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agasfer [191]
3 years ago
14

When gases, liquids, or solids are in contact with a moving object, the flow of

Physics
1 answer:
Gwar [14]3 years ago
5 0

Answer:

Heat

Friction is what causes heat.

Brainliest always helps.

You might be interested in
A particle has a charge of -4.25 nC.
SpyIntel [72]

Answer:

-611.32 N/C

0.43723 m

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

q = Charge = -4.25 nC

r = Distance from particle = 0.25 m

Electric field is given by

E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -4.25\times 10^{-9}}{0.25^2}\\\Rightarrow E=-611.32\ N/C

The magnitude is 611.32 N/C

The electric field will point straight down as the sign is negative towards the particle.

E=\dfrac{kq}{r^2}\\\Rightarrow r=\sqrt{\dfrac{kq}{E}}\\\Rightarrow r=\sqrt{\dfrac{8.99\times 10^9\times 4.25\times 10^{-9}}{13}}\\\Rightarrow r=1.71436\ m

The distance from the electric field is 1.71436 m

4 0
3 years ago
Read 2 more answers
8 N to the left , and 4 N to the right. Find the net force. Is this balanced?
alukav5142 [94]

Explanation:

12N by first law of newton is net force after colloision

7 0
2 years ago
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun
AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and  i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561  × 10¹⁹ electrons per second = 1.561  × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

8 0
3 years ago
an object is placed 20cm from a converging lens. If the real image is formed 80cm from the object, what is the focal length of t
masya89 [10]

Explanation:

Here,

object distance(u)=20cm

image distance(v)=80cm-20cm=60cm

focal length(f)=?

we know ,

(1/f)=(1/v)+(1/u)

1/f = 1/60 + 1/20

1/f = (1+3)/60

60 =4f

f=15cm

8 0
2 years ago
A man starts walking north at 2 ft/s from a point p. five minutes later a woman starts walking south at 4 ft/s from a point 500
icang [17]
Refer to the diagram shown below.

After 5 minutes (300 seconds):
The man travels north by (2 ft/s)*(300 s) = 600 ft
The woman, located at q, 500 east of p, begins walking south at 4 ft/s.
The distance separating them is
d₁ = √(600² + 500²) = 781.025 ft

After 20 minutes:
The man has traveled for 20 minutes (1200 s).
The woman has traveled for 15 minutes (900 s).
The man has moved (2 ft/s)*(1200 s) = 2400 ft north of p.
The woman has moved (4 ft/s)*(900 s) = 3600 ft south of q.
The distance separating them is
d₂ = √(6000² + 500²) = 6020.8 ft

The separation from d₁ to d₂ occurs in 15 minutes (900s).
Therefore the rate of separation is
Rate = (d₂ - d₁ ft)/(900 s) = (6020.8 - 781.025)/900 = 5.822 ft/s
or
Rate = (5.822 ft/s)*(60 s/min) = 349.32 ft/min

Answer: 349.32 ft/min (or 5.82 ft/s)

4 0
3 years ago
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