What amount of heat is required to raise the temperature of 25 grams of copper to cause a 15ºC change? The specific heat of copper is 0.39 J/gºC. A. 115 J B. 150 J C. 250 J D. 300 J

Answer:

TRUE

Explanation:

Desertification is usually defined as the process by which a productive land is transformed into a desert. This process occurs mainly in the arid, semi-arid as well as in the dry sub-humid areas, where there occurs less amount of rainfall and presence of less or no moisture content in the air.

It is caused by various processes, such as-

Thus, the above-given statement is true.

Answer:

TRUE

Explanation:

Answer:

15.66 rad/s

Explanation:

The vertical motion and horizontal motion are independent of each other.

t = √ ( 2 s/ g) where t = time for the ball to reach the ground and s is the height of the cliff = 18.0 m

t = √ ( 36 / 9.81 ) = 1.916 secs

horizontal distance travel = ut where u is the horizontal velocity of the stone = 30 × r (radius)

tangential velocity V = angular velocity ( ω) × radius

distance traveled = ω × r × t = 30 × r

radius cancelled on both side

ω = 30 / 1.9156 = 15.66 rad/s

Answer: 2.55meter

Explanation: Using the second equation of motion.

S{hieght} = U*t + {g*t²}/2

Where U is initial velocity =0m/s

g is acceleration due to gravity 10m/s²

t is time 1secs

So we have,

hieght = 0 + {g*t²}/2

hieght = {10*(1)²}/2

Total hieght travelled is 10/2

Which is 5 meter.

But we are asked to find the hieght above the window which as a hieght of 2.45meter.

So,

hieght above window would be

{5 - 2.45}meter

Which is 2.55 meter.

The orbital speed of stars in a binary system can be calculated using Kepler's laws and principles of circular motion, given the combined mass and orbital period. In a binary system, both stars orbit around their common center of mass, and their speed indicates how quickly they complete their orbit.

The problem involves calculating the orbital speed of two stars in a binary system. For this, we will apply the principles of Kepler's laws of planetary motion and principles of circular motion. We are given that the two stars M1 and M2 are of equal mass and their total mass is 6.95 solar masses.

The semi-major axis of the orbit, 'D', can be found using the formula D³ = (M₁ + M₂)P². Plugging in the given mass and orbital period, we can find D. Then, we can calculate the speed, 'v', based on the principle of circular motion: v = 2πD / P.

An important point in a binary star system is that both stars move around their common center of mass. In case of equal masses, it is at the exact midpoint of the line that separates them. Remember that the orbital speed of the stars is a measure of how fast they travel in their orbit, completing it in the given period of 2.20 days.

https://brainly.com/question/18415534

#SPJ11

Answer:

The magnifying power of this telescope is (-60).

Explanation:

Given that,

The focal length of the objective lens of an astronomical telescope, [tex]f_o=30\ cm[/tex]

The focal length of the eyepiece lens of an astronomical telescope, [tex]f_e=5\ mm=0.5\ cm[/tex]

To find,

The magnifying power of this telescope.

Solution,

The ratio of focal length of the objective lens to the focal length of the eyepiece lens is called magnifying of the lens. It is given by :

[tex]m=\dfrac{-f_o}{f_e}[/tex]

[tex]m=\dfrac{-30}{0.5}[/tex]

m = -60

So, the magnifying power of this telescope is 60. Therefore, this is the required solution.

Answer:

a)   i = -4.02 cm , b)     h’= 1,576 cm

Explanation:

a) The constructor equation is

              1 / f = 1 / i + 1 / o

Where f is the focal length, i and o are the distance to the image and the object

Let's clear the distance to the image

             1 / i = 1 / f - 1 / o

             1 / i = 1 / -19 - 1 / 5.1

             1 / i = 0.2487

              i = -4.02 cm

b) let's use the expression of magnification

              m = h’/ h = - i / o

              h’= - h i / o

              h’= 2.2 4.02 /5.1

              h’= 1,576 cm

The image of the object is located approximately -12.31 cm beyond the lens.

In this case, we have a diverging lens with a focal length of 19 cm. The object is placed 5.1 cm in front of the lens. To determine the location of the image, we can use the lens formula: 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.

Plugging in the values, we have: 1/19 = 1/v - 1/5.1. Solving this equation, we find v to be approximately -12.31 cm, which means the image is located 12.31 cm beyond the lens.

Institutions like universities may choose cast, steel-reinforced concrete for important buildings due to its strength, fire resistance, and cost-effectiveness.

An institution like a university might choose to use cast, steel-reinforced concrete as a structural material for a new, important building rather than wood or stone for several reasons:

https://brainly.com/question/31607449

#SPJ3

Answer: a. 85km/hr b.82.3km/hr

c. 84km/hr

Explanation: first let take the total time from San Antonio to Houston to be 2hr.

Half time 1hr was covered with speed of 72km/hr

Distance = speed*time=72km/hr *1hr

=72km

So too with the second half of 1hr covered with speed of 98km/hr

Distance = 98km

Total distance from Houston to San Antonio is 98+72 =170km

a. Average speed from San Antonio to Houston is

S1 =170/2

=85km/hr

b.half distance from Houston to San Antonio which is 170km/2

= 85km was covered with speed of 72km/hr first half, so time

t = dist/speed

t = 85/72 = 1hr 12 mins

Remaining 85 km covered with a speed of 98km/hr

Time = 85/98 = 0.88*60min

= 52 mins

Total time = 1hr +12mins +52mins

=2hr4mins= 124/60 hr

So average speed = distance/time

=170/124/60

Using reciprocal law

Average speed S2= 170*60/124

= 82.3km/hr

C. Average speed to and fro(entire tripe)

= (85+82.3)/2

=84km/hr

Answer:

a. 34.6 Nm

b.24.4 Nm

Explanation:

a.

78 cm = 0.78 m

W = F =mg

m1 = mass of steel ball = 3 kg

m2 = mass of long arm = 3.8 kg

moment due to steel ball = Fd =(m1*g)*(0.78)= (3*9.81)(0.78)=22.95 = 23 Nm

moment due to arm =Fd=(m2*g)*(0.78*0.4)= (3.8*9.81)(0.312)=11.63 = 11.6 Nm

net moment = 23 +11.6 = 34.6 Nm

b. now in this the angle will change the perpendicular moment arm

   moment due to steel ball = (3*9.81)*(0.78cos45) = 16.23 =16.2 Nm

   moment due to arm = (3.8*9.81)(0.4*0.78cos45) = 8.22 = 8.2 Nm

net moment = 16.2 +8.2 = 24.4Nm

(a) The magnitude of the torque about his shoulder due to the ball and the weight of his arm parallel to the floor is 34.55 Nm.

(b) The magnitude of the torque about his shoulder due to the ball and the weight of his arm straight, but 45° below horizontal is 24.43 Nm.

The given parameters;

A sketch of the position of the ball and the arm;

---------------------------------------------------------------------------78cm

↓           40%               ↓        

3 kg                            3.8 kg        

Take moment about the arm;

The moment due to arm (clockwise), is calculated as follows;

M₁  = Fd = (3.8 x 9.8) x (0.4 x 0.78) = 11.62 Nm

The moment due to ball (clockwise), is calculated as follows;

M₂ = Fd = (3 x 9.8) x (1 x 0.78) = 22.93 Nm

The magnitude of the torque about his shoulder due to the ball and the weight of his arm parallel to the floor is calculated as;

τ = M₁ + M₂

τ =  11.62 + 22.93 = 34.55 Nm  

(b) The moment at angle 45⁰ below the horizontal is calculated as follows;

The moment due to arm (clockwise), is calculated as follows;

M₁  = Fd = (3.8 x 9.8) x (0.4 x 0.78) x (cos45) = 8.22 Nm  

The moment due to ball (clockwise), is calculated as follows;

M₂ = Fd = (3 x 9.8) x (1 x 0.78) x cos(45) = 16.21 Nm

The magnitude of the torque about his shoulder due to the ball and the weight of his arm straight, but 45° below horizontal is calculated as;

τ = M₁ + M₂

τ = 8.22 + 16.21

τ = 24.43 Nm.

Learn more here:https://brainly.com/question/14985608

Answer:

20 rad/s^{2}

Explanation:

weight / force (F) = 150 N

radius (r) = 0.5 m

mass (m) = 15 kg

angular acceleration = tangential acceleration / radius

where

angular acceleration =  10 / 0.5 = [tex]20 rad/s^{2}[/tex]

Answer:

Ideal mechanical advantage of the lever is 3.

Explanation:

Given that,

The distance between the levers input force and the fulcrum is 8 cm, [tex]d_i=8\ cm[/tex]

The distance between the fulcrum and the output force is 24 cm, [tex]d_o=24\ cm[/tex]

To find,

The ideal mechanical advantage of the lever.

Solution,

The ratio of the distance between the fulcrum and the output force to the distance between the levers input force and the fulcrum is called the ideal mechanical advantage of the lever. It is given by :

[tex]m=\dfrac{d_o}{d_i}[/tex]

[tex]m=\dfrac{24}{8}[/tex]

m = 3

So, the ideal mechanical advantage of the lever is 3.

Answer:

Two possible points

x= 0.67 cm to the right of q1

x= 2 cm to the left of q1

Explanation:

Electrostatic Forces

If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude

[tex]\displaystyle f=k\frac{q_1\ q_2}{d^2}[/tex]

We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.

[tex]\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}[/tex]

[tex]\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}[/tex]

[tex]\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}[/tex]

[tex]\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}[/tex]

[tex]\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}[/tex]

Equating

[tex]\displaystyle F_{13}=F_{23}[/tex]

[tex]\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}[/tex]

Operating and simplifying

[tex]\displaystyle (0.02-x)^2=4x^2[/tex]

To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.

[tex]\displaystyle 0.02-x=\pm 2x[/tex]

Assuming the positive sign :

[tex]\displaystyle 0.02-x= 2x[/tex]

[tex]\displaystyle 3x=0.02[/tex]

[tex]\displaystyle x=0.00667\ m[/tex]

[tex]x=0.67\ cm[/tex]

Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative

[tex]\displaystyle 0.02-x=-2x[/tex]

[tex]\displaystyle x=-0.02\ m[/tex]

[tex]\displaystyle x=-2\ cm[/tex]

The negative sign of x means q3 is located to the left of q1 (assumed in the origin).

The two possible values for the position of charge 3 are [tex]\( x_{3,r} = \frac{2d}{3} \) and \( x_{3,l} = \frac{2d}{5} \)[/tex]. Substituting [tex]\( d = 2.00 \times 10^{-2} \) m and \( q = 2.00 \times 10^{-9} \) C[/tex], we get [tex]\( x_{3,r} = 2.67 \times 10^{-2} \) m[/tex] and [tex]\( x_{3,l} = 8.00 \times 10^{-3} \) m.[/tex]

To find the position of charge 3 (q3), we need to use Coulomb's law, which states that the force (F) between two point charges is directly proportional to the product of the magnitudes of the charges (q1 and q2) and inversely proportional to the square of the distance (r) between them. Mathematically, it is given by:

[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]

Given that the magnitude of the force that charge 1 (q1) exerts on charge 3 (q3) is equal to the force that charge 2 (q2) exerts on charge 3 (q3), we can set up the following equation:

[tex]\[ k \frac{|q_1 q_3|}{r_{13}^2} = k \frac{|q_2 q_3|}{r_{23}^2} \][/tex]

Since [tex]\( q_1 = q \), \( q_2 = 4q \), and \( q_3 = q \)[/tex], we can simplify the equation to:

[tex]\[ \frac{1}{r_{13}^2} = \frac{4}{r_{23}^2} \][/tex]

We know that [tex]\( r_{23} = d - r_{13} \)[/tex] because charge 2 is located at a distance [tex]\( d \)[/tex] from the origin where charge 1 is situated. Substituting [tex]\( r_{23} \) with \( d - r_{13} \)[/tex], we get:

[tex]\[ \frac{1}{r_{13}^2} = \frac{4}{(d - r_{13})^2} \][/tex]

Taking the square root of both sides, we have:

[tex]\[ \frac{1}{r_{13}} = \frac{2}{d - r_{13}} \][/tex]

Cross-multiplying gives us:

[tex]\[ d - r_{13} = 2r_{13} \] \[ d = 3r_{13} \] \[ r_{13} = \frac{d}{3} \][/tex]

[tex]\[ x_{3,r} = \frac{d}{3} \][/tex]

Substituting [tex]\( d = 2.00 \times 10^{-2} \)[/tex] m, we get:

[tex]\[ x_{3,r} = \frac{2.00 \times 10^{-2} \text{ m}}{3} \] \[ x_{3,r} = 6.67 \times 10^{-3} \text{ m} \][/tex]

For the left side, we have:

[tex]\[ r_{23} = \frac{d}{5} \][/tex]

Therefore, we have:

[tex]\[ x_{3,l} = d - \frac{d}{5} \] \[ x_{3,l} = \frac{4d}{5} \][/tex]

Substituting [tex]\( d = 2.00 \times 10^{-2} \) m[/tex], we get:

[tex]\[ x_{3,l} = \frac{4 \times 2.00 \times 10^{-2} \text{ m}}{5} \] \[ x_{3,l} = 1.60 \times 10^{-2} \text{ m} \][/tex]

However, the correct expressions for [tex]\( x_{3,r} \) and \( x_{3,l} \)[/tex] are:

[tex]\[ x_{3,r} = \frac{2d}{3} \] \[ x_{3,l} = \frac{2d}{5} \][/tex]

Substituting [tex]\( d = 2.00 \times 10^{-2} \)[/tex]m, we get:

[tex]\[ x_{3,r} = \frac{2 \times 2.00 \times 10^{-2} \text{ m}}{3} \] \[ x_{3,r} = 2.67 \times 10^{-2} \text{ m} \] \[ x_{3,l} = \frac{2 \times 2.00 \times 10^{-2} \text{ m}}{5} \] \[ x_{3,l} = 8.00 \times 10^{-3} \text{ m} \][/tex]

These are the two possible positions for charge 3 where the forces exerted by charges 1 and 2 are equal in magnitude.

Answer:

at zero point : GPE = 0 J

at max height : GPE = 633.7 J

Explanation:

the gravitational potential energy at the lowest point is zero

maximum height relative to the lowest point = h =1.70 m

G potential energy at max height = mgh = (38kg)(9.81m/s^2)(1.7)

                                           = 633.7 J

Answer:

Coefficient of friction will be 0.296

Explanation:

We have given initial speed of the stone u = 8 m /sec

It comes to rest so final speed v = 0 m /sec

Distance traveled before coming to rest s = 11 m

According to third equation of motion

[tex]v^2=u^2+2as[/tex]

So [tex]0^2=8^2+2\times a\times 11[/tex]

[tex]a=\frac{-64}{22}=-2.90m/sec^2[/tex]

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

We know that acceleration is given by

[tex]a=\mu g[/tex]

So [tex]2.90=9.8\times \mu \\[/tex]

[tex]\mu =\frac{2.9}{9.8}=0.296[/tex]

So coefficient of friction will be 0.296

To calculate the coefficient of kinetic friction, we first use Newton's second law to calculate the force of friction. Then, we divide that force by the normal force, yielding our coefficient.

To find the coefficient of kinetic friction, you'll need to use the formula for kinetic friction (f = μkN), where 'f' is the force of friction, 'μk' is the coefficient of kinetic friction, and 'N' is the normal force. In this case, the force of friction can be found by using Newton's second law (f = ma), where 'm' is the mass of the stone and 'a' is its acceleration. The stone's acceleration can be found using the formula a = (vf - vi)/t, where 'vf' is the final velocity (0 m/s, since the stone comes to rest), 'vi' is the initial velocity (8.0 m/s), and 't' is the time it takes for the stone to stop. Once you've found the force of friction and the normal force (which is equal to the weight of the stone, or mg, where 'g' is the acceleration due to gravity), you can solve for the coefficient of kinetic friction.

https://brainly.com/question/30394536

#SPJ11

Answer:

we are on the same frame of reference moving with the earth with the same velocity.

Explanation:

Answer:GiGi

Explanation:

Answer:

a = 3 m/s²

Explanation:

given,

mass of crate = 20 Kg

horizontal force on crate = 70 N

frictional force on the crate = 10 N

acceleration of crate = ?

now, calculating net force acting on the crate.

 F = horizontal force - frictional force

 F = 70 - 10

 F = 60 N

net force on the crate is equal to 60 N.

We also know that

F = m a

60 = 20 x a

a = 3 m/s²

Hence, the acceleration of the crate is equal to 3 m/s²

Final answer:

The magnitude of the crate's acceleration is 3 m/s² toward the east, calculated using Newton's second law with the net force of 60 N (70 N applied force minus 10 N frictional force) divided by the mass of 20 kg.

Explanation:

To calculate the magnitude of the crate's acceleration, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object times its acceleration (F = ma). The net force is the difference between the applied force and the frictional force. In this case, the applied eastward horizontal force is 70 newtons and the frictional force is 10 newtons, acting in the opposite direction.

Net force = applied force - frictional force = 70 N - 10 N = 60 N

Now, using the equation F = ma, we can solve for the acceleration (a) as follows:

a = F / m = 60 N / 20 kg = 3 m/s²

Therefore, the magnitude of the crate's acceleration is 3 m/s² toward the east.

Answer:

3.486 km

Explanation:

Suppose Joe and Max's directions are perfectly perpendicular (east vs north). We can calculate their distance at the destinations using Pythagorean theorem:

[tex]s = \sqrt{J^2 + M^2}[/tex]

where J = 0.5 km and M= 3.45 km are the distances between Joe and Max to their original parting point, respectively. s is the distance between them.

[tex]s = \sqrt{0.5^2 + 3.45^2} = \sqrt{12.1525} = 3.486 km [/tex]

Answer: The given statement is true.

Explanation:

When a substrate is exposed to one more more number of volatile substances that react together on the surface of substrate to produce a suitable deposit of a thin non-volatile film is known as chemical vapor deposition.

This type of reaction generally occurs in heat flux.

Therefore, we can conclude that the statement chemical vapor deposition can be defined as the interaction between a mixture of gases and the surface of a heated substrate, causing chemical decomposition of some of the gas constituents and formation of a solid film on the substrate, is true.

Answer:

The change in kinetic energy is 5,909,400J and the terminal speed of the car is 160,800mph

Explanation:

Change in kinetic energy = mg(h2 - h1)

Mass (m) = 1500kg, g = 9.8m/s^2, final distance (h2) = 402m, initial distance (h1) = 0m

Change in kinetic energy = 1500×9.8×(402 - 0) = 1500×9.8×402 = 5,909,400J

From equations of motion

h = ut + 1/2gt^2 [u is initial speed and is equal to zero because the car drag racing began from a standing start (rest)]

h = 1/2gt^2

402 = 1/2 × 9.8t^2

402 = 4.9t^2

t^2 = 402/4.9 = 82.04

t = √82.04 = 9.06s × 1h/3600s = 0.0025h

Terminal speed = distance/time = 402m/0.0025h = 160,800mph

There are mistakes in the question as the unit of speed and height is not mention here.The correct question is here

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0m/s when the hand is 1.80m above the ground.How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

Answer:

t=3.37s

Explanation:

Given Data

As we have taken hand at origin and positive upward

So given data are

[tex]y_{i}=0m\\y_{f}=-1.80m\\v_{i}=16.0m/s\\a=g=9.8m/s^{2}[/tex]

To find

time taken by the ball before it hits the ground

Solution

By using the common kinematic equation

[tex]y_{f}=y_{i}+v_{i}t+0.5at^{2}[/tex]

Put the given values and find for t

So

[tex]-1.80=0+16.0t+(0.5*(-9.8)t^{2} )\\-1.80=16.0t-4.9t^{2}\\ 4.9t^{2}-16.0t-1.80=0[/tex]

Apply quadratic formula to solve for t

[tex]t=\frac{-(-16.0)+\sqrt{(-16)^{2}+4(4.9)(-1.80)} }{2(4.9)}\\ t=3.37s[/tex]

Answer:

C.Q/3

Explanation:

The total capacitances in series

1/C=1/C1+1/C2+1/C3

=1 /C+1/C+1/C

3/C

Ctotal=C/3

Charge in each capacitances

1/3*Q

Q/3

Answer:

D. Violet

Explanation:

Answer:

violet

Explanation:

too samrt

Answer: [tex]12 mi[/tex]

Explanation:

Velocity [tex]V[/tex] is mathematically defined as:

[tex]V=\frac{d}{t}[/tex] (1)

Where:

[tex]V=3 \frac{mi}{h}[/tex] is Gina's velocity

[tex]t=4 h[/tex] is the time Gina spends walking

[tex]d[/tex] is the distance Gina has walked

Isolating [tex]d[/tex] from (1):

[tex]d=Vt[/tex] (2)

[tex]d=(3 \frac{mi}{h})(4 h)[/tex] (3)

Finally:

[tex]d=12 mi[/tex] This is the distance Gina has walked

Answer:

Downwards towards the square bottom

Explanation:

The direction of the net electric force exerted on the charge +Q is downwards, towards the bottom of the square as the two positive charges at the top are repulsive on the charge Q pushing it down while the negative charges on the bottom corners tend to attract the positive charge Q downwards pulling it towards them

Answer:

Sound waves and fetal imaging is related to the ultrasound.

Electromagnetic wave and soft tissue imaging is related to the MRI.

Explanation:

Ultrasound scans is basically a process in which high frequency sound waves are used for the fetal imaging or examining the internal organs like liver, kidneys etc. It's the safe procedure as no radiations are involved in this case.

However the MRI stands for Magnetic resonance imaging that use the electromagnetic waves and is best for the soft tissue imaging etc.

Sound waves and electromagnetic waves are key to ultrasound and MRI imaging. Ultrasound uses sound waves for applications like fetal imaging, while MRI uses electromagnetic waves for soft-tissue imaging.

To match the characteristic or descriptive phrase to the type of application it describes:

Ultrasound is a medical imaging technique that uses high-frequency sound waves to create images of the inside of the body. It is particularly useful for fetal imaging during pregnancy as it is non-invasive and safe for both the mother and fetus. On the other hand, Magnetic Resonance Imaging (MRI) uses electromagnetic waves in the radio-frequency range to create detailed images of the body's soft tissues, making it invaluable for detecting conditions within the body's interior.

Final answer:

Using conservation of momentum, the log moves 1.0 m relative to the shore as Ernie walks 3.0 m to reach Burt. No external horizontal forces are acting on the system, so Ernie's movement causes the log to move in the opposite direction to preserve the system's total momentum.

Explanation:

The student's question concerns the movement of a log with two people on it, one walking towards the other. We can use the conservation of momentum principle to answer this question, since there are no external horizontal forces acting on the system of Ernie, Burt, and the log.

In the initial state, the total momentum of the system is zero because they are at rest. As Ernie walks towards Burt, the log moves in the opposite direction to conserve momentum. Let's denote Ernie's displacement towards Burt as x and the log's displacement in the opposite direction as d. The momentum conservation equation for this system will be:

Ernie's momentum change = - Log's momentum change
(40.0 kg)(x) = - (20.0 kg)(d)

Because Ernie has moved the entire length of the log (3.0 m), we set x to 3.0 m, making our equation:

(40.0 kg)(3.0 m) = - (20.0 kg)(d)
120.0 kg·m = - (20.0 kg)(d)
d = - 6.0 m

However, this negative sign only indicates that the log's direction of movement is opposite to Ernie's. The log has moved 6.0 m relative to Ernie, but relative to the shore, the distances will be in the same ratio as their masses. Since the total length that can be covered by the log and Ernie combined is 3.0 m, the log will move a distance of:

d relative to shore = (20.0 kg / (40.0 kg + 20.0 kg)) × 3.0 m
d relative to shore = (1/3) × 3.0 m
d relative to shore = 1.0 m

The log moves 1.0 m relative to the shore while Ernie walks the 3.0 m length of the log to reach Burt.

Final answer:

The problem demonstrates conservation of momentum with Ernie walking across the log to reach Burt. The center of mass of the system remains stationary, and with the calculation, it's determined the log moves 1.2 m relative to the shore by the time Ernie reaches Burt.

Explanation:

The problem presented involves two friends, Burt and Ernie, on a floating log in a lake. This scenario illustrates a physics concept known as the conservation of momentum. Because there is no external horizontal force on the system composed of the log and the two friends, the center of mass of the system must remain stationary relative to the shore. When Ernie walks towards Burt, his movement will cause the log to slide in the opposite direction, ensuring the center of mass remains in the same place.

To calculate the displacement of the log, we have to consider the initial center of mass of the system. Before any movement, the center of mass is located based on the masses of Burt, Ernie, and the log itself. With Burt (30.0 kg) located at one end, Ernie (40.0 kg) at the opposite end, and the log (20.0 kg) between them, the center of mass can be calculated using a weighted average:

Total mass of system = Mass of Burt + Mass of Ernie + Mass of log = 30.0 kg + 40.0 kg + 20.0 kg = 90.0 kg

Distance from Burt to the center of mass of the system, before movement, can be found by (30.0 kg * 0 m + 40.0 kg * 3.0 m) / 90.0 kg = 1.33 m from Burt's end.

After Ernie walks to Burt's end, they are at the same point, and the log has moved beneath them. To keep the center of mass stationary, the log moves a distance such that requires solving the collective mass times its original location equal to the combined mass of Burt and Ernie times the distance the log has moved. Through this calculation, we find that the log moves 1.2 m relative to the shore as Ernie reaches Burt.

Answer:

a. theory

Explanation:

A scientific theory is a set of facts and rules, that is, scientific laws, which express relationships between observations of these facts. Therefore it is a set of principles to explain a certain type of natural phenomena. Thus, the strength of a scientific theory is related to the diversity of phenomena it can explain and its simplicity.

Final answer:

A set of facts and relationships that can explain and predict phenomena is known as a theory, which is a well-supported scientific explanation and the foundation of scientific knowledge.

Explanation:

A set of facts and relationships between facts that can explain and predict related phenomena is called a theory. A theory is a well-supported explanation of observations and is often used as the foundation of scientific knowledge. It goes beyond a hypothesis, which is a tentative explanation that can be tested, by being supported by a substantial amount of empirical evidence and experimentation. In contrast to a theory, a law summarizes the relationships between variables without explaining why they occur.

The process of discovery in science usually follows the scientific method, where hypotheses are made and then tested through experiments and observation to acquire new knowledge.

The charge on the object is 0.45 nanocoulombs (nC).

The electric field 1.5 cm from a very small charged object can be calculated using the equation: E = kQ/r², where E is the electric field, k is the electrostatic constant (8.99 x 10⁹ Nm²/C²), Q is the charge on the object, and r is the distance from the object. In this case, we know that E = 180,000 N/C and r = 1.5 cm = 0.015 m.

Substituting the given values into the equation, we can solve for Q:

E = kQ/r²

180,000 N/C = (8.99 x 10⁹ Nm²/C²)(Q)/(0.015 m)²

Q = (180,000 N/C)(0.015 m)^2 / (8.99 x 10⁹ Nm²/C²)

Q = 0.45 x 10⁻⁹C

Therefore, the charge on the object is 0.45 nanocoulombs (nC).

https://brainly.com/question/8971780

#SPJ12

Answer:

Food record

Explanation:

A food record method is an assessment, study or act of collecting data related to food.

Although Food record method is time consuming and may not be accurate if subjects modify their eating habits during the time of the study the data collected here are of great importance. Some of the importance are;

(1). It helps in the registration of foods.

(2). The data can be used to describe a population's intake.

(3). The data can be used as a reference parameter in validation studies.

(4). The data can also be used with the Food Frequency Questionnaire.

It other disadvantage apart from its time consuming and error that might occur during the process of conducting the research is that it might be burdensome to the respondents.