A boy pulls on a wagon with a force of 100 N [E]. The wagon pulls on the boy with a
force of
A). zero
B) less than 100 N [E]
C) 100 N [W]
D) Less then 100 N [W]
E) greater than 100 N [W]

Answers

Answer 1
The correct answer is F) 69 because it just is trust the expert which is me

Related Questions

How does the area of triangle rst compare to the area of triangle lmn? the area of △rst is 2 square units less than the area of △lmn. the area of △rst is equal to the area of △lmn. the area of △rst is 2 square units greater than the area of △lmn. the area of △rst is 4 square units greater than the area of △lmn.

Answers

The area of △rst is equal to the area of △lmn.

The answer is option B.

Problem-solving is the act of defining a problem; figuring out the purpose of the trouble; identifying, prioritizing, and selecting alternatives for an answer; and imposing an answer.

Problem-solving starts with identifying the issue. As an example, a trainer may need to parent out a way to improve a scholar's overall performance on a writing talent test. To do that, the instructor will overview the writing exams looking for regions for improvement.

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Which of the following graphs represents the equation y-2 = 3(x-1)?

Answers

Answer:

Graph b

Explanation:

I used Desmond calculator which is a calculator where you can graph an equation! :)

Hello!

The slope-intercept form is :

y - 2 =  3(x - 1)

y - 2 = 3x - 3

y = 3x - 1

Two points that are found to be on the line :

(0, -1) and (1, 2)

∴ The graph which contains both points is Graph B.

Calculate the gravitational force of the Earth and Moon. The Earth has a mass of 5.972x 1024 kg and the Moon has a mass of 7.348x 1022 kg. They are an average of 384 million meters apart.

Answers

Answer:

The force of gravitation is 1.98×10^(20) m.

Explanation:

Note: The value of m1, m2, r, and G is taken to be 5.97×10^(24) kg, 7.348×10^(22) kg, 3.84×10^(8) m, and 6.67×10^(-11) N m^2/ kg^2 respectively.

The force of gravitation F acting between two masses is calculated using the formula,

F=Gm1m2/ r^2

where G is the gravitational constant.

Hence force F is

F=6.67×10^(-11)×5.97×10^(24)×7.348×10^(22)/ (3.84×10^(8))^2

F=1.98×10^(20) N

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What are two forces on earth that could change the motion of the object?

Answers

Answer:

Force makes things move or, more accurately, makes things change their motion. Two natural forces that we have experienced are the force of gravity and magnetic forces magnetic forces. These two forces act at a distance and do not require direct contact between the objects to function..

A toy dart gun uses a spring with a spring constant of 35 N/m. To use the dart gun, you compress the spring by pushing in a dart of mass 0.002 kg.
a. If you compress the spring by 0.02 m, what force is the spring exerting on the dart?

b. With the spring compressed 0.02 m, how much elastic potential energy is stored in the spring?

c. If you release the spring, it pushes the dart forward. What is the kinetic energy of the dart when it reaches the natural length of the spring? Explain how you found the answer.

d. What is the speed of the dart when it reaches the natural length of the spring?

Answers

(a) The force is the spring exerting on the dart is 0.7 N.

(b) The elastic potential energy stored in the spring is 7 x 10⁻³J.

(c) The kinetic energy of the dart when it reaches the natural length of the spring is 7 x 10⁻³J.

(d) The speed of the dart when it reaches the natural length of the spring is 2.65 m/s.

Force the spring exerting on the dart

F = kx

where;

k is spring constantx is compression of the spring

F = 35 N/m x 0.02 m

F = 0.7 N

Energy stored in the spring

U = ¹/₂kx²

U = ¹/₂(35)(0.02)²

U = 7 x 10⁻³ J

Kinetic energy of the dart

The elastic potential energy of the spring will be converted into kinetic energy of the dart, with a magnitude of  7 x 10⁻³ J when it reaches natural length of the spring.

Speed of the dart

K.E = ¹/₂mv²

v² = (2K.E)/m

where;

v is the speed of the dartm is mass of the dart

v² = (2 x 7 x 10⁻³)/(0.002)

v² = 7

v = 2.65 m/s

Thus, the force is the spring exerting on the dart is 0.7 N.

The elastic potential energy stored in the spring is 7 x 10⁻³J.

The kinetic energy of the dart when it reaches the natural length of the spring is 7 x 10⁻³J.

The speed of the dart when it reaches the natural length of the spring is 2.65 m/s.

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lamp 1 and lamp 2 are in a series circuit
lamp 1 stops working
explain why lamp 2 also stops working​

Answers

Lamps 1 and 2 are connected in series. lamp 1 stops working because the circuit is not complete.

What is a series circuit?

In a series circuit, the amount of current flowing through any component is the same and the sum of the individual resistances equals the overall resistance of any series circuit.

The voltage in a series circuit is equal to the total of the individual voltage drops.

For the bulb to glow the circuit should be complete. If the circuit is not completed the bulb will not glow.

Lamps 1 and 2 are connected in series. lamp 1 stops working because the circuit is not complete.

Hence, lamp 1 stops working.

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A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The dragster's final speed is 130 m/s. Neglecting friction, what average power was needed to produce this final speed

Answers

The average power needed to produce this final speed is 1069.1 hp.

Mass of the dragster,  m = 500.0 kg,

Displacement travelled by the dragster,  s = 402 m,

Time taken in this travel,  t = 5.0 s,

Final velocity of the dragster,  v = 130 m/s.

Let the initial velocity of the dragster be u and acceleration be a.

Using kinematical equation,  s = ut + (1/2)at^2.

402  =  u*5  + (1/2)*a*5^2

10*u + 25*a  = 804.      ...........(1)

Using kinematical equation, v = u +at.

130 = u + 5*a

5*u + 25*a = 650.       .............(2)

Solving (1)and (2), we get,

u =  30.8 m/s.

According to work-energy theorem,

Work done = change in kinetic energy

W  = (1/2)*m*(v^2 - u^2)

W = (1/2)*500*(130^2 - 30.8^2)

W  =  3987840. J

Therefore power rating of the dragster is given by,

P  ⇒  W/t. =  3987840/5 = 797568 watt.

P  ⇒ 797568/746 =  1069.1 hp.

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The answer is 1069.1 hp.  

Mass of the dragster,  m = 500.0 kg,

Displacement travelled by the dragster,  s = 402 m,

Time taken in this travel,  t = 5.0 s,

Final velocity of the dragster,  v = 130 m/s.

Let the initial velocity of the dragster be u and acceleration be a.

Using kinematical equation,  s = ut + (1/2)at^2.

 402  =  u*5  + (1/2)*a*5^2

 10*u + 25*a  = 804.      ...........(1)

Using kinematical equation, v = u +at^.

 130 = u + 5*a^

5*u + 25*a = 650.       .............(2)

Solving (1)and (2), we get,

u =  30.8 m/s.

According to work -energy theorem,

Work done = change in kinetic energy

 W  = (1/2)*m*(v^2 - u^2)

,  W = (1/2)*500*(130^2 - 30.8^2)

,  W  =  3987840. J

therefore Power rating of the dragster is given by,

P  =  W/t. =  3987840/5 = 797568 watt.

,  P  = 797568/746 =  1069.1 hp.        

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A mass m is placed on an inclined plane (m > 0) and slides down the plane with constant speed. If a similar block (same m) of mass 2m were placed on the same incline, it would:

Answers

A mass m is placed on an inclined plane (m > 0) and slides down the plane with constant speed. If a similar block (same m) of mass 2m were placed on the same incline, it would

For 2m, the gravitational force pressing downward on the plane is twofold. However, the normal force is also doubled, and as a result, so is the friction force (the same factor). This indicates that the two forces continue to cancel and produce a net force of zero.

Gravitational forceThe force that binds all masses in the universe together, particularly the pull of the earth's mass on objects that are close to its surface.The attracting force that the earth exerts on all other objects in the cosmos is known as gravitational force. A ball being flung high and then falling back to the earth as a result of gravitational force is an example of gravitational force.

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A mass is undergoing simple harmonic motion. When its displacement is 0, it is at its equilibrium position. At that moment, its speed is _______ and its acceleration is _______.

Answers

When displacement is 0, an object is at its equilibrium position, at that moment, its speed is maximum and its acceleration is zero.

What is speed?

Speed is a scalar quantity used in physics to indicate the distance of change of an object that travels in any direction.

Conversely, displacement and acceleration are vector quantities because they express magnitudes but not a direction.

In conclusion, when an object is at its equilibrium position its speed is maximum and its acceleration is zero.

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On the way to school, the bus speeds up from 20 m/s to 36 m/s in 4 seconds. What distance
does the bus cover in this time frame?
A. 64 m
B. 112 m
C. 80 m
D. 144 m

Answers

Answer B. 112 m



Step-by-Step Explanation

initial velocity u = 20 m /s
final velocity v = 36 m /s
time taken t = 4 s
acceleration = (v - U) / t
= (36 - 20) / 4
a=4m/s2
from the formula
7-u2=2as , sis distance covered
putting the values
362-202=2×4×s
1296 - 400 = 8 x S
S= 112 m

The speed that a tsunami can travel is modeled by the equation , where s is the speed in kilometers per hour and d is the average depth of the water in kilometers. What is the approximate depth of water for a tsunami traveling at 200 kilometers per hour?.

Answers

0.32 km

How to reach the answer:

The formula below provides the tsunami's speed in response to the query.

s = 356√d

when the tsunami's speed is 200 km/h, the equation's "s" is changed to "200," and the equation of speed is then expressed as follows.

200 = 356√d

The equation can be changed to compute the depth of the water as,

√d = 200÷356

√d = 0.562

Now on squaring both sides:

(√d)² = (0.562)²

d = (0.562)² = 0.316 or 0.32

Therefore the approximate depth of water for a tsunami travelling at 200km/h is 0.32 km

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A typical muscle fiber is 2.0 cm long and has a cross-section area of 3.1 * 10-9 m2. When the muscle fiber is stimulated, it pulls with a force of 1.2 mN. What is the work done by the muscle fiber as it contracts to a length of 1.6 cm

Answers

= F.d

F = 1.2 x 10-3N

d= (2-1.6)10-2=0.004m

W=1.210-3(0.004)=4.8 x 10 -6 J

A typical muscle fiber is 2.0 cm long and has a cross-section area of 3.1 * 10-9 m2. When the muscle fiber is stimulated, it pulls with a force of 1.2 mN.

Please check the attached file for a brief answer.

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The work done by the muscle fiber is 4.80x 10 J

The work done by the muscle fiber as it contracts to a length is

W = FAL

= (1.20x103 N) (0.4x10-2 m)

= 4.80x 10 J

Muscle tissues contain something known as muscle fibers. Muscle fibers consist of a single muscle cellular.

They assist to control the physical forces in the body. When grouped collectively, they are able to facilitate organized movement of your limbs and tissues.

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A 5.0 kg box slides down a 4.0 m long ramp that makes a 25 angle with the ground. If the coefficient of kinetic friction is 0.65, how much thermal energy was produced?

Answers

The thermal energy was produced is 116J

What is the thermal energy produced?

Now we know that the frictional force produces the energy that is lost as heat as the body slides down the incline. The magnitude of the frictional force is obtained from;

Ff= μmgcosθ

Ff =  0.65 *  5.0 kg * 9.8 m/s^2 * cos 25

Ff = 29 N

Hence, the thermal energy is;

29 N * 4.0 m = 116J

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A 18 kg box slides from rest down a ramp inclined at 25° to the horizontal onto a spring with a spring constant 740 N/m
as shown in the diagram. The spring is compressed 0.32 m before the box stops. Determine how far along the ramp the
box will slide before it stops. Assume friction is negligible.

Answers

The distance traveled by the box along the ramp is 0.51 m.

Distance traveled by the box

Apply the principle of conservation of energy;

Potential energy of the box at top of incline = Elastic potential energy at bottom

mgh = ¹/₂kx²

where;

h is the height of the incline m is mass of the boxx is the compression of the springk is spring constant

h = L sinθ

where;

L is length of the incline = distance traveled by the box

mg(L sinθ) = ¹/₂kx²

(18)(9.8)(L sin25) = ¹/₂(740)(0.32)²

74.55 L = 37.89

L = 37.89 / 74.55

L = 0.51 m

Thus, the distance traveled by the box along the ramp is 0.51 m.

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A satellite orbits the earth at a speed of 8.0x 10^3m/s . Calculate the period of orbit of the satellite if the distance between the centre of the earth and the satellite is 6400km?​

Answers

Hi there!

The period of an orbit can be found by:

[tex]T = \frac{2\pi r}{v}[/tex]

T = Period (? s)
r = radius of orbit (6400000 m)

v = speed of the satellite (8000 m/s)

This is the same as the distance = vt equation. The total distance traveled by the satellite is the circumference of its circular orbit.

Let's plug in what we know and solve.

[tex]T = \frac{2\pi (6400000)}{8000} = \boxed{5026.55 s}[/tex]

A solid disk is rolling without slipping on a level surface at a constant speed of 2.50 m/s. (a) if the disk rolls up a 30.0o ramp, how far along the ramp will it move before it stops? (b) explain why your answer in part (a) does not depend on either the mass or the radius of the disk.

Answers

(a) By using the law of conservation of energy, the distance moved by the solid disk along the ramp is 0.96 m.

(b) The answer does not depend on the mass and radius of the disc as these values are canceled when the law of conservation of energy is applied.

What is the law of conservation of energy?

The law of conservation of energy states that the total energy of an isolated system is conserved.

The total initial energy Ei of the disc rolling down an inclined plane is given by the formula,

Ei=1/2*mv^2 + 1/2*Iω^2

where m is the mass of the disc, v is the velocity of the disc, I is the moment of inertia and ω is the angular velocity of the disc.

For a solid disc, I=1/2mr^2, and since it is rolling without slipping, the rolling velocity of the disc will be equal to its translational velocity, that is,

v=ωr or ω=v/r

So using ω=v/r and I=1/2mr^2, it can be written,

Ei=1/2*m(v)^2 + 1/2*(1/2mr^2)(v/r)^2

Ei=1/2*m(v)^2*(1+1/2*)

Ei=3/4*mv^2

If the height covered by the disc is h before stopping, then its final total  energy Ef will be equal to the potential energy, that is,

Ef=m*g*h

From the law of conservation of energy, it can be written,

Ei=Ef

3/4*mv^2=m*g*h

h=3v^2/(4g)

The length of the ramp is then given by the formula,

l=hcosecθ

where θ is the inclination angle. So

l= 3v^2/(4g)*cosecθ

Here g=9.8 m/s^2, v=2.5 m/s  and θ=30.0 degree. Using these values,

l= 3*(2.5 )^2/(4*9.8)*cosec( 30)

l= 0.96 m

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The primary coil of a transformer is connected to a 120 V wall outlet. The secondary coil is connected to a lamp that dissipates 60W. What is the rms current in the primary coil

Answers

The rms current in the primary coil of a transformer is 0.5 Ampere .

Energy/power is not gained or lost going through an ideal transformer.

Consequently, the transformer in this situation doesn't really matter. The entire device receives 60 watts of power from the wall outlet if the lamp consumes energy at a rate of 60 watts.

∴ Power = (voltage) x (current)

60 watts = (120 v) x (current)

Current = (60 watts) / (120 v)

Current = 0.5 Ampere

Therefore, the rms current in the primary coil of a transformer is 0.5 ampere.

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A projectile is launched with a velocity of 13.2 m/s at an angle of 37.0° above the horizontal.

How high above the point from which it was fired will the projectile be 1.0 s after firing?

a.
3.0 m


b.
12.8 m


c.
5.6 m


d.
13.2 m

Answers

The height reached by the projectile from the point of projection is determined as 12.84 m.

Height reached by the projectile

H = vt + ¹/₂gt²

where;

v is vertical velocityt is time of motiong is acceleration due to gravity

H = (13.2 x sin37)(1) + (0.5)(9.8)(1)²

H = 12.84 m

Thus, the height reached by the projectile from the point of projection is determined as 12.84 m.

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Please answer number 3!! High-speed stroboscopic photographs show that the head of a 0.2kg golf club is traveling
at 55 m/s just before it strikes a 0.046kg golf ball at rest on a tee. After the collision, the
golf club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball
just after impact.

Answers

The velocity of the golf ball is 65.2 m/s

What is the principle of conservation of momentum?

The principle of conservation of momentum is that the momentum before collision is equal to the momentum after collision.

Given that;

(0.2kg *  55 m/s) + (0.046kg * 0 m/s) = (0.2kg * 40 m/s) + (0.046kg  * v)

11 = 8 + 0.046v

v = 11 - 8/0.046

v = 65.2 m/s

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A local FM radio station broadcasts at an energy of kJ/photon. () Calculate the frequency at which it is broadcasting. Frequency

Answers

The frequency is 94.2 Hz

We  are given the energy of a photon = 6.24x10-29 kJ

We  are asked to find the frequency (ν),

We use the equation E = hν where E is energy, h is Planck's constant and ν is the frequency

The quantity of events per unit of time that occur when an event repeats itself at regularly spaced intervals is known as its frequency.

6.24x10-29 kJ = 6.626x10-34 J-sec * ν

ν = 6.24x10-29 kJ x 1000 J/kJ  / 6.626x10-34 Jsec

ν = 6.24x10-26 J / 6.626x10-34 Jsec

ν = 0.942x108 s-1 = 9.42x107 s-1 = 9.42x107 Hz

9.42x107 Hz x 1 MHz/106 Hz = 94.2 MHz

Hence the frequency is 94.2MHz

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A toroid has a 4.63 cm square cross section, an inside radius of 15.0 cm, 527 turns of wire, and a current of 0.725 A. What is the magnetic flux through the cross section

Answers

Using the equation, B = (μoIN/2πr)

The inner radius is r = 16.2 cm,

so the field there is

B = (4π 10-7).(0.813).(535)/2π(0.162)

= 5.37 × 10-4 T

The outer radius is r = 16.2 + 5.2 = 21.4 cm.

The field there is B = (4π 10-7).(0.813).(535)/2π(0.214)

= 4.06 × 10-4 T.

A toroid having a square cross-section, 5.20cm on edge and an inner radius of 16.2cm has 535 turns and carries a current of 813mA to calculate the magnetic.

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A spring is hung from the ceiling. A 2.0-kg mass suspended hung from the spring extends it by 6.0 cm. A downward external force applied to the mass extends the spring an additional 10 cm. What is the work done by the force

Answers

The work done by force on a spring hung from the ceiling will be 1.67 J

Any two things with mass are drawn together by the gravitational pull. We refer to the gravitational force as attractive because it consistently seeks to draw masses together rather than pushing them apart.

Given that a spring is hung from the ceiling with a 2.0-kg mass suspended hung from the spring extends it by 6.0 cm and a downward external force applied to the mass extends the spring an additional 10 cm.

We need to find the work done by the force

Given mass is of 2 kg

So let,

F = 2 kg

x = 0.1 m

Stiffness of spring = k = F/x

k = 20/0.006 = 333 n/m

Now the formula to find the work done by force will be as follow:

Workdone = W = 0.5kx²

W = 0.5 x 333 x 0.1²

W = 1.67 J

Hence the work done by force on a spring hung from the ceiling will be 1.67 J

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A box having a mass of 7.3 kg is lifted 0.60 m from ground level to the top of a desk. What is the gravitational potential energy in J of the box on the desktop?

Answers

Considering the definition of potential energy, the potential energy of the box is 42.9678 J.

Potential energy

Potential energy is the energy that measures the ability of a system to do work based on its position. In other words, this is the energy that a body located at a certain height above the ground has.

Gravitational potential energy is the energy associated with the gravitational force. This will depend on the relative height of an object to some reference point, the mass, and the strength of gravity.

So for an object with mass m, at height h, the expression applied to the gravitational energy of the object is:

Ep= m×g×h

Where:

Ep is the potential energy in joules (J).m is the mass in kilograms (kg).h is the height in meters (m).g is the acceleration of fall in m/s² (approximately 9.81 m/s²).

Potential energy in this case

In this case, you know:

m= 7.3 kgg= 9.81 m/s²h= 0.60 m

Replacing in the definition of potential energy:

Ep= 7.3 kg × 9.81 m/s² × 0.60 m

Solving:

Ep= 42.9678 J

Finally, the potential energy of the box is 42.9678 J.

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Suppose that the engine of a 1,500 kg automobile has a maximum power output of 30 hp. What is the maximum grade (in percent) that the automobile can climb at 41 km/h if the drag force on it is 410 N

Answers

The maximum grade (in percent) is 14%.

Suppose angle be theta.

Power = force * velocity

35*746 = [1300*9.8* sin theta + 410]* (43*1000/3600)

sin theta = (35*746/(43*1000/3600) - 410)/(1300*9.8) = 0.139

grade = tan ( arcsin 0.139 )

= 0.140

= 14%

Drag is a mechanical pressure. it is generated via the interplay and speaks to a stable body with a fluid (liquid or fuel). It is not generated via a force area, within the feel of a gravitational field, or an electromagnetic field, wherein one item can have an effect on some other item without being in physical touch.

For instance drag on a ship moving in water or drag on an aircraft shifting in the air. therefore drag pressure is the resistance pressure resulting from the motion of a body thru a fluid like water or air. This drag force acts opposite to the path of the oncoming float velocity.

Drag is generated via the distinction in speed between the strong item and the fluid. There needs to be motion among the object and the fluid. If there's no movement, there's no drag. It makes no difference whether or not the item moves via a static fluid or whether or not the fluid movements beyond a static stable item.

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The maximum grade (in percent) is 14%.

Suppose angle be theta.

Power = force * velocity

35*746 = [1300*9.8* sin theta + 410]* (43*1000/3600)

sin theta = (35*746/(43*1000/3600) - 410)/(1300*9.8) = 0.139

grade = tan ( arcsin 0.139 )

= 0.140

= 14%

Drag is a mechanical pressure. it is generated via the interplay and speaks to a stable body with a fluid (liquid or fuel). It is not generated via a force area, within the feel of a gravitational field, or an electromagnetic field, wherein one item can have an effect on some other item without being in physical touch.

For instance drag on a ship moving in water or drag on an aircraft shifting in the air. Therefore drag pressure is the resistance pressure resulting from the motion of a body thru a fluid like water or air. This drag force acts opposite to the path of the oncoming float velocity.

Drag is generated via the distinction in speed between the strong item and the fluid. There needs to be a move between the object and the fluid. If there's no movement, there's no drag. It makes no difference whether or not the item moves via a static fluid or whether or not the fluid moves beyond a static stable item.

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Suppose a car traveling at 8m/s is brought to rest in a distance of 20m.what is it's deceleration and time taken.

Answers

The deceleration is -8 m/s² and the time taken by the car is 2.5 s.

It is given that,

Speed, v = 8 m/s

Distance, d = 20m

We need to find the time taken by the car and deceleration.

As we know speed is defined as

v = d/t

Plugging in the values,

t =  20/8

t = 2.5s

Now from the deceleration formula,

a =  - v/ t

a is the deceleration, v is the speed and t is the time taken.

a = - 20/ 2.5

a = - 8 m/s²

Thus, the time taken by the car and deceleration is 2.5 s and -8 m/s² respectively.

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7. Suppose a force of 15 N is required to stretch and hold a spring 0.25 m from its equilibrium position. a. Assuming the spring obeys Hooks law, find the spring constant k. b. How much work is required to compress the spring 0.2 m from its equilibrium position

Answers

Spring constant k = 75 N/m

Hooke's law for this spring F(x) = 75x

Work = 1.5 J

What is work?When an object is moved over a distance by an external force, at least a portion of that force must be applied in the direction of the displacement. This is known as work.The amount of work required to move an object down a line from point a to point b in the direction of a variable force F is

                                          [tex]$W=\int_{a}^{b} F(x) d x$[/tex]

First solution

According to Hooke's law, the spring must be held stretched at x=0.2m by a force of 15N.

F (0.2) = k (0.2 m) = 15 N

Solve for k

k = 75 N/m

Hence, Hooke's law is F(x) = 75x

Second solution

The spring is compressed from x = 0 to X = 0.2,

[tex]$W=\int_{0}^{0.2} 75 x d x=\left.\left(\frac{75 x^{2}}{2}\right)\right|_{0} ^{0.2}$[/tex]

W = 1.5 J

So, the work is required to compress the spring 0.2 m from its equilibrium position W = 1.5 J

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Use the graph to complete the sentences about two runners.

Runner A is
than runner B.

Runner
has a head start.

Answers

Runner A is faster than runner B and Runner B has a head start.

What is a velocity time graph?

The velocity time graph shows us how fast that an object can be. The slope of the graph is the acceleration.

If we look at the graph as shown in the image attached to this answer, we can see that Runner A is faster than runner B and Runner B has a head start.

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Answer:

Runner A is

✔ faster

than runner B.

Runner

✔ B

has a head start.

Explanation:

I Explain why you feel cold when tap or well water in winter. state with reason.

Answers

Answer:

We feel cold when tap or well water in winter because heat flows from hot body to cold body.

Explanation:

Our body is in optimal status is a hot body and tap or well water is a cold body. Therefore we feel cold.

When we drink tap or well water in the winter, we feel chilled because heat transfers from a hot body to a cold body.

What is convection?

Convection, often known as a heat transfer mechanism, is a method of moving heat. When a fluid, such as air or water, is moving, this heat transfer takes place. Temperature differences within the fluid cause convection.

Due to the increased kinetic energy of the particles, a fluid loses density at the area nearest to the heat source as it is heated. The less dense portion of the fluid then rises, while the denser portion sinks.

The process keeps happening because the less dense fluids sink as they move away from their heat source while the denser fluids rise as they move toward the heat source. Convection currents are produced by this.

Therefore, When we drink tap or well water in the winter, we feel chilled because heat transfers from a hot body to a cold body.

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A gas cylinder contains only the gases radon, nitrogen, and helium. The radon has a pressure of 222 torr while the nitrogen has a pressure of 446 torr. If the total pressure inside the cylinder is 771 torr, what is the pressure that is due to the helium?.

Answers

Answer:

103 torr

Explanation:

The remainder is due to the helium

771 - 222 - 446  = 103 torr due to helium

What is typically responsible for forming volcanic island chains in the oceans?
A. Shield volcanoes
B. Fissures
Composite volcanoes
Hot spot volcanism

Answers

Hot spot volcanism is responsible for the formation of volcanic island chains in the oceans and is denoted as option D.

What is a Volcano?

This is a rupture in the earth crust thereby giving rise to the release of hot lava, gases etc from below the surface.

The pushing of the lava through the cracks partially melts the plate at a subduction zone thereby forming the volcanic island chains.

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