Please help which one is correct

Please Help Which One Is Correct

Answers

Answer 1

Answer:

option (i) is correct

Explanation:

as there is no air resistance, no force is acting on the object horizontally, but gravitational acceleration will obviously act, regardless of the air resistance... option (i) is correct


Related Questions

a tiger leaps with an initial velocity of 55 km/hr at an angle of 13° with respect to the horizontal. what are the components of the tigers velocity?

Answers

Answer:

vₓ = 53.6 km/h

vy = 12.4 km/h

Explanation:

if we define two axis perpendicular each other with origin in the point represented by the tiger leaping (assuming we can treat it as a point mass) coincidently with the horizontal (x-axis) and vertical (y-axis) directions, we can obtain the components of the velocity in both independent directions.We can do it simply getting the projections of the velocity vector on both axes, using simple trigonometry, as follows:

       [tex]v_{x} = v_{o} * cos \theta = 55 km/h * cos 13 = 53.6 km/h[/tex]

       [tex]v_{y} = v_{o} * sin\theta = 55 km/h * sin 13 = 12.4 km/h[/tex]

If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answers

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is [tex]u  =  7.73 \  m/s [/tex]

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      [tex]s = h+  u t +  0.5 gt^2[/tex]

Here s  is the distance covered by the bag at sea level which is zero

      [tex]0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2[/tex]

=>    [tex]0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2[/tex]

=>   [tex]u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}[/tex]

=>   [tex]u  =  7.73 \  m/s [/tex]

     

Which statements about potential energy are true?
▫ Gaining potential energy is always associated with a force field.
▫ A change in position always means that an object gains potential energy.
▫ There's only one kind of potential energy.
▫ Some kinds of potential energy are related to electric forces exerted by atoms and molecules.​

Answers

Answer:

the answer is 1 and 4

Explanation

Plato users

For the potential energy, statement 1 and statement 4 are correct.

The potential energy of the object the energy of the object in its steady position. When the object is at rest, the energy of the object in that condition is called potential energy.

Let us consider an electron having charge [tex]e[/tex] is moving the distance [tex]d[/tex] in uniform electric field E.

Its potential energy can be written as,

[tex]P = eEd[/tex]

Where P is the potential energy and E is the electric field.

Hence, the potential energy of the electron is associated with the electric field.

The electric force can be written as,

[tex]F =eE[/tex]

Where [tex]F[/tex] is the electric force, [tex]E[/tex] is the electric field and [tex]e[/tex] is the charge on the electron.

So, the potential energy can be written as,

[tex]P=Fd[/tex]

Hence, the potential energy is related to electric force.

For more information, follow the link given below.

https://brainly.com/question/1413008.

Help solve these two problems im having trouble trying to start these problems?​

Answers

Answer:

25.  Approximately 8.1 meters

26. North 1.31 km, and East 2.81 km

Explanation:

25.

Notice that the displacements: 6 meters east and 5.4 south create the legs of a right angle triangle. The hypotenuse of that triangle will be the distance (d) needed to cover in order to get the ball in the hole in one putt. That is:

[tex]d=\sqrt{6^2+5.4^2} =\sqrt{65.16} \approx 8.072\,\,\,m[/tex]

which can be rounded to 8.1 m.

26.

Notice that the 3.1 km at an angle of 25 degrees north of east, is the hypotenuse of a right angle triangle that has for legs the east and north components of that distance.

We can find the leg corresponding to the east displacement using the cosine function (that relates adjacent side with hypotenuse):

[tex]east\,\, comp=3.1 * cos(25^o)\approx 2.809\,\,km[/tex]

and we can calculate the north component using the sine function that relates the opposite side to the angle with the hypotenuse.

[tex]north\,\,component = 3.1 * sin(25^o) \approx 1.31 \,\,km[/tex]

An alligator crawls 25\,\text m25m25, start text, m, end text to the left with an average velocity of -1.2\,\dfrac{\text m}{\text s}−1.2
s
m

minus, 1, point, 2, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction.

Answers

Answer:

21 s.

I hope this helps!

Answer:21s

Explanation:

What is the ratio of the displacement amplitudes of two sound waves given that they are both5.0 kHz but have a 3.0 dB intensity level difference?

Answers

Answer:

The ratio of the displacement amplitudes of two sound waves is 1.16.

Explanation:

Given that,

Frequency = 5.0 kHz

Intensity level difference = 3.0 dB

We know that,

The sound intensity is inversely proportional to the square of distance.

[tex]I\propto\dfrac{1}{r^2}[/tex]

The sound intensity for first wave is

[tex]\beta_{1}=10\log\dfrac{I_{1}}{I_{0}}[/tex]...(I)

The sound intensity for second wave is

[tex]\beta_{2}=10\log\dfrac{I_{2}}{I_{0}}[/tex]...(II)

We need to calculate the ratio of intensity

From equation (I) and (II)

[tex]\beta_{2}-\beta_{1}=10\log\dfrac{I_{2}}{I_{0}}-10\log\dfrac{I_{1}}{I_{0}}[/tex]

[tex]\Delta \beta=10\log(\dfrac{I_{2}}{I_{1}})[/tex]

Put the value into the formula

[tex]3.0=10\log(\dfrac{I_{2}}{I_{1}})[/tex]

[tex]\dfrac{I_{2}}{I_{1}}=e^{\dfrac{3.0}{10}}[/tex]

[tex]\dfrac{I_{2}}{I_{1}}=1.34[/tex]

We need to calculate the ratio of the displacement

Using formula of displacement

[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{\dfrac{I_{2}}{I_{1}}}[/tex]

Put the value into the formula

[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{1.34}[/tex]

[tex]\dfrac{r_{1}}{r_{2}}=1.16[/tex]

Hence, The ratio of the displacement amplitudes of two sound waves is 1.16.

The gravitational force acting on a lead ball is much larger than that acting on a wooden ball of the same size. Which statement is true about when they are dropped

Answers

Complete question is;

The gravitational force acting on a lead ball is much larger than that acting on a wooden ball of the same size. Which statement is true about when they are dropped?

A) The acceleration due to gravity is the same for both balls.

B) The lead ball’s acceleration due to gravity is larger than the wooden ball’s.

C) The lead ball’s acceleration due to gravity is less than the wooden ball’s.

Answer:

Option A) The acceleration due to gravity is the same for both balls.

Explanation:

Density of Lead is 11.34 g/cm³ while density of hard wood is around 1.3 g/cm³

Thus, it means lead is the heavier object when both have same masses.

We are told that the wood and lead both have same size but from the density given above, it's clear that the mass of the lead ball is more than that of the wooden ball and thus, the force from would be more for the lead ball than the wooden ball.

Now, If we were to measure the position of both balls when falling, we will observe that they do not fall fall with a constant speed. This is because they fall with a constant acceleration and as such the speed increases.

Therefore both balls will hit the ground at the same time because they both start from a position of rest and both will have the same acceleration.

It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.

Answers

Answer:

102 m

Explanation:

Given that It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.

Let the stopping distance be equal to S.

According to the definition of speed,

Speed = distance / time.

make time the subject of the formula

Time = distance / speed

then, the equivalent time is:

48.96 / 12 = S / 25

Cross multiply

12S = 48.96 x 25

12S = 1224

S = 1224 / 12

S = 102 m

Therefore, the stopping distance is 102 m

Derek is watching steam form swirling patterns above the boiling water in his beaker.
Which type of thermal energy transfer is he observing?
A.Conduction
B.Translation
C.Convection
D.Radiation

Answers

Answer:

Convection

Explanation:

Answer:

C. Convection

Explanation:

I just did it

How will the motion of an object that is moving to the right change, if it is pushed in the opposite direction with a greater force?
Оооо
The object will move at a constant speed in the same direction for a while and then slow down and stop.
The object will slow down for a while and then move at a slower constant speed in the same direction.
The object will slow down and then begin to move faster and faster in the opposite direction.
0 The object will speed up and then begin to move at a slower speed in the opposite direction.

Answers

ويويوسوسوسميىيىطنطنظنسنسنمسم ستسري ستي تس ستي تس ستي تس ستي سر س

When adding the components of a vector, the resultant magnitude is always the sum of the two component magnitudes.
True or false?

Answers

Answer:False

Explanation:

Answer:

False

Explanation:

calculate sound intensity

Answers

Answer:

I dont know sorry,I hope you find what are you looking for.

Please help me with this

Answers

Answer:

7 Newton's East

Explanation:

when the force is going in the same direction in this case east, you add the forces.

A spring has natural length 16 cm. A force of 3 N is required to holdthe spring compressed compressed to 11 cm. Find the amount ofwork instretching the spring from 17 cm to 19 cm.

Answers

Answer:

W = 0.012 J

Explanation:

For this exercise let's use Hooke's law to find the spring constant

         F = K Δx

         K = F / Δx

         K = 3 / (0.16 - 0.11)

         K = 60 N / m

Work is defined by

         W = F. x = F x cos θ

in this case the force and the displacement go in the same direction therefore the angle is zero and the cosine is equal to 1

         W = ∫ F dx        

         W = k ∫ x dx

we integrate

         W = k x² / 2

          W = ½ k x²

let's calculate

         W = ½ 60 (0.19 -0.17)²

         W = 0.012 J

A 6.13-g bullet is moving horizontally with a velocity of 361 m/s, where the sign indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1233 g, and its velocity is 0.741 m/s after the bullet passes through it. The mass of the second block is 1646 g. (a) What is the velocity of the second block after the bullet imbeds itself

Answers

Answer:

v₃ = 0.786 m/s

Explanation:

Here, we will use the law of conservation of momentum, which states the following:

Total Momentum of System Before Collision =

Total Momentum of System After Collision

m₁u₁ + m₂u₂ + m₃u₃ = m₁v₁ + m₂v₂ + m₃v₃

where,

m₁ = mass of bullet = 6.13 g = 0.00613 kg

m₂ = mass of 1st block = 1233 g = 1.233 kg

m₃ = mass of 2nd block = 1646 g = 1.646 kg

u₁ = speed of first bullet before collision = 361 m/s

u₂ = speed of first block before collision = 0 m/s

u₃ = speed of 2nd block before collision = 0 m/s

v₁ = speed of bullet after collision

v₂ = speed of 1st block after collision = 0.741 m/s

v₃ = speed of 2nd block after collision = ?

Therefore,

(0.00613 kg)(361 m/s) + (1.233 kg)(0 m/s) + (1.646 kg)(0 m/s) = (0.00613 kg)(v₁) + (1.233 kg)(0.741 m/s) + (1.646 kg)(v₃)

2.2129 kg m/s + 0 kg m/s + 0 kg m/s - 0.9136 kg m/s = (0.00613 kg)(v₁) + (1.233 kg)(0.741 m/s) + (1.646 kg)(v₃)

1.2992 kg m/s = (0.00613 kg)(v₁) + (1.646 kg)(v₃)

since, the bullet is embedded in 2nd block after collision. Thus, there velocities will become same. (v₁ = v₃)

Therefore,

1.2992 kg m/s = (0.00613 kg)(v₃) + (1.646 kg)(v₃)

v₃ = (1.2992 kg m/s)/(1.6521 kg)

v₃ = 0.786 m/s

To begin your activity, open this Tracker experiment: Car Round Trip. Click play to watch the video. The other video controls allow you to rewind the video or step forward or backward one frame at a time.

Answers

Answer:

The car appears to have a constant, positive acceleration for most of the video clip.

The period of a wave is equal to the time it takes for one wavelength to pass by a fixed point. You stand on a pier watching water waves and see 10.9 waves pass by in a time of 28 seconds.What is the period of the water waves? Round to nearest .01 and do not include units in your answer.

Answers

Answer:

T = 2.57 s

Explanation:

The period of a wave is equal to the time it takes for one wavelength to pass by a fixed point.

No of waves observed = 10.9

Time taken, t = 28 s

We need to find the period of the water waves. Number of waves per unit time is called frequency. Let it is f. So,

[tex]f=\dfrac{n}{t}\\\\f=\dfrac{10.9}{28}\\\\f=0.389\ Hz[/tex]

If T is period of the wave. It is equal to the reciprocal of frequency.

[tex]T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.389}\\\\T=2.57\ s[/tex]

So, the period of the water waves is 2.57 seconds.

At a distance of 10.0 m from a loudspeaker, the sound intensity level is measured to be 70 dB. At what distance from the source will the intensity be 40 dB?

Answers

Answer:

At a distance of 100 m from the source the intensity will be 40 dB.

Explanation:

Sound intensity is the acoustic power transferred by a sound wave per unit area normal to the direction of propagation.

The sound intensity depends on the power of the sound source, where the higher the power the greater the intensity, the distance to the sound source, the greater the distance being the lower the intensity, and the nature of the transmission medium.

The conversion between intensity and decibels corresponds to:

[tex]L=10*log\frac{I}{I0}[/tex]

where I0 = 10⁻¹² W/m² and corresponds to a level of 0 decibels therefore.

In this case, you can apply the following relationship between two intensities and distance, considering that the intensity of the sound level decreases with distance:

[tex]L1 - L2=10*log\frac{I1}{I0} - 10*log\frac{I2}{I0}[/tex]

[tex]L1 - L2=10*(log\frac{I1}{I0} - *log\frac{I2}{I0})[/tex]

[tex]L1 - L2=10*[log(\frac{I1}{I0}\frac{I0}{I2})][/tex]

[tex]L1 - L2=10*[log(\frac{I1}{I2})][/tex]

Being L1= 70 dB and L2= 40 dB

[tex]70 dB - 40 dB=10*[log(\frac{I1}{I2})][/tex]

[tex]30=10*[log(\frac{I1}{I2})][/tex]

[tex]\frac{30}{10} =log(\frac{I1}{I2})[/tex]

[tex]3=log(\frac{I1}{I2})[/tex]

[tex]10^{3} =\frac{I1}{I2}[/tex]

[tex]1,000=\frac{I1}{I2}[/tex]

The intensity is inversely proportional to the square of the distance to the source. The relationship between the intensities I1 and I2 at distances d1 and d2 respectively is:

[tex]\frac{I1}{I2} =\frac{d2^{2} }{d1^{2} }[/tex]

Then:

[tex]1,000=\frac{d2^{2} }{d1^{2} }[/tex]

Being d1= 10 m

[tex]1,000=\frac{d2^{2} }{10^{2} }[/tex]

[tex]1,000=\frac{d2^{2} }{100}[/tex]

1,000*100= d2²

10,000= d2²

√10,000= d2

100 m= d2

At a distance of 100 m from the source the intensity will be 40 dB.

Blood is 92% water. Blood is
drawn using a capillary tube.
Write 1 sentence explaining how it
illustrates this characteristic of
water. Use the terms adhesion
and capillary action.

Answers

Answer:

Plasma, which constitutes 55% of blood fluid, is mostly water (92% by volume), and contains proteins, glucose, mineral ions, hormones, carbon dioxide (plasma being the main medium for excretory product transportation), and blood cells themselves.

Explanation:

Capillary motion is crucial for circulating water. Your body's cells would not hydrate without this flow, and crucial communication between your brain and body would slack off, hence blood contains water 92% water helps in capillary action.

What is Capillary action?

Capillary action, also known as capillary effect or motion, is the process by which liquid moves through constricted areas without the aid of external forces like gravity but rather with the help of intermolecular forces that exist between the liquid and solid surface (s).

Adhesion: The attraction of two different molecules, such as the hydrogen and oxygen molecules found in water and plastic drinking straws.

Proteins, glucose, mineral ions, hormones, carbon dioxide (plasma is the principal medium for excretory product transfer), and blood cells themselves are contained in plasma, which makes up 55% of blood fluid. Plasma is mostly water (92% by volume) and contains water as well as several other substances.

Therefore, the water contains in the blood helps in capillary action.

To know more about Capillary action:

https://brainly.com/question/13228277

#SPJ2

Hollywood and video games often depict the bad guys being "blown away" when they’re shot by a bullet (i.e. once hit, their feet leave the ground and they fly backwards). Assuming that even if a handgun cartridge did generate enough momentum for the bullet to do this, why is it still nonsense on-screen?

Answers

Answer:

Taking a look at Newton's third law of motion which states "for every force exerted, their is an opposite force equal in magnitude and opposite in direction on the first force".

Similarly if a bullet had enough forces behind it to hurl someone through the air when they were hit, a similar force would act on the person holding the gun that fired the bullet.  

What we load into the gun is called a 'cartridge' Each piece is composed of four basic substance the casing, the bullet, the primer, and the powder.  

The primer explodes lighting the powder which causes a buildup of pressure behind the bullet. This powder can be used in rifle cartages because the bullet chamber is designed to withstand greater pressures.  

It is difficult in practice to measure the forces within a gun bagel, but the one easily measured parameter is the velocity with which the bullet exits muzzle velocity, therefore assuming that even if a handgun cartridge which generate enough momentum for the bullet to do this,  it is still nonsense on screen in Hollywood and video.  

             

What is the mass of an object that has a volume of 56 ml and a density of 1.24 g/ml

Answers

Answer:

The answer is 69.44 g

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

mass = Density × volume

From the question

volume = 56 ml

density = 1.24 g/ml

We have

mass = 56 × 1.24

We have the final answer as

69.44 g

Hope this helps you

a 12000.0-kg car is traveling at 19 m/s. the driver suddenly slams on the brakes and skids to a stop. The coefficient of kinetic friction between the tires and the road is 0.80. for what length of time is the car skidding?

Answers

What I don’t get it ?

A system has the same velocity profile but a depth of 10 feet. The average velocity of the stream with a depth of 10 feet is __________ the stream with a depth of 6 feet.

a. Greater than
b. Less tharn
c. The same as
d. The answer cannot be determined with the given information.

Answers

Answer:

The average velocity of the stream with a depth of 10 feet is greater.

(a) is correct option.

Explanation:

Given that,

Depth [tex]h_{1}= 10\ feet[/tex]

Depth [tex]h_{2}=6\ feet[/tex]

We need to calculate the average velocity of the stream

According to question,

[tex]h_{1} > h_{2}[/tex]

The velocity for first case,

[tex]v_{1}=u_{1}\dfrac{x_{1}}{h_{1}}[/tex]

[tex]\dfrac{v_{1}}{x_{1}}=\dfrac{u_{1}}{h_{1}}[/tex]

The velocity for second case,

[tex]v_{2}=u_{2}\dfrac{x_{2}}{h_{2}}[/tex]

[tex]\dfrac{v_{2}}{x_{2}}=\dfrac{u_{2}}{h_{2}}[/tex]

For the same velocity profile,

[tex]\dfrac{dv}{dx}=\dfrac{v_{1}}{x_{1}}=\dfrac{v_{2}}{x_{2}}[/tex]

Then,

[tex]\dfrac{u_{1}}{h_{1}}=\dfrac{u_{2}}{h_{2}}[/tex]

Put the value into the formula

[tex]\dfrac{u_{1}}{10}=\dfrac{u_{2}}{6}[/tex]

[tex]u_{1}=\dfrac{5}{3}u_{2}[/tex]

[tex]u_{1}=1.67u_{2}[/tex]

The velocity is [tex]u_{1} > u_{2}[/tex]

We need to calculate the average velocity for first case

Using formula of average velocity

[tex]v_{avg}_{1}=\dfrac{0+u_{1}}{2}[/tex]

Put the value into the formula

[tex]v_{avg}_{1}=\dfrac{0+u_{1}}{2}[/tex]

[tex]v_{avg}_{1}=\dfrac{u_{1}}{2}[/tex]

We need to calculate the average velocity for second case

Using formula of average velocity

[tex]v_{avg}_{2}=\dfrac{0+u_{2}}{2}[/tex]

Put the value into the formula

[tex]v_{avg}_{2}=\dfrac{0+u_{2}}{2}[/tex]

[tex]v_{avg}_{2}=\dfrac{u_{2}}{2}[/tex]

If [tex]u_{1} > u_{2}[/tex] then [tex]\dfrac{u_{1}}{2} >\dfrac{u_{2}}{2}[/tex]

So, we can say that the average velocity of the stream with a depth of 10 feet will be greater than the stream with a depth of 6 feet.

Hence, The average velocity of the stream with a depth of 10 feet is greater.

(a) is correct option.

my heart strike him to dead.what figure of speech is that?​

Answers

Answer:

Hyperbole

Explanation:

this is an extreme exaggeration or overstatement/ magnification

g A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the tube has magnitude f. You attach the upper end of a lightweight vertical spring of force constant k to the cap at the top of the tube, and attach the lower end of the spring to the top of the cylinder. Initially the cylinder is at rest and the spring is relaxed. You then release the cylinder. What vertical distance will the cylinder descend before it comes momentarily to rest

Answers

Answer:

The vertical distance is  [tex]d = \frac{2}{k} *[mg + f][/tex]

Explanation:

From the question we are told that

   The mass of the cylinder is  m

    The kinetic frictional force is  f

Generally from the work energy theorem

    [tex]E  =  P +  W_f[/tex]

Here E the the energy of the spring which is increasing and this is mathematically represented as

       [tex]E =  \frac{1}{2} * k  *  d^2[/tex]

Here k is the spring constant

        P is the potential energy of the cylinder which is mathematically represented as

     [tex]P  = mgd[/tex]

And

     [tex]W_f[/tex]  is the workdone by friction which is mathematically represented as

      [tex]W_f  =  f *  d[/tex]

So

    [tex] \frac{1}{2} * k  *  d^2 =  mgd +  f *  d [/tex]

=>    [tex] \frac{1}{2} * k  *  d^2 =  d[mg +  f    ][/tex]

=>  [tex] \frac{1}{2} * k  *  d =  [mg +  f    ][/tex]

=> [tex]d = \frac{2}{k} *[mg + f][/tex]

A wire loop with 3030 turns is formed into a square with sides of length ss . The loop is in the presence of a 1.20 T1.20 T uniform magnetic field B⃗ B→ that points in the negative yy direction. The plane of the loop is tilted off the x-axisx-axis by θ=15∘θ=15∘ . If i=1.10 Ai=1.10 A of current flows through the loop and the loop experiences a torque of magnitude 0.0256 N⋅m0.0256 N⋅m , what are the lengths of the sides ss of the square loop, in centimeters?

Answers

Answer:

2.59 cm

Explanation:

The torque τ on a current carrying loop of wire is given by τ = NiABsinθ where N = number of turns of loop, i = current in loop, A = area of loop and B = magnetic field.

Now, given that τ = 0.0256 Nm, i = 1.10 A, B = 1.20 T,N = 30 and since the loop is tilted 15° off the x-axis and the magnetic field points in the negative y- direction, the angle between the normal to the loop and the magnetic field is thus 90° - 15° = 75°. So, θ = 75°.

We now find the area of the loop A from

τ = NiABsinθ

A = τ/NiBsinθ

substituting the values of the variables, we have

A = 0.0256 Nm/30 × 1.10 A × 1.20 T × sin75°

A = 0.0256 Nm/38.25

A = 6.69 × 10⁻⁴ m²

Since the loop is a square, with length of side L, its area A = L² and

L = √A

= √(6.69 × 10⁻⁴ m²)

= 2.59 × 10⁻² m

converting to cm, we have

L = 2.59 × 10⁻² m × 100 cm/m

L = 2.59 cm

So, the lengths of sides of the loop is 2.59 cm

A loop of area 0.250 m^2 is in a uniform 0.020 0-T magnetic field. If the flux through the loop is 3.83 × 10-3T· m2, what angle does the normal to the plane of the loop make with the direction of the magnetic field?

Answers

Answer:

40.0⁰

Explanation:

The formula for calculating the magnetic flux is expressed as:

[tex]\phi = BAcos\theta[/tex] where:

[tex]\phi[/tex] is the magnetic flux

B is the magnetic field

A is the cross sectional area

[tex]\theta[/tex] is the angle that the normal to the plane of the loop make with the direction of the magnetic field.

Given

A = 0.250m²

B = 0.020T

[tex]\phi[/tex] = 3.83 × 10⁻³T· m²

3.83 × 10⁻³ = 0.020*0.250cosθ

3.83 × 10⁻³ = 0.005cosθ

cosθ = 0.00383/0.005

cosθ = 0.766

θ = cos⁻¹0.766

θ = 40.0⁰

Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰

The seasons of the year result from Earths movement, or____, around the sun.

Answers

Answer:

rotation

Explanation:

The seasons of the year result from Earths movement, or rotation around the sun. The answer for the blanks is rotation .

What is the earth's orbit?

The Earth orbits the Sun in a counterclockwise pattern above the Northern Hemisphere at an average distance of 149.60 million km.

The Earth revolves around its axis once every 24 hours and orbits the sun once every 365 days. The Earth's rotation on its axis, not its orbit around the sun, causes day and night.

The term 'one day' refers to the time it takes the Earth to rotate once on its axis, which includes both day and night.

A season is a division of the year that is based on variations in weather, ecology, and the number of daylight hours in a certain place.

Seasons on Earth are caused by Earth's orbit around the Sun and its axial tilt relative to the ecliptic plane.

The seasons of the year result from Earths movement, or rotation around the sun.

Hence, the answer for the blanks is rotation .

To learn more about the earth's orbit, refer to the link;

brainly.com/question/15022652

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) A 1000-gallon tank currently contains 100.0 gallons of liquid toluene and a gas saturated with toluene vapor at 85°F and 1 atm. (a) What quantity of toluene (lbm) will enter the atmosphere when the tank is filled and the gas displaced? (b) Suppose that 90% of the displaced toluene is to be recovered by compressing the displaced gas to a total pressure of 5 atm and then cooling it isobarically to a temperature T(°F). Calculate T.

Answers

Answer:

A) m[tex]_{T}[/tex] = 0.3025 * 0.0476 * 92.13 = 1.327 Ibm

B) T= 63.32°F

Explanation:

Given data:

1000 gallon tank currently contains 100.0 gallons of liquid toluene

and A gas saturated with toluene vapor at 85°F and 1 atm

A) Calculate quantity of toluene ( Ibm ) that will enter the atmosphere when the tank is filled

m[tex]_{T}[/tex] = [tex]n_{gas} * Y_{T} * M_{T}[/tex]

[tex]n_{gas}[/tex] (total mole of gas) = 0.3025 Ib-mole  ( calculated using : [tex]\frac{PV}{RT}[/tex] )

[tex]Y_{T}[/tex] (mole fraction of toluene) = 0.0476 ( calculated using [tex]\frac{P_{T} }{P}[/tex] )

M[tex]_{T}[/tex] = 92.13 Ibm/Ib-mole

therefore:  m[tex]_{T}[/tex] = 0.3025 * 0.0476 * 92.13 = 1.327 Ibm

B) using Antoine equation to solve for T

Antoine equation : [tex]log_{10} (P_{T} ) = A - \frac{B}{T+C}[/tex]

PT( partial pressure ) = 18.95 ( calculated using : [tex]y_{tb} * P[/tex] )

A = 6.95805

B = 1346.773

T = ?

C = 219.693

to calculate T make T the subject the subject of the equation

T + 219.693 = 1346.773 / 5.68044

∴ T = 17.40°C

convert T to Fahrenheit

T = 1.8 * 17.40 +32

  = 63.32°F

A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal component of the force is 4.5 N. At what angle (in degrees) above the horizontal is the force directed?

Answers

Answer:

Fy = 14.3 [N]

Explanation:

To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:

When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.

[tex]F = \sqrt{F_{x}^{2} +F_{y}^{2} }[/tex]

where:

F = 15 [N]

Fx = horizontal component = 4.5 [N]

Fy = vertical component [N]

[tex]15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\ F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\ F_{y}=14.3 [N][/tex]

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