17. A force of 150N is applied at an angle of 60°

Ag,Au and Cu are called coinage metals why plzzzz hurry its urgent plzzz

Ronch [10]

Answer:

This is because these metals are used for minting (making) coins.

8 0

3 years ago

Feng and Isaac are riding on a merry-ground. Feng rides on a horse at the outer rim of the circular platform, twice as far from

kogti [31]

Answer: The question is incomplete or missing details. here is the remaining part of the question ;

1. impossible to determine

2. half of Isaac’s

3. the same as Isaac’s

4. twice Isaac’s

The angular speed of feng will be the same as that of Isaac. Hence the answer is option 3

Explanation:

Since we have been told that both feng and isaac are riding on a merry go round i.e in a circular motion, irrespective of how fast one ride above the other, the angular speed will be constant since they are riding on a merry go round, as such both feng and isaac will maintain equal angular speed, hence the angular speed of feng will be the same as that of Isaac.

4 0

3 years ago

Referring to the above diagram, how high will the ball rise on the right-hand incline?

Marina CMI [18]

I think it’s 15cm
Might be 7cm

3 0

3 years ago

A child on a bridge throws a rock straight down to the water below. The point where the child released the rock is 82 m above th

eimsori [14]

Answer:

40 m/s

Explanation:

given,

height of the fall, h = 82 m

time taken to fall, t = 1.3 s

rock velocity, v = ?

acceleration due to gravity, g = 9.8 m/s²

rock is released initial velocity, u = 0 m/s

using equation of motion

v² = u² + 2 a s

v² = 0 + 2 x 9.8 x 82

v² = 1607.2

v = 40 m/s

hence, rock's velocity is equal to 40 m/s

6 0

2 years ago

Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The ga

Radda [10]

Answer:

a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm

Explanation:

For isothermal expansion PV = constant

So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,

So, P₁V₁ = P₂V₂

P₂ = P₁V₁/V₂

Since V₂/V₁ = 0.19,

P₂ = P₁V₁/V₂

P₂ = 1 atm (1/0.19)

P₂ = 5.26 atm

For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas

So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,

So, P₂V₂ⁿ = P₃V₃ⁿ

P₃ = P₂V₂ⁿ/V₃ⁿ

P₃ = P₂(V₂/V₃)ⁿ

Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19

1/0.19,

P₃ = P₂(V₂/V₃)ⁿ

P₃ = 5.26 atm (0.19)⁽⁵/³⁾

P₃ = 5.26 atm × 0.0628

P₃ = 0.33 atm

Using the ideal gas equation

P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)

P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

T₃ = (0.33 atm/1 atm)(1)273 K

T₃ = 90.1 K

So,

a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

d. The lowest pressure attained by the gas is P₃ = 0.33 atm

6 0

3 years ago