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2 years ago

If you cook a marshmallow a few feet above a camp fire, it will cook mostly

Physics

1 answer:

nasty-shy [4]2 years ago

6 0

Answer:

A) radiation.

Explanation:

Hello,

In this case, heat transfer from a body with a high temperature to a body with a lower temperature may occur when the bodies are not in direct physical contact with each other or when they are separated in space and it is is called heat radiation. Solids, liquids, or gaseous substances are able to emit energy via a process of electromagnetic radiation because of vibrational and rotational movement of their molecules and atoms.

Hence, since the camp fire is not in contact with the marshmallow and it still cooks it, we can infer that the mechanism of heat transfer is A) radiation. Moreover, both conduction and convection occur when the bodies are in contact.

Regards.

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3) A tolley of mass 4kg, moving with a velocity

Leokris [45]

Answer:

3.57 m/s

Explanation:

The sum of the 2 momentums Is equal the finale momentums. so if momentums Is q, v Is velocity and m Is Mass, q3=m1*v1+m2**v2=16+9=25 m*kg/s

q3=m3*v3

v3=q3/m3=25/(4+3)=3.57m/s

8 0

2 years ago

Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

2 years ago

Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l

Masja [62]

Answer:

a) I2 = 3 (o-10) / (o- 30) , b) h ’/h= 3 (o-10) / o (o-30)

Explanation:

The builder's equation is

1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

1 / i1 = 1 / f - 1 / o

1 / i1 = 1/15 - 1 / o

i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

o2 = d-i1

We write the builder equation

1 / f2 = 1 / o2 + 1 / i2

1 / i2 = 1 / f2 -1 / o2

1 / i2 = 1 / f2 - 1 / (d-i1)

1 / i2 = 1/20 - 1 / (d-i1) (1)

Let's evaluate the last term

d-i1 = d - 15 o / (o-15)

d-i1 = (d (o-15) - 15 o) / (o-15)

d- i1 = (30 or -30 15 -15 o) / (o-15)

d-i1 = (15 or - 450) / (o- 15)

d-i1 = = (15 or -450) / (o-15)

replace in 1

1 / i2 = 1/20 - (or - 15) / (15 or -450)

1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

1 / i2 = (-5 or -150) / (15 or -150)

1 / i2 = (or -30) / (3 or - 30)

I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

m = h ’/ h = - i / o

h ’= - h i / o

In our case

h ’= h i2 / o

h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

5 0

2 years ago

In which of the following situations or places would you want to reduce the

sashaice [31]

Answer:

Explanation:

<em>The correct answer would be in the axle of the wheels while you ride your bicycle.</em>

Options A, B, and C requires that the forces of friction is increased in order to have more control.

However, option D requires that there is a minimal frictional force in the axle of the wheels of a bicycle while riding so that a little effort would be required to keep the bicycle moving.

<u>The lesser the friction, the lower the effort that would be needed to keep the bicycle moving and vice versa.</u>

4 0

2 years ago

Which of the following is a nonferromagnetic material? (a) aluminum (b) iron (c) cobalt (d) gadolinium

ivolga24 [154]

Answer:

Explanation:

Iron and gadlinium are both very easily made into magnetic substances. Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but

Aluminum by itself is not able to be magnetized.

5 0

2 years ago