Help me plz on the one i just put i neeed it uhg plz anter

Which of the following statements is correct?

Mumz [18]

Answer: c. Generally, metals are ductile.

Explanation:

From the options given in the question, the correct statement is that"Generally, metals are ductile.

Ductility of a metal simply means that a metal can be plastically deform before it is then fractured. It implies that metals can be drawn to thin wires. The only exception we have in this case is mercury.

7 0

2 years ago

In a Broadway performance, an 84.5-kg actor swings from a R = 4.30-m-long cable that is horizontal when he starts. At the bottom

Arada [10]

Answer:

1.57772 m

Explanation:

M = Mass of actor = 84.5 kg

m = Mass of costar = 55 kg

v = Velocity of costar

V = Velocity of actor

$h_i$ = Intial height of actor = 4.3 m

g = Acceleration due to gravity = 9.81 m/s²

As the energy of the system is conserved

$\frac{1}{2}MV^2=Mgh_i\\\Rightarrow V=\sqrt{2gh_i}\\\Rightarrow V=\sqrt{2\times 9.81\times 4.3}\\\Rightarrow V=9.18509\ m/s$

As the linear momentum is conserved

$MV=(m+M)v\\\Rightarrow v=\frac{MV}{m+M}\\\Rightarrow V=\frac{84.5\times 9.18509}{84.5+55}\\\Rightarrow v=5.56372\ m/s$

Applying conservation of energy again

$\frac{1}{2}(m+M)v^2=(m+M)gh_f\\\Rightarrow h_f=\frac{v^2}{2g}\\\Rightarrow h_f=\frac{5.56372^2}{2\times 9.81}\\\Rightarrow h_f=1.57772\ m$

The maximum height they reach is 1.57772 m

3 0

2 years ago

As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of

kykrilka [37]

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution decrease and the vapor pressure of the solution decrease .

Explanation:

Depression in freezing point :

$\Delta T_f=K_f\times m$

where,

$\Delta T_f$ =depression in freezing point =

$K_f$ = freezing point constant

m = molality ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

If molality of the solution in high the depression in freezing point of the solution will be more.
If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

$\frac{p_o-p_s}{p_o}=\chi_{solute}$

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of solution

$\chi_{solute}$ = Mole fraction of solute

$p_s\propto \frac{1}{\chi_{solute}}$

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.

8 0

2 years ago

What is a net force on an object that has a mass of 20.0 kg, an applied force of 100 n moving on a surface with a friction coeff

sergiy2304 [10]

The net force on the object as described is; 58.84N

Two forces acting on the object are;

The applied force and the frictional force.

In essence; the frictional force can be evaluated as;

Frictional force; = coefficient × Weight of object.

Frictional force = 0.21 × 20 × 9.8.

Frictional force = 41.16N

The Net force = Applied force - frictional force

Net force = 100 - 41.16N

Net Force = 58.84 N.

Read more:

brainly.com/question/94428

5 0

2 years ago

¿Cuál es la aceleración centrípeta de un móvil que recorre una

Lady bird [3.3K]

Answer:

a) a = 4.57 m/s², b) a = 6.48 m / s² , c) a = 1.42 m / s²,d) r = 82.3 m

Explanation:

The centripetal acceleration is the acceleration responsible for the change of direction of the acceleration vector and occurs in circular movements, the expression is

a = v² / r

let's apply this precaution to our cases

a) let's calculate

a = 8²/14

a = 4.57 m/s²

b) an automobile at v = 65 km / h (1000 m / 1km) (1 h / 3600 s) =18,055 m/s

let's reduce feet to meters

1 ft = 0.3048 m

r = 165 ft (0.3048 m / 1 ft) = 50.292 m

a = 18,055 2 / 50,292

a = 6.48 m / s²

c) we calculate

a = 1.25²2 / 1.1

a = 1.42 m / s²

d) we look for the radius

a = v² / r

r = v² / a

we reduce

v = 80 km / h (1000 m / 1km) (1h / 3600s) = 22.22 ms

r = 22.22²/6

r = 82.3 m

e) the cenripeta acceleration is used to take the curves on the highway,

Used in centrifuges to separate compounds

It is used in the games of the park of atraccio

Used in CD players and computer hard drives

5 0

2 years ago