* WILL GIVE BRAINLIEST TO CORRECT ANSWER *

Determine the heat energy required to vaporize 13.9 grams of liquid water at 100° C. O 2,006 cal O 47.8 cal O 7,506 cal O 24.9 c

Harlamova29_29 [7]

Answer:

7506 cal

Explanation:

Given: Mass of water = m = 13.9 grams.

Heat energy required to determine the conversion of 13.9 grams of water to steam or vapor is m L,

where L = Latent heat of vaporization of water=540 cal/g.

Q = mL = (13.9)(540) = 7,506 cal.

6 0

4 years ago

A 1200-kg SUV is moving alone a straight highway at 12.0 m/s. Another car, with mass 1800 kg and speed 20.0 m/s, has its center

andreev551 [17]

Answer:

A) d = 24 m

B) 50400 kg.m/s

C) v₀ = 16.8 m/s

D) 50400 kg.m/s. It's equal to the momentum found in part B.

Explanation:

We are given;

Mass of station wagon;m1 = 1200 kg Velocity; V = 12 m/s

Mass of car; m2 = 1800 kg

Velocity of car; v2 = 20 m/s

a ) Let centre of mass of car and station wagon be at a distance d from wagon

Thus,

If we take moment of weight about it, we have;

1200 x d = 1800 x ( 40 - d )

Where, d is the position of the center of mass of the system consisting of the two cars

Thus,

1200d = 72000 - 1800d

1200d + 1800d = 72000

3000 d = 72,000

d = 72,000/3000

d = 24 m

b ) Total momentum= m1•v1 + m2•v2

= (1200 x 12) + (1800 x 20)

= 14400 + 36000

= 50400 kg.m/s

c ) Let speed of centre of mass be v₀

Thus,

v₀ = (m1•v1 + m2•v2)/(m1 + m2)

v₀ = 50400/(1200 + 1800)

v₀ = 50400/3000

v₀ = 16.8 m/s

d) System Total momentum = velocity of centre mass x total mass

Thus,

Total momentum = v₀(m1 + m2)

= 16.8(3000) = 50400 kg.m/s .

This value is equal to what was calculated in part b

8 0

3 years ago

Consider n equal positively charged particles each of magnitude Q/n placed symmetrically around a circle of radius a. Calculate

Sonja [21]

Answer: Magnitude of electric field =p × Ke × Q/d^3

Explanation: Using Coulombs law of point charge,each charge on the circle would exert a fieldEc at point given by:

Ec= Ke × (Q/n)/d^2

Where Ke= Coulomb's constant

d= distance between the charges and the point of measurement, P with d^2=a^2+p^2

(Q/n)= Magnitude of the charge.

For charges in a circle,all the force components and direction(x,z) are cancelled by the symmetry,leaving only the vertical force(y-direction)

The Resultant vector will be

Ecy=Eq × sin(theta)

Ecy=Ke × (Q/n)/d^2 ×(p/d)

Adding the forces from all the charges,the magnitude of electric field Ey=n×Ecy

/Ey/= n ×[Ke ×(Q/n)/d^2 ×(p/d)]

/Ey/= p × Ke × (Q/d^3)

3 0

3 years ago

Two boxes are on opposite ends of a plank that is 3.0 m long. The plank is supported in the middle by a fulcrum. The box on the

zysi [14]

For a given condition above, use the concept of equal moments. Moment is the product of the force and its distance from a reference point. Let x be the distance of the 25 kg body from the fulcrum and 3 - x be the distance of the right mass, 15 kg, from the fulcrum. Equating the moments give,
(25 kg)(9.8 m/s²)(x) = (15 kg)(9.8 m/s²)(3 - x)
The value of x is 1.125 m. Therefore, the fulcrum should be 1.875 m from the right mass.

8 0

3 years ago

if a book of mass 50 kg is placed on the table of mass 200 kg what will be the weight of table and reaction of table on the book

ollegr [7]

Answer:

See the answer below.

Explanation:

The weight of the table should be equal to the weight of the table plus the weight of the book.

$w_{table}=m_{table}*g\\w_{table}=200*9.81\\w_{table}=1962[N]$

And for the book

$w_{book}=m_{book}*g\\w_{book}=50*9.81\\w_{book}=490.5[N]$

And the total weight

$w = w_{table}+w_{book}\\w=1962+490.5\\w = 2452.5 [N]$

For the system to be in balance, the reaction force exerted by the table on the book must be equal to the weight of the book. This force in physics is known as the normal force.

$N_{force} = w_{book}\\N_{force}=490.5[N]$

4 0

3 years ago