. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a
distance of 110 m. Determine the acceleration of the car.

The force the car will use to hit the tree, given it has a mass of 3000 Kg and accelerating at 2 m/s² is 6000 N

From the question given above, the filling data were obtained:

Force is related to acceleration according to the following formula:

Force = mass × acceleration

Thus, we can obtain the force as illustrated below:

Force = mass × acceleration

Force = 3000 × 2

Force = 6000 N

From the calculation made above, we can conclude that the force the car will use to hit the tree is 6000 N

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The partial pressure of gas expressed after equilibrium is 409 torr.

What is the partial pressure of gas expressed after equilibrium:

1. Write out the units given.

V1= 1 L

Ptot= 818 torr

PPB= 328 torr

V2= 1L

2. Remember= PaVa= PtotVtot

3. Add together volumes for total volumes.

1 L + 1 L= 2 liters

4. Plug in values and solve for PP of gas A.

PPA= Ptot x VA/Vtot

818 torr x 1/2 L= 409 torr

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Answer:

the 2nd one and the 3rd one

Answer:

1,3,4,5

Explanation:

just got the answer correct on edge

Answer:

momentum = mass x velocity

1×50=50_

mu=mv

mu+mv=(m1+m2)v

50×1+20×0.5=v(1+0.5)

v=60÷1.5

v=40

A horse that can run at a speed of 12 m/s will travel a distance of 52 m in a period of 4.33 seconds

We know that,

v = d / t

where,

v = Velocity

d = Distance

t = Time

Given that,

v = 12 m / s

d = 52 m

t = d / v

t = 52 / 12

t = 4.33 s

Velocity is the displacement of an object in a given amount of time. It is denoted by V.

Therefore, the time taken by the horse is 4.33 seconds

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The rate of deceleration of a racing car that reaches a speed of 42m/s is 7.64m/s.

Deceleration is the amount by which a speed or velocity decreases.

The rate of deceleration can be calculated using the following formula:

d = (v - u)/t

Where;

According to this question, a racing car reaches a speed of 42m/s. It then begins to decelerate, using its parachute and braking system, and comes to rest 5.5s later. The deceleration can be calculated as follows:

d = (42 - 0)/5.5

d = 42/5.5

d = 7.64m/s

Therefore, the rate of deceleration of a racing car that reaches a speed of 42 m/s is 7.64m/s.

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3N would bring about an extension of 0.02m.

A spring force is the force required or applied to compress or stretch a spring on an object attached to it. When an object exerts a force on a spring, the spring exerts an equal and opposite force on the object. It always acts to return the mass to its equilibrium position.

so here it is given that a spring has a natural length of 0.64m.It extended 0.024 m when a load of 3.6 N is applied.

formula

F₂/F₁=x₂/x₁

so

F₂=x₂/x₁*F₁

=0.02/0.024*3.6

=3N

So from the given data and using the equation we solved the proble and got 3N as the final answer

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The average speed for the two events is 25.2 m/s

Average speed is the ratio of the total distance travelled by a body to the total time. It is measured in m/s

If you drive in a straight line at 20.0 m/s for 10.0 miles, then at 34.0 m/s for another 10.0 miles. To calculate the average speed, first convert the mile to meter

1 mile = 1609.34 m

10 miles = 10 × 1609.34 = 16093.4 m

Calculate the time for each event.

speed = distance / time

time = distance / speed

T1 = 16093.4 / 20

T1 = 804.67 s

T2 = 16093.4 / 34

T2 = 473.34 s

Total time t = T1 + T2

t = 804.67 + 473.34

t = 1278.01 s

Total distance s = 16093.4 + 16093.4

s = 32186.8 m

Average speed = s / t

Average speed = 32186.8 / 1278.01

Average speed = 25.19 m/s

Therefore, the average speed for the two events is 25.2 m/s approximately.

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The concentration of (OH −) Hydroxide ions increase when the excessive base is added to the solution of Sodium Hydroxide as the base itself dissociates to give out OH −  ions which increase their concentration per unit volume.

when sodium hydroxide NaOH dissolves in water it separates into positively charged sodium ions cations and negatively charged hydroxide ions anions. these ions move round in the water, free and unbiased of every other, although cations have a tendency to be surrounded greater intently with the aid of anions and vice versa.

Sodium while sodium is added to water, the sodium melts to form a ball that moves around at the surface. It fizzes unexpectedly, and the hydrogen produced may also burn with an orange flame before the sodium disappears. The resulting huge concentration of OH- makes the solution greater basic and results in a dramatic growth inside the pH.

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The correct answer is 378000 km.

The speed of light is the rate at which light waves move through distinct substances. In instance, it has now been established that the speed of light in a vacuum is 299,792,458 metres per second.

Using the equation

distance = speed × time,

the total distance travelled is calculated.

Consequently, the overall distance travelled is

Distance = ( 3.00 × 10^8 m/s ) ( 2.52 )

= 756000000 m

The total distance travelled divided by two determines the distance between the Earth and the Moon.

Distance between earth to moon = 756000000/2

= 378000000 meters

= 378000 km.

Approximately 384 thousand kilometres separate the Earth and the Moon.

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The force of gravity between two bodies is 6.673 x 10-11 N m²/kg²

Greater gravitational forces will be used to attract heavier things since the force of gravity is directly related to the mass of both interacting components. Therefore, the gravitational pull between two objects grows as their respective masses do as well. The gravity force between two objects is equal to their respective masses multiplied by two. If one of the items triples in mass, the gravitational pull between them will also triple. The gravitational force between two objects is multiplied by two when their masses are each doubled, and so on. More distance will lead to decreased gravitational forces since the force of gravity is inversely related to the square of the difference between two interacting objects. Therefore, the gravitational pull between two things weakens as they go farther apart.

So any two items will be attracted to one another by a gravitational force of 6.673 x 10-11 N m² / kg²

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Since the magnification is equal to 1, the image height is​ the same as the object height which is 5 cm

Magnification is when the ratio of the size of the image to the size of the object is greater than 1

Given that an object of height 5cm is placed at 20cm from a concave mirror of focal length 10cm.

To find the image height, let us first find the image distance by using the formula below

1/f = 1/u + 1/v

Substitute all the parameters

1/10 = 1/20 + 1/v

1/v = 1/10 - 1/20

1/v = (2 - 1)/20

1/v = 1/20

v = 20 cm

Magnification = v/u = H/h

20/20 = H/5

H = 5 cm

Therefore, the image height is​ the same as the object height which is 5 cm

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The electric field along the [tex]z[/tex] axis is [tex]E\left( z \right) = \frac{{kqz}}{{{{\left( {{z^2} + {a^2}} \right)}^{\frac{3}{2}}}}}[/tex]

The radius of the ring is: [tex]a[/tex]

The charge on the ring is: [tex]q[/tex]

The coulomb's constant is: [tex]k = \frac{1}{{4\pi {\varepsilon _0}}}[/tex]

The expression of the electric field is: [tex]E = \frac{{kq}}{{{r^2}}}[/tex] .......(i)

Here,

[tex]r[/tex] ⇒ distance = [tex]\sqrt {{z^2} + {a^2}}[/tex]

Now, putting [tex]r[/tex] in equation (i), we get,

[tex]\begin{array}{c}\\E = \frac{{kq}}{{{{\left( {\sqrt {{z^2} + {a^2}} } \right)}^2}}}\\\\ = \frac{{kq}}{{\left( {{z^2} + {a^2}} \right)}}\\\end{array}[/tex]

The electric field along the z-direction is

[tex]dE\left( z \right) = dE\cos \theta[/tex] ........(ii)

here, [tex]\cos \theta = \frac{z}{{\sqrt {{z^2} + {a^2}} }}[/tex]

Integrating equation (ii), we get,

[tex]\begin{array}{c}\\\int {d{E_z}} = \int {dE\cos \theta } \\\\E\left( z \right) = E\cos \theta \\\end{array}[/tex].......(iii)

Now substituting the values in equation (iii), we get,

[tex]\begin{array}{c}\\E\left( z \right) = \left( {\frac{{kq}}{{\left( {{z^2} + {a^2}} \right)}}} \right)\left( {\frac{z}{{\left( {\sqrt {{z^2} + {a^2}} } \right)}}} \right)\\\\ = \frac{{kqz}}{{{{\left( {{z^2} + {a^2}} \right)}^{\frac{3}{2}}}}}\\\end{array}[/tex]

Therefore, the electric field along the [tex]z[/tex] axis is [tex]E\left( z \right) = \frac{{kqz}}{{{{\left( {{z^2} + {a^2}} \right)}^{\frac{3}{2}}}}}[/tex]

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Consider a uniformly charged ring in the [tex]xy[/tex] plane, centered at the origin. The ring has radius [tex]a[/tex] and positive charge [tex]q[/tex] distributed evenly along its circumference. What is the magnitude of the electric field along the positive [tex]z[/tex] axis? Use k in your answer, where [tex]k = \frac{1}{4 \pi \epsilon_0}[/tex]

It is thought that oxidation/reduction reactions take place in an atom's valence shell. Therefore, the existence or absence of additional valence electrons is related to the formal charge.

Any of the fundamentally negative charged atom-external particles called valence electrons participate in the synthesis of chemical bonds. The atomic number (Z) of oxygen is 8. The oxygen atom's nucleus has 8 protons (this is what makes it an oxygen atom).

Therefore, the neutral atom should have 8 electrons, but 2 of them are thought to be in the inner core. We typically represent oxygen in a molecule as R- OR, with 2 lone pairs on the oxygen. While the O center is thought to have half of the 2 electrons that make up the C-O bond or 1 electron, these 2 lone pairs are thought to have a charge that is totally connected with the O atom.

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The "arm" of the lever is the handle or bar; it is the portion that you push or pull against. The point on which the lever rotates or balances is known as the "fulcrum.

The pivotal point of the beam is known as the fulcrum. A load is applied at the other end of a lever when an effort is exerted to one end of the lever. A mass will be raised as a result.

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Answer:

The graph is a velocity-time graph, showing velocity as an f(t), (function of time).

The graph depicts the following:

For 6 seconds, the mailman travels at the constant rate of 3m/s. At second 6, he abruptly stops (when he stops for the mailbox which he overshot) which is represented by the dotted vertical line. He then stays at the constant rate of 0 m/s (stationary) for 2 seconds before he goes into reverse for 2 seconds at the rate of 2m/s before arriving back at the mailbox.

Some items to note:

In a real-world situation, the abrupt stop would be unrealistic and therefore may be a slope for about 1 second.

A dotted line represents a break in the graph, for our situation, it means the velocity has dropped.

A negative velocity, like in his graph, is just going in reverse.

To find the position of the mailbox:

Change the graph to a position-time graph which shows position as an f(t). Using this new graph, you can find the position of the mailbox using time at the 10th second.

Just a quick note, changing between types of graphs is an essential part of physics and you should be able to do it with no reference. It is in your best interest to learn this to be able to apply it in the future.

Hope that helps

Using time and velocity relationship, we concluded that he horse will be in air for 0.80 seconds.

velocity-time. In evenly accelerated, straight-line motion, the relationship between velocity and time is straightforward. The magnitude of the velocity change increases with lengthier acceleration. When acceleration is constant, the change in velocity is exactly proportional to time.

Both the horse and the human have constant accelerations, which are 0 for the horse and g downward for the human.

(A) Take into account the man's vertical motion after he leaves the limb (with v₀ = 0, yi = 3m).

Finding the man's time of fall from y = y₀ + vt + 1/2 a t² requires modeling him as a particle with continuous acceleration.

when v₀ = 0

t = √(2(y - y₀)/a)

= √ (2(0 - 3) / -9.28)

= √0.64

= 0.80s

A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 10.0m/s, and the distance from the limb to the level of the saddle is 3.00m. (a) For what time interval is he in the air?

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Answer: B

Explanation: I am 99% sure that is correct I remember doing something very simular

An interior designer is choosing a paint color for a room and has two samples that are both blue, however one blue is described as brighter than the other. Both would have the same wavelength but would differ due to the amplitude property of light.

It is the amplitude of light which makes one sample brighter than other.

The amplitude plays an important part in differentiating the same colors of the same wavelength. It tells us about the brightness or intensity of a certain light wave as compared to others of the same wavelength. So, whenever two same colors look different due to brightness, it is due to the amplitude property of the light.

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Time taken to slow down the speed will be 5 seconds.

Average acceleration of the bus, a = 1.8 m/s²

Initial velocity of the bus, u = 9.0 m/s

Final velocity of the bus, v = 0 m/s

According to the First Equation of the motion which relates the acceleration, velocity and time we can say that

v = u + at

v - u = at

0 - 9 = (-1.8) * t

9 = 1.8 * t

t = [tex]\frac{9}{1.8}[/tex]

t = 5 s

Therefore 5 seconds will be required by the bus to slow down its speed from 9.0 m/s to 0.0 m/s

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The final velocity would be 20m/s.

What is its final velocity?

Using the first equation of motion

V = U + at

where V = Final Velocity

U = Initial Velocity

a = Acceleration

t = Time

V = 10 + 1*10

V = 20 m/s.

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A finds a field that is the same as the field found by B.

The physical field that surrounds electrically charged particles and acts to either attract or repel all other charged particles in the field is known as an electric field (or E-field). Additionally, it describes the physical environment of a system of charged particles.

No matter how large the test charge is, stationary charges always produce the same electric field at a certain area.

At a distance "r" from it, the electric field created by a stationary charge "q" is given as

[tex]E = K_{q} /r^{2}[/tex]

Thus, the appropriate option is e) the same as the field found by B.

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I must fly about 62.56 km in the Southwest direction to reach a private landing strip that is 32.0 km due west of the college station airport.

This can be understood through concept of Vectors. Vectors are defined as entity which has both direction and magnitude.

According to the question, 27 km in a direction 34.0° south of east hence angle is -34°.

Along x axis = 27cos (-34) ° and along y axis = 27sin (-34) °

Along x axis ≈ 22.4 km and along y axis ≈ -15.1 km.

Now, fly 46 km due north.

Along y axis=46, along x axis = 0

Next, along x axis ≈ 22.4 km, Along y axis ≈ -15.1 + 46 ≈ 30.9 km

Hence (22.4, 30.9)

Final Position is 32 km due west is that is ( -32, 0)

Distance between (22.4, 30.9) and ( -32, 0)

√ {(22.4 - (-32)) ² + (30.9 - 0) ²}

≈ 62.56 km Direction Southwest

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Answer:According to the context, its nucleus breaks apart is an event that occurs when an atom undergoes nuclear fission.

Explanation:

The length of the second pendulum on earth is 1m and on the moon 1/6m.

Given:

The time period of a second pendulum T = 2s

By using the formula T = 2π √l/g we can calculate the length of the pendulum on earth and the moon.

Here g = acceleration due to gravity and

l = the length of the pendulum.

Now

l = g(T/ 2π)^2

l = 9.8 m/s^2(2s / 2 x 3.14)^2

le = 1 m

Now calculate the length of the pendulum on the moon

T =  2π √lm/gm      ∴ gm = ge/6

Now putting the value of g we get.

or 2 = 2 π √lm / ge/6

2 = 2 π √6lm / π ^2le

lm = le/6

now given le = 1m

lm = 1/6 m

Thus the length of the second pendulum on earth is 1m and on the moon 1/6m.

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Answer: Panama

Explanation:

The statement is wrong and the correct average speed of the student will be 3/2 or 1.5 miles per hour.

Given in the question,

The student climbs the mountain at a speed of 1 mile per hour, then the distance traveled by the student uphill is given by

Distance = Speed × Time,

Put in the values, we get

Distance = 2 × 1 = 2 miles

So the student travels 2 miles uphill and turns around and climbs back at 3 miles per hour.

So the time taken by the student to climb back is given by

Time = Distance/Speed

Time = 2/3 hr

So Now

Total distance traveled by the student = Uphill climb  +  Downhill Climb

Total distance traveled by the student = 2 + 2

Total distance traveled by the student = 4 miles

Total time taken = Time to climb uphill +  time to climb downhill

Total time taken = 2 +  2/3 hr

Total time taken = 8/3 hr

Now average speed is given by the formula,

Average Speed  =  Total Distance traveled/Total Time taken

Average Speed  = 4 / (8/3)

Average Speed  = (4 × 3) / 8

Average Speed  = 12 / 8

Average Speed  = 3/2 miles per hour

Therefore, the statement that the average speed will be 3 miles per hour is wrong and the correct average speed of the student will be 3/2 or 1.5 miles per hour.

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Two protons that are very far apart are hurled straight at each other, each with an initial kinetic energy of 0.16 MeV.

The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body holds onto the kinetic energy it acquired during its acceleration until its speed changes. The body exerts the same amount of effort when slowing down from its current pace to a condition of rest. Formally, a kinetic energy is any term that includes a derivative with respect to time in the Lagrangian of a system.

In classical physics, an object of mass m moving at a speed of v has kinetic energy.

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The purpose of the motion lab is to collect experimental data on two constant-speed cars with different constant speeds and a dynamic car showing acceleration.

The laboratory report must document the experiments and raw data included in the report,  provide sufficient information to reproduce or extend the data analyze the data, and draw conclusions. It has three main functions presenting and creating. Recommendations based on experimental work

In this lab, we were able to study projectile motion by determining the ball's initial velocity and finally predicting the projectile's range. In this projectile motion lab, we used the equations of motion for a projectile to predict distance. It combines the examination of a person's anatomy with the measurement of function during activity.

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