Physics College

## Answers

**Answer 1**

For this problem, all of a skater's energy is either kinetic or potential, and the amount of energy has to stay conserved. When a skater has the lowest potential energy, it has the highest kinetic energy. Potential energy is lowest at the lowest point of the track, so the **lowest point of the track has the highest kinetic energy**.

**Answer 2**

The lowest part so the dot right next to the skater

## Related Questions

A light bulb acts as a _____ in a circuit, transforming electrical energy into light and heat.A.switchB.resistorC.conducting wireD.energy source

### Answers

When the current passes through the filament of the light bulb, due to the resistance, the filament gets heated up.

This heating of the filament causes the emission of light from the light bulb.

In order to avoid oxidization of filament or for the emission of different colors, inert gases are filled in the bulb.

**Thus, the light bulb act as a resistor in a circuit, transforming electrical energy into light and heat.**

**Hence, option B is the correct answer.**

At a carnival, you ride the "Tilt a Whirl" that achieves an angular speed of 0.76 radians per second. If the ride has a radius of 5 meters, what linear speed in meters per second of a person standing on the wall of the ride achieve?

### Answers

**Given data**

*The given angular speed is

[tex]\omega=0.76\text{ rad/s}[/tex]

*The radius of the ride is **r **= 5 m

The formula for the linear speed in meters per second of a person standing on the wall of the ride is given as

[tex]v=\omega r[/tex]

Substitute the known values in the above expression as

[tex]undefined[/tex]

Problem 3 A small ball is launched at an angle of 30.0 degrees above the horizontal. It reaches a maximum height of 2.5 m with respect to the launch position. Find (a) the initial velocity of the ball when it’s launched and (b) its range, defined as the horizontal distance traveled until it returns to his original height. As always you can ignore air resistance.(a) Initial velocity[Hints: How is v0 related to vx0 and vy0. How can you use the information given to calculate either or both of the components of the initial velocity?](b) Range[Hints: This problem is very similar to today’s Lab Challenge except that for the challenge the ball will land at a different height.]

### Answers

a)

In order to find the initial velocity of the ball, we can use the formulas below:

[tex]\begin{gathered} v_{y0}=v_0\cdot\sin (\theta) \\ v^2_y=v^2_{y0}+2\cdot g\cdot d \end{gathered}[/tex]

At the maximum height, the vertical speed is zero. So, using theta = 30°, d = 2.5 m and vy = 0, we have:

[tex]\begin{gathered} v^2_y=v^2_{y0}+2\cdot g\cdot d \\ 0^2=v^2_{y0^{}}+2\cdot(-9.8)\cdot2.5 \\ v^2_{y0}=49^{} \\ v_{y0}=7 \\ \\ v_{y0}=v_y\cdot\sin (\theta) \\ 7=v_y\cdot\frac{1}{2} \\ v_y=14\text{ m/s} \end{gathered}[/tex]

b)

To find the range, let's calculate the horizontal component of the velocity and the time of flight:

[tex]\begin{gathered} v_x=v\cdot\cos (\theta)_{} \\ v_x=14\cdot\frac{\sqrt[]{3}}{2} \\ v_x=12.124\text{ m/s} \\ \\ v_y=v_{y0}+g\cdot t \\ 0=7-9.8\cdot t \\ t=\frac{7}{9.8} \\ t=0.714 \\ \\ t_f=2t=1.43\text{ s} \\ \\ d_x=v_x\cdot t \\ d_x=12.124\cdot1.43 \\ d_x=17.34\text{ m} \end{gathered}[/tex]

In the figure below, A is a 44 N block and B is a 22 N block. (a) Determine the minimum weight (block C) that must be placed on A to keep it from sliding, if the coefficient of static friction between A and the table is 0.20. (b) Block C suddenly is lifted off A. Calculate the acceleration of block A, if the coefficient of kinetic friction between A and the table is 0.15.

### Answers

Given data:

* The weight of the block A is 44 N.

* The weight of the block B is 22 N.

* The coefficient of static friction between A and table is 0.20.

Solution:

(a). The weight of the A and C block is balanced by the weight of the block B.

Thus, the friction force acting on the Block A in terms of the weight of the Block B is,

[tex]W_B=\mu_sW_{AC}[/tex]

This describe the static friction force acting on the Block A ( with block C above it) in contact with the table.

where W_AC is the combined weight of block A and C, W_B is the weight of the block B, and

[tex]\mu_s\text{ is the static friction}[/tex]

Substituting the known values,

[tex]\begin{gathered} 22\text{ = }0.2\times W_{AC} \\ W_{AC}=\frac{22}{0.2} \\ W_{AC}=110\text{ N} \end{gathered}[/tex]

Thus, the minimum weight of the block C in terms of combined weight is,

[tex]\begin{gathered} W_C=W_{AC}-W_A \\ W_C=110-44 \\ W_c=66\text{ N} \end{gathered}[/tex]

**Hence, the minimum weight of the block C is 66 N.**

(I) Calculate the mass m needed in order to suspend the leg shown in Fig. 9–47. Assume the leg (with cast) has a mass of 15.0 kg, and its cg is 35.0 cm from the hip joint; the cord holding the sling is 78.0 cm from the hip joint.

### Answers

The **mass m** needed in order to suspend the leg is 6.73 kg if the leg (with cast) has a mass of 15.0 kg, and its cg is 35.0 cm from the hip joint; the cord holding the sling is 78.0 cm from the hip joint.

**τ = r F sin θ**

τ = Torque

r = Radius

F = Force

θ = Angle between r and F

θ = 90°

F = m g

τ = r m g

Counter clockwise torque is positive and clockwise torque is taken as negative. So torque due to center of gravity is negative and the torque due to cast is positive.

m1 = 15 kg

r1 = 35 cm = 0.35 m

r2 = 78 cm = 0.78 m

Since the whole system is at rest,

**∑ τ = 0**

- τ 1 + τ 2 = 0

- r1 m1 g + r2 m2 g = 0

( - 0.35 * 15 * 9.8 ) + ( 0.78 * m2 * 9.8 ) = 0

m2 = 51.45 / 7.64

m2 = 6.73 kg

Therefore, the mass m needed in order to suspend the leg is **6.73 kg**

To know more about **torque**

https://brainly.com/question/28220969

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Two part question need help with homework1.)What can you say about snowboards kinetic energy as he jumps off a rail? What can you say about the potential energy of the snowboarder? Based on your observations what can you say about your prediction above concerning the potential and kinetic energy?

### Answers

**Explanation**

**Step 1**

The potential energy of a body is due to its mass, the height at which it is, and the acceleration due to gravity.

it is given by the expression:

[tex]PE=mgh[/tex]

hence

at top

[tex]\begin{gathered} PE=maximum \\ KE=0 \end{gathered}[/tex]

at b

[tex]\begin{gathered} PE=0 \\ KE=\left(maximum\right) \end{gathered}[/tex]

so, as the potential energy depends on on the heigth, the maximum potential energy is when he is on top(A)

Kinetic energy is a form of energy that an object or a particle has by reason of its motion.

[tex]KE=\frac{1}{2}mv^2[/tex]

As the snowboarder loses height, he gains speed, so all the potential energy is converted into kinetic energy, a little energy is lost due to the friction of the wind and the friction of the board with the snow (although it is minimal).

I hope this helps you

What magnitude charge creates a 19.59 N/C electric field at a point 9.52 m away?

### Answers

**Given:**

The magnitude of the electric field is: E = 19.59 N/C

The distance at which the electric field is measured is: r = 9.52 m

**To find:**

The magnitude of the charge.

**Explanation:**

The expression for the electric field intensity E is given as:

[tex]E=k\frac{q}{r^2}[/tex]

Here, k = 4πε₀ ≈ 8.99 × 10⁹ N.m²/C², q = the magnitude of the charge.

Rearranging the above equation, we get:

[tex]q=\frac{Er^2}{k}[/tex]

Substituting the values in the above equation, we get:

[tex]\begin{gathered} q=\frac{19.59\text{ N/C}\times\text{ \lparen}9.52\text{ m\rparen}^2}{8.99\times10^9\text{ N.m}^2\text{/C}^2} \\ \\ q=\frac{19.59\text{ N/C}\times90.6304\text{ m}^2}{8.99\times10^9\text{ N.m}^2\text{/C}^2} \\ \\ q=1.975\times10^{-7}\text{ C} \\ \\ q=0.1975\text{ }\times10^{-6}\text{ C} \\ \\ q\approx0.1975\text{ }\mu\text{C} \end{gathered}[/tex]

**Final answer:**

**The magnitude of the charge is 0.1975 microCoulombs.**

Stephen is looking through some design diagrams created for a specific application.He spots a diagram which uses a parallelogram component.What component could it be?

### Answers

In the image you can see it is part of a flow chart, it can be used to describe a process showing its steps, output, inputs, and the desicions that can be taken during the process

Attribut in an ER diagram

In each part calculate the kinetic energy of the given objects in joules c) an electron (mass 9.11x10^-31kg) moving at a speed of 2.74x10^7m/s

### Answers

The kinetic energy of an object with mass m that moves with speed v is given by:

[tex]K=\frac{1}{2}mv^2[/tex]

Plugging the mass and speed given in the equation above we have that:

[tex]\begin{gathered} K=\frac{1}{2}(9.11\times10^{31})(2.74\times10^7)^2 \\ K=3.42\times10^{-16} \end{gathered}[/tex]

**Therefore, the kinetic energy of the electron is:**

[tex]3.42\times10^{-16}\text{ J}[/tex]

A wire carrying 19.6 amps of current has a length of 29.5 centimeters within a magnetic field of strength 0.946 teslasa. What is the force on the wire if it makes an angle with the magnetic field of 90.0 degrees? b. What is the force on the wire if it makes an angle with the magnetic field of 29.6 degrees?

### Answers

Given:

Length , L = 29.5 cm

Current, I = 19.6 A

B = 0.946

Let's find the following:

(a). Force on the wire if it makes an angle with the magnetic field of 90.0 degrees.

To find the force, apply the formula

[tex]F=BILsin\theta[/tex]

Where:

B is the magnetic field strength = 0.946 T

I is the current = 19.6 A

L is the length of the wire in meters.

Here, the length is in cm, let's convert from cm to m.

We have:

1 cm = 0.01 m

29.5 cm = 29.5 x 0.01 = 0.295 m

Plug in the values and solve for the force, F.

[tex]\begin{gathered} F=0.946*19.6*0.295sin90 \\ \\ F=5.47\text{ N} \end{gathered}[/tex]

**Therefore, the force when the angle is 90 degrees is **** 5.47 N.**

• (b). The force on the wire when the angle is 29.6 degrees.

[tex]\begin{gathered} F=0.946*19.6*0.295sin29.6 \\ \\ F=2.7\text{ N} \end{gathered}[/tex]

**The force when it makes an angle of 29.6 degrees is **** 2.70 N.**

**ANSWER:**

**(a). 5.47 N**

**(b). 2.70 N**

Which causes the more severe burn hot water or steam? Why?

### Answers

**Answer:**

steam

**Explanation:**

Steam can cause worse burns than hot water. This is because when steam touches your skin, it turns back into liquid. When this happens, it releases energy. That energy, along with the heat itself, contributes to how bad the burn is.

calculate the gravitational potential Energy of the ball relative to the ground before being thrown

### Answers

Given,

The mass of the ball, m=0.11 kg

The height of the ball before it was thrown, h₁=0.24 m

The height of the ball after it was caught by Donald, h₂=0.82 m

The gravitational potential energy is the energy stored in an object due to its position above the ground.

The gravitational potential energy of the ball before being thrown is given by,

[tex]E=\text{mgh}_1[/tex]

On substituting the known values,

[tex]\begin{gathered} E=0.11\times9.8\times0.24 \\ =0.26\text{ J} \end{gathered}[/tex]

**Thus the gravitational potential energy of the ball relative to the ground, before being thrown is 0.26 J**

A gymnast falls from a height onto a trampoline. For a moment, both the gymnast’s kinetic energy and gravitational potential energy are zero. How is the gymnast’s mechanical energy stored for that moment?1) rest energy2) chemical energy3) elastic energy4) thermal energy

### Answers

**ANSWER**

**3) elastic energy**

**EXPLANATION**

When the gymnast reaches the trampolin, both the gymnast and the trampolin move down because of the elasticity of the trampolin, until the trampolin is at its maximum point of which it can stretch. At this point the gravitational energy of the gymnast is zero because of zero height and its kinetic energy is zero too, because of zero energy. Therefore all the energy is stored as **elastic energy.**

8. An object of mass 4 kg moving with a speedof 29 m/s to the right collides with an objectof mass 11.5 kg moving with a speed of 18 m/sto the left. After collision, the 11.5 kg object movesto the right with a speed of 4 m/s to the right. Calculatethe velocity of the 4 kg object after collision. (1 point)A. O-53.475 m/sB. O-13.341 m/sC. O-34.25 m/sD. -19.579 m/sE. O-27.7 m/s9. An object of mass 2 kg moving with a speed

### Answers

**Given**

Mass of one object, m=4 kg

Velocity of the object, u=29 m/s towards the right

Mass of the other object, m'=11.5 kg

Velocity of the other object, u'=18 m/s towards left

After collision

The object with mass 11.5 kg moves at a speed, v'=4 m/s towards right

**To find**

The velocity of the object with mass 4 kg after collision

**Explanation**

Let the direction towards right be positive

By conservation of momentum

[tex]\begin{gathered} mu+m^{\prime}u^{\prime}=mv+m^{\prime}v^{\prime} \\ \Rightarrow4\times29-11.5\times18=4v+11.5\times4 \\ \Rightarrow v=-34.25\text{ m/s} \end{gathered}[/tex]

**Conclusion**

The velocity of the object is C**.-34.25 m/s**

A long jumper with a speed of 120m/s at an angle of 30 degrees with respect to the horizontal. how far did he land in the sand from his jump spot?

### Answers

Given data:

* The initial velocity of the jumper is,

[tex]u=120ms^{-1}[/tex]

* The angle between the initial velocity and the horizontal line is,

[tex]\theta=30^{\circ}[/tex]

Solution:

The horizontal range of the projectile motion by the jumper is,

[tex]H=\frac{u^2\sin (2\theta)}{g}[/tex]

where g is the accleration due to gravity, and H is the horizontal range.

Substituting the known values,

[tex]\begin{gathered} H=\frac{120^2\times\sin (2\times30^{\circ})}{9.8} \\ H=1272.53\text{ m} \end{gathered}[/tex]

**Thus, the distance of the jump from its spot is 1272.53 m.**

True or false - Objects do NOT weigh anything when placed in a vacuum. Identify your answer, then explain your answer selection.

### Answers

First, we need to understand what the weight is. Mathematically, the weight is given by:

[tex]\begin{gathered} W=m\cdot g-Fb \\ where: \\ m=mass \\ g=gravity \\ Fb=Force_{\text{ }}of_{\text{ }}bouyancy \end{gathered}[/tex]

In vacuum, there is no force of buoyancy, so in the vacuum the weight reaches its maximum value.

In this sense, the statement is absurd, so we can conclude it is** False**

3. How much of a 75 mg sample of will remain after 75 days if the sample has a half-life of 7.5days?

### Answers

Answer:

0.073 mg sample will remain

Explanations:

The final amount of a substance after t years given the initial amount and the half life is expressed as:

[tex]A(t)=A_0(\frac{1}{2})^{\frac{t}{x}}[/tex]

where A(t) is the amount after time t

x is the half life

A₀ = initial amount = 75 mg

t = 75 days = 75/365 = 0.2055 years

Half life, x = 7.5 days = 0.02055 years

Substitute these values into the formula given

[tex]\begin{gathered} A=75(0.5)^{\frac{0.2055}{0.02055}} \\ A\text{ = 75(0.5)}^{10} \\ A\text{ = }0.073\text{ mg} \end{gathered}[/tex]

0.073 mg sample will remain

8.A satellite is 100 miles above the earth. It has a mass of 100 kg. If the mass ofthe satellite were doubled the force of gravity of the satellite on the earthwouldA) DoubleB) QuadrupleC) Stay the sameD) Be half as much

### Answers

The distance of satellite is r=100 miles and the mass of satellite is m=100 kg.

The force of gravity which acts on the satellite is,

[tex]F=\frac{GMm}{r^2}[/tex]

Here, F is the force of gravity, G isthe gravitational constant, M is the mass of earth, m is the mass of satellite and r is the distance of satellite.

As the force of gravity depends directly on the mass of satellite, therefore if the mass of the satellite were doubled then the force of gravity would also be doubled which means option (A) is correct.

Given the Rotational Inertia equation for a solid sphere (see question 9) and thefollowing equation for a hollow sphere, which sphere would have the greateracceleration if they both have equal mass? Defend your answer.Hollow Sphere Rotational Inertia: I = m r2

### Answers

The formula for calculating the moment of inertia of a solid sphere is

expressed as

I = 2/5mr^2

where

m is the mass

r is the radius

The formula for calculating the moment of inertia of a hollow sphere is

expressed as

I = 2/3mr^2

By comparing both equatins, we can see that the rotational inertia of the hollow sphere is larger than that of the solid sphere. Recall, inertia is the tendency of a body to resist changes in motion. This means that it is more difficult to speed up the hollow ball or slow down a spinning hollow ball. Acceleration is the rate of change of velocity. Thus,** the solid ball would have the greater acceleration**

An ice block on a frictionless path starts at rest and is released on the path. How high will it go (vertically) as it appoaches point C?

### Answers

The high reached by the block when it approaches point C, depends of the potential energy at the beginning of the motion.

The potential energy is:

**U = m·g·h **

due to the path is frictionless, the energy of the system is constant. Then, while the block approaches to point C, all its energy becomes potential energy and this potential energy must be the same that the initial potential energy.

Due to the mass m and g are constant, and U must be the same, then, it is necessary that h, the height, is the same that at the beginning of the motion.

**Hence, the hight is 4.5 m (the block does not reache point C) as the block approaches point C.**

while standing on a cliff a boy tosses a stone into a river below. If he throws the ball horizontally with a velocity of 3.0 m/s and it strikes the water 4.5 m away, how high above the water is the boy

### Answers

Given data:

* The horizontal velocity of the stone is 3 m/s.

* The range of the stone is 4.5 m.

* The acceleration of the stone in the horizontal direction is zero.

Solution:

By the kinematics equation, the horizontal motion of the stone in terms of range and time is,

[tex]R=u_xt+\frac{1}{2}a_xt^2_{}[/tex]

where u_x is the horizontal velocity, a_x is the horizontal acceleration, t is the time at which the stone strikes the water, and R is the horizontal range,

Substituting the known values,

[tex]\begin{gathered} 4.5=3\times t+0 \\ t=\frac{4.5}{3} \\ t=1.5\text{ s} \end{gathered}[/tex]

By the kinematics equation, the vertical motion of the stone in terms of the time is,

[tex]H=u_yt+\frac{1}{2}gt^2_{}[/tex]

where u_y is the vertical velocity of the stone, g is the acceleration due to gravity, t is the time take to strike the water and H is the height of the stone,

[tex]\begin{gathered} H=0+\frac{1}{2}\times9.8\times(1.5)^2 \\ H=11.025\text{ m} \end{gathered}[/tex]

**Thus, the height of the cliff is 11.025 m.**

Three vectors are shown in the figure. Their magnitudes aregiven in arbitrary units. Determine the sum of the three vectors.Give the resultant in terms of•components•magnitude•angle with the +x axis(Figure 1)

### Answers

**Components:**

First, calculate the cartesian components of each vector, as follow:

[tex]\begin{gathered} A_x=44.0\cdot\cos (28.0)=38.85 \\ A_y=44.0\cdot\sin (28.0)=20.65 \\ B_x=-26.5\cdot\cos (56.0)=-14.82 \\ B_y=26.5\cdot\sin (56.0)=21.97 \\ C_x=0 \\ C_y=-31.0 \end{gathered}[/tex]

Next, consider that the components of the resultant vector R, are given by the sum of the x components and y components of all vectors A, B and C:

[tex]\begin{gathered} R_x=A_x+B_x+C_x=38.5-14.82+0=23.68 \\ R_y=A_y+B_y+C_y=20.65+21.97-31.0=11.62 \end{gathered}[/tex]

**Magnitude:**

The magnitude is calculated as follow:

[tex]R=\sqrt[]{R^2_x+R^2_y}=\sqrt[]{(23.68)^2+(11.62)^2}=26.38[/tex]

**Angle with x axis:**

The angle related to the x axis is obtained as follow:

The tangent of the angle related to the x axis is:

[tex]\tan \theta=\frac{R_y}{R_x}[/tex]

which is basically, the quotient between the opposite site and adjacent side of a right triangle formed by the components of the vector.

To obtain the angle you apply tan^-1 to cancel out the tangent, as follow:

[tex]\theta=\tan ^{-1}(\frac{R_y}{R_x})=\tan ^{-1}(\frac{11.62}{23.68})=26.14\degree[/tex]

at what speed is the top of the ladder along w the electrician sliding down the wall at that instant

### Answers

The given problem can be exemplified in the following diagram:

We are given that a ladder is against a wall. According to the diagram, the ladder, the wall, and the floor form a right triangle, therefore, if "T" is the distance from the top to the floor and "B" is the distance from the bottom to the wall we can apply the Pythagorean theorem and we get:

[tex]T^2+B^2=14^2[/tex]

Now, since we want to know the speed, we will derivate implicitly with respect to time on both sides of the equation:

[tex]2T\frac{dT}{dt}+2B\frac{dB}{dt}=0[/tex]

Now we solve for the value of the speed of the top of the ladder, this is dT/dt:

[tex]2T\frac{dT}{dt}=-2B\frac{dB}{dt}[/tex]

The 2 cancels out:

[tex]T\frac{dT}{dt}=-B\frac{dB}{dt}[/tex]

Now we divide both sides by "T":

[tex]\frac{dT}{dt}=-\frac{B}{T}\frac{dB}{dt}[/tex]

Now, since we determine the value of "T" from the Pythagorean theorem, we get:

[tex]T^2+B^2=14^2[/tex]

Subtracting B squared from both sides:

[tex]T^2=14^2-B^2[/tex]

Taking the square root:

[tex]T=\sqrt[]{14^2-B^2}[/tex]

Now we replace these values in the formula for the velocity:

[tex]\frac{dT}{dt}=-\frac{B}{\sqrt[]{14^2-B^2}}\frac{dB}{dt}[/tex]

Now we have an expression for the velocity of the top of the ladder. Replacing the given values:

[tex]\frac{dT}{dt}=-\frac{6ft}{\sqrt[]{14^2-(6ft)^2}}(1\frac{ft}{s})[/tex]

Solving the operations we get:

[tex]\frac{dT}{dt}=-0.47\frac{ft}{s}[/tex]

Therefore, the speed of the top of the ladder is -0.47 feet per second.

Electricity and magnetism grade 12 ye just have to explain only one of the following

### Answers

**Doorbell:**

The doorbell consists of an electromagnet.

The electromagnet is a coil wound on an iron core.

This electromagnet behaves like a magnet when electric current flows through the coil.

This electromagnet attracts the iron strip near it, causing the sound.

The diagram is shown below

Here, iron strip is attracted to the electromagnet.

Multiple choiceeeeee homework A 70kg boy is standing at rest on a ice skating rink (assume it's a frictionless surface) gets hit in the face by ahuge 3.5 kg snowball moving at 15 m/s. The snowball sticks to his face. At what speed is the snowball coveredboy traveling after the collision?

### Answers

**Given:**

The mass of the boy, M=70 kg

The mass of the snowball, m=3.5 kg

The speed of the snowball before hitting the boy, v=15 m/s

**To find:**

The speed of the snowball-covered boy after the collision.

**Explanation:**

From the law of conservation of momentum, the total momentum of a system always remains the same. Thus the total momentum of the boy and the snowball before the collision must be equal to the total momentum of the boy and the snowball after the collision.

Thus,

[tex]\begin{gathered} mv=(m+M)u \\ \Rightarrow u=\frac{mv}{(m+M)} \end{gathered}[/tex]

Where u is the velocity of the snowball-covered boy after the collision.

On substituting the known values,

[tex]\begin{gathered} u=\frac{3.5\times15}{3.5+70} \\ =0.71\text{ m/s} \end{gathered}[/tex]

**Final answer:**

**The speed of the snowball-covered boy after the collision is 0.71 m/s**

**Thus the correct answer is option B.**

Old-fashioned alarm clocks have springs that you wind up". What type of energy does the spring have after it iswound? As the alarm clock runs and performs its various functions the spring loses this energy - what becomes of it? Inother words, the energy initially in the spring is changed into what other forms?

### Answers

Let's determine the various forms of energy used by an old-fashioned alarm clock with springs.

The type of energy the spring has after it is wound can be called the elastic potential energy.

The potential energy can be defined as the energy possessed by an object due to its position due to some zero position.

**Therefore, after the spring is wound, it will have the**** potential energy ****due to the arrangement/position of the spring.**

As the alarm clock runs and performs its various functions, the spring loses energy.

Since the spring is in motion, it loses energy and the potential energy lost becomes **kinetic energy.**

Kinetic energy can be said to be the energy of an object due to its motion.

Also, due to friction, the some of its potential energy is converted to heat energy.

**Therefore, in other words, the energy initially in the spring which is the **** ****potential energy****, **** is changed into the **** ****kinetic energy and the heat energy****.**

**ANSWER:**

**The initial energy which is the potential energy is converted into the kinetic energy and heat energy.**

Convert 70 in2 to SI units. Work across the line and show all steps in the conversion. Use scientific notation and apply the proper use of significant figures. Round your answer to 2 significant figures.

### Answers

To convert squared inches to squared centimeters we need to remember that one inch is equal to 2.54 cm.

Now that we know we can do the conversion:

[tex]70in^2\cdot\frac{2.54\text{ cm}}{1\text{ in}}\cdot\frac{2.54\text{ cm}}{1\text{ in}}=\frac{451.612in^2\cdot cm^2}{in^2}=451.612cm^2[/tex]

Therefore 70 squared inches is equal to 451.612 squared cm (this is the exact result). If we use two significant figures then this will be equal to:

[tex]4.5\times10^2cm^2[/tex]

Now, if we want to convert this to squared mm we need to remember that one cm is equal to 10 mm then we would have:

[tex]4.5\times10^2cm^2\cdot\frac{10\text{ mm}}{1\text{ cm}}\cdot\frac{10\text{ mm}}{1\text{ cm}}=4.5\times10^4\operatorname{mm}[/tex]

**Therefore we conclude that 70 squares inches is**

[tex]4.5\times10^4\operatorname{mm}[/tex]

A ship is travelling 10 km due west from the harbor and then 7 km in N50.Question: How far is theship from the port?

### Answers

If the boat moves this way, then the 10km and 7km lines, and the direct distance d form a right triangle. From here we can calculate d using the pythagorean theorem.

Which of the following is true for the direction of an electric field?It is same as the direction of the force exerted on a negative test charge.It is opposite to the direction of the force exerted on a positive test charge.It is same as the direction of the force exerted on a positive test charge.It is opposite to the direction of the force exerted on a neutral test charge

### Answers

Take into account that by definition, electric force over a positive charge has the same direction that electric field. For negative charge, electric force is opposite to the electric field.

Then, based on the previous description, the following statement is true:

**It is same as the direction of the force exerted on a positive test charge.**

It was the same sound as produced on a hot day of 40°C what is the wave length in millimeters of the wave in the water temperature

### Answers

Given:

Frequency, f = 20 kHz.

Speed of sound in air, v = 331 m/s

Let's find the wavelength for the following:

• (a). When the temperature is 0 degrees Celsius.

To find the wavelength, apply the formula:

[tex]\lambda=\frac{v}{f}[/tex]

Where:

v is the speed

f is the frequency in Hz = 20 x 1000 = 20000 Hz.

To find the speed of sound on a day with 0 C, we have:

[tex]\begin{gathered} v=331*\sqrt{1+\frac{0}{273}} \\ \\ v=331*\sqrt{1} \\ \\ v=331\text{ m/s} \end{gathered}[/tex]

Now, to find the wavelength, input the values into the formula:

[tex]\begin{gathered} \lambda=\frac{v}{f} \\ \\ \lambda=\frac{331}{20000} \\ \\ \lambda=0.01655\text{ m}\approx16.55\text{ mm} \end{gathered}[/tex]

**The wavelength when the temperature is 0 C is **** 16.55 mm.**

• (b). Wavelength when the temperature is 40 degrees Celsius.

The speed, v, on a day when the temperature is 40 C will be:

[tex]\begin{gathered} v=331*\sqrt{1+\frac{40}{273}} \\ \\ v=331*\sqrt{1.1465} \\ \\ v=331*1.0708 \\ \\ v=354.42\text{ m/s} \end{gathered}[/tex]

Hence, to find the wavelength, we have:

[tex]\begin{gathered} \lambda=\frac{v}{f} \\ \\ \lambda=\frac{354.42}{20000} \\ \\ \lambda=0.01772\text{ m}\approx17.72\text{ mm} \end{gathered}[/tex]

**The wavelength on a day the temperature is 40 C is **** 17.72 mm.**

**ANSWER:**

• (a). 16.55 mm

• (,b). 17.72 mm.