each season is a distinct time of year that can be recognized by special climatic characteristics. Every year, the seasons of spring, summer, fall, and winter follow one another. Each experiences cyclical patterns of light, climate, and weather every year.

Why are the seasons so important?

Soil saturation, precipitation patterns, river flows, lake concentrations, and snow cover are all impacted by seasonal variations in temperature and precipitation. Plants deteriorate and leaves fall as the wet and cold seasons approach. These alterations in the flora have an impact on the type and availability of food for humans and other animals.

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Explanation:

Object is under the effect of the acceleration of gravity

v = 1/2 a t^2       a = 9.81 m/s^2   t = 2.3

v = 25.9 m/s^2 = ~ 26 m/s   ( two significant digits)

Answer:

Frequency of oscillation will increase

Explanation:

Because The period of a pendulum is inversely proportional to the square root of the gravitational acceleration, so as. Gravity increases on the moon period of oscillation decreases whereas period of oscillating pendulum is inversely proportional to frequency thus frequency is directly proportional to gravity so as gravity increases on the moon the frequency will increase

Answer:

a) Therefore the period of oscillation in the moon will increase.

\(T_{E}<T_{M}\)

b) The frequency in the moon will decrease

\(\omega_{E}>\omega_{M}\)

Explanation:

The period of the pendulum is given by this equation:

\(T=2\pi\sqrt{\frac{L}{g}}\) (1)

As we can see, T is proportional to L and inversely proportional to the gravity vector. We know that g in the earth is grader than g in the moon.

\(g_{E}>g_{M}\) (2)

Therefore the period of oscillation in the moon will increase.

\(T_{E}<T_{M}\) (3)              

Now, the frequency of oscillation is:

\(\omega=\sqrt{\frac{g}{L}}\)

In this case, ω is proportional to g, then using (1) we can conclude that:

\(\omega_{E}>\omega_{M}\)

Therefore the frequency in the moon will decrease.

I hope it helps you!

Answer:

A and C I think

Explanation:

The two statements describe behaviors of particles that are related to the physical properties of the materials are A and C.

Physical properties are measurable property whose value represents a condition of an object.

The two statements describe behaviors of particles that are related to the physical properties of the materials are ;

A. They allow electrons to move freely between them.

c. They change their positions relative to one another.

The two statements describe behaviors of particles that are related to the physical properties of the materials are A and C.

Hence, option A and C are correct.

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Answer:

hi

Explanation:

hiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

Answer:

5.00

Explanation:

Answer:

90% of people marry there 7th grade love. since u have read this, u will be told good news tonight. if u don't pass this on nine comments your worst week starts now this isn't fake. apparently if u copy and paste this on ten comments in the next ten minutes you will have the best day of your life tomorrow. you will either get kissed or asked out in the next 53 minutes someone will say i love you

Explanation:

plz do so it is true

The increase in the temperature of the soil when bag fall from the given height is 0.065 ⁰C.

The given parameters:

Apply the principle of conservation of energy as follows;

\(mgh = mC \Delta T\\\\gh = C \Delta T\\\\\Delta T= \frac{gh}{C}\)

Convert the value of C in kcal/kg ⁰C  to  J/kg ⁰C

1 kcal = 4180 J

\(0.200 \ kcal/kg ^0 C= 0.2 \times 4180 (J/kg ^0C) = 836 \ J/kg^0 C\)

The increase in the temperature of the soil is calculated as follows;

\(\Delta T= \frac{gh}{C}\\\\\Delta T= \frac{9.8 \times 5.5}{836} \\\\\Delta T= 0.065 \ ^0C\)

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Answer:

it depends on wether the + and - are facing eachother

or away from eachother

Explanation:

Answer:

Yes, they are correct, you need to include all parts of the question, don't just copy and paste the words. But in the diagram that was supposed to be there, the magnets' north poles are facing each other. Therefore they would push away from each other if they were to get any closer.  So, the magnetic potential energy would decrease due to their movement away from each other. So your answer is (C. It will decrease.

Explanation:

A. The height the motorcycle can coast up the hill if friction is neglected is 1.63m.

B. Energy lost to friction before coming to rest is 56595J.

A. The height the motorcycle can coast up the hill, neglecting friction, is determined by the initial kinetic energy of the motorcycle. Using the equation for kinetic energy, KE = 1/2mv^2, we can calculate the initial kinetic energy of the motorcycle:

KE = 0.5 * 175kg * 32m/s^2 = 2800 J

The gravitational potential energy gained by the motorcycle is equal to the initial kinetic energy. Using the equation for gravitational potential energy, PE = mgh, we can calculate the height, h, the motorcycle can coast up the hill:

PE = 2800J = mgh

h = 2800J / (175kg * 9.8m/s^2) = 1.63m

Therefore, the motorcycle can coast up the hill to a height of 1.63m, neglecting friction.

B. The amount of energy lost to friction can be calculated using the equation for work, W = Fd. If the motorcycle only gains an altitude of 33m before coming to rest, then the force of friction, F, must be equal to the weight of the motorcycle, and the distance traveled by motorcycle, d, is 33m.

W = Fd = (175kg * 9.8m/s^2) * 33m = 56595J

Therefore, the amount of energy lost to friction is 56595J.

Therefore,

A. The height the motorcycle can coast up the hill if friction is neglected is 1.63m.

B. Energy lost to friction before coming to rest when altitude gained is 33m is 56595J.

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Answer:

Explanation:

From the given information if we assume that the runner didn't  come back to her original position.

Her initial velocity = 0 m/s

Final Velocity = 9.6 m/s

Time Taken = 2 seconds

She started from rest therefore her initial time would be 0 seconds.

We know that,

Average acceleration = (final velocity - initial velocity)/(final time - initial time)

=(v-u)/t1-t0

=(9.6 - 0)/(2-0) m/s^2

=4.8 m/s^2

Answer:

\(\triangle v=1.57*10^-^3^5miles/hour\)

Explanation:

From the question we are told that

Mass of of automobile M= 1500 kg

Generally the equation of the uncertainty principle is  given by

\(\triangle p=\frac{h}{ \triangle x}\)

\(\triangle p=\frac{1.054*10^-^3^4}{ 10^-^2}\)

\(\triangle p=1.054*10^-^3^2\)

Generally the uncertainty in velocity is mathematically given by

\(\triangle v=\frac{\traingle p}{m}\)

\(\triangle v=\frac{1.054*10^-^3^2}{1500}\)

\(\triangle v=7*10^-^3^6m/s\)

Therefore the minimum uncertainty in the velocity (in miles per hour)

\(\triangle v=7*10^-^3^6*\frac{3600}{1609.3}\)

\(\triangle v=1.57*10^-^3^5miles/hour\)

The correct option is A. which states that in the same liquid, the pressures are equal at all points that are at the same distance below the liquid surface.

This is because the pressure is determined by the weight of the liquid column above it, which is the same for all points at the same depth. The pressure along the vertical walls of the container is not equal because the liquid is in hydrostatic equilibrium which means that the pressure is not the same at all points in the liquid, and since the walls are vertical, the pressure is the same along them. The pressure at any point not touching any immersed objects is not equal to the pressure at any other point not touching any immersed objects because the pressure is determined by the weight of the liquid column above it, which is the same for all points not touching any immersed objects. The pressure is not affected by factors such as sunlight since the pressure at any point  in the sunlight is equal to the pressure at any other point not in the sunlight.

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Answer:

The correct solution is "11.51 mA".

Explanation:

Given:

Time average power,

\(P_{avg}=0.05 \ W/m^2\)

n = 377

As we now,

⇒ \(P_{avg}=\frac{E_0^2}{n}\)

or,

⇒ \(E_0^2=0.05\times 377\)

⇒       \(=4.341 \ V\)

hence,

⇒ \(H_0=\frac{E_0}{n}\)

By putting the values, we get

         \(=\frac{4.341}{377}\)

         \(=11.51 \ mA\)

Answer:

"a certain compact disc (CD) contains 783.216 megabytes of digital information 4"

(a)  the horizontal distance between the saddle and limb when the man makes his move is approximately 11.386 meters.

(b)  the man is in the air for approximately 0.843 seconds.

To determine the horizontal distance between the saddle and limb when the man makes his move, we need to consider the horizontal velocity of the man when he drops from the tree limb.

Given:

Speed of the horse (constant velocity), v = 13.5 m/s

Vertical distance between the limb and saddle, h = 3.55 m

a) To find the horizontal distance, we can use the formula:

horizontal distance = horizontal velocity × time

Since the man drops vertically, his initial horizontal velocity is zero. The only horizontal velocity he will have is due to the motion of the horse.

The time taken by the man to fall can be determined using the equation for free fall:

h = (1/2) × g × t²

Where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time.

Rearranging the equation, we get:

t = √(2h / g)

Substituting the given values:

t = √(2 × 3.55 / 9.8) ≈ 0.843 s

Now, we can find the horizontal distance:

horizontal distance = v × t

horizontal distance = 13.5 × 0.843 ≈ 11.386 m

Therefore, the horizontal distance between the saddle and limb when the man makes his move is approximately 11.386 meters.

b) The time the man is in the air can be calculated using the same equation for free fall:

t = √(2h / g)

Substituting the given value of h:

t = √(2 × 3.55 / 9.8) ≈ 0.843 s

Thus, the man is in the air for approximately 0.843 seconds.

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