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3 years ago

Friction occurs when the and of two surfaces stick to each other

Physics

2 answers:

Lilit [14]3 years ago

7 0

Friction occurs when the end of two surfaces stick to each other. Yes correct

Misha Larkins [42]3 years ago

3 0

Answer:

yes, kinda like when babies are made, there is friction

Explanation:

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A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of

ehidna [41]

Answer:

Explanation:

Given

Speed of block at bottom is $v=3.8\ m/s$

Distance traveled $s=6\ m$

initial velocity is zero

using equation of motion

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$(3.8)^2-0=2\times a\times 6$

$a=1.203\ m/s^2$

when it is 3 m from top then

$v^2-u^2=2as$

$v^2-0=2\times 1.203\times 3$

$v=2.68\ m/s$

5 0

3 years ago

You prepare tea with 0.250 kg of water at 85.0 ºC and let it cool down to room temperature (20.0 ºC) before drinking it. Essenti

vladimir2022 [97]

Answer:

232 J/K

Explanation:

The amount of heat gained by the air = the amount of heat lost by the tea.

q_air = -q_tea

q = -mCΔT

q = -(0.250 kg) (4184 J/kg/ºC) (20.0ºC − 85.0ºC)

q = 68,000 J

The change in entropy is:

dS = dQ/T

Since the room temperature is constant (isothermal):

ΔS = ΔQ/T

Plug in values (remember to use absolute temperature):

ΔS = (68,000 J) / (293 K)

ΔS = 232 J/K

5 0

3 years ago

A train travels 55 kilometers in 2 hours, and then 52 kilometers in 5 hours. What is its average speed?

bulgar [2K]

The formula for working out speed is distance ÷ time.

55 km ÷ 2 hours = 27.5 km/h (average speed for first part of journey)

52km ÷ 5 hours = 10.4 km/h (average speed for second part of journey)

(27.5 + 10.4) ÷ 2 = 18.95 km/h (average speed throughout the journey)

4 0

3 years ago

Read 2 more answers

PLS HELP I HAVE EXAM ILL GIVE U BRAINLIST

stiv31 [10]

Explanation:

False

electron and proton attract each other

3 0

3 years ago

An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields a

Lapatulllka [165]

Answer:

The speed of the electron is $2.55\times 10^3\ m/s$ .

Explanation:

Given that,

The magnitude of electric field, $E=1.32\ kV/m=1.32\times 10^3\ V/m$

The magnitude of magnetic field, B = 0.516 T

Both the magnetic and electric fields are acting on the moving electron. Then, the magnitude of electric field and magnetic field is balanced such that :

$evB=eE\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.32\times 10^3}{0.516}\\\\v=2558.13\ m/s$

$v=2.55\times 10^3\ m/s$

So, the speed of the electron is $2.55\times 10^3\ m/s$ . Hence, this is the required solution.

3 0

3 years ago