A body of mass 20kg initially at rest is subjected to a force of 40m for 15sec. Calculate the change in k.e

1. 412.9 g of dry ice sublimes at room temperature. a. What’s changing? --- sublimation b. What constant will you use? ----- 25.

hoa [83]

1. 236 kJ

a. The phase (or state of matter) of the substance: from solid state to gas state (sublimation)

b. The enthalphy of sublimation, given by: $\lambda=571 J/g$

c. The equation to use will be $Q=m\lambda$ , where m is the mass of dry ice and $\lambda$ is the enthalpy of sublimation

d. The energy is being absorbed, because the heat is transferred from the environment to the dry ice: as a consequence, the bonds between the molecules of dry ice break and then move faster and faster, and so the substance turns from solid into gas directly.

e. The amount of energy being transferred is

$Q=m\lambda=(412.9 g)(571 J/g)=2.36\cdot 10^5 J=236 kJ$

2. 165 kJ

a. The phase (or state of matter) of the substance: from gas state to liquid state (condensation)

b. The latent heat of vaporisation of water, given by $\lambda=2260 J/g$

c. The equation to use will be $Q=m\lambda$ , where m is the mass of steam that condenses and $\lambda$ is the latent heat of vaporisation

d. The energy is being released, since the substance turns from a gas state (where molecules move faster) into liquid state (where molecules move slower), so the internal energy of the substance has decreased, therefore heat has been released

e. The amount of energy being transferred is

$Q=m\lambda=(72.9 g)(2260 J/g)=1.65\cdot 10^5 J=165 kJ$

3. 3.64 kJ

a. Only the temperature of the substance (which is increasing)

b. The specific heat capacity of silver, which is $C_s = 0.240 J/gC$

c. The equation to use will be $Q=m C_s \Delta T$ , where m is the mass of silver, Cs is the specific heat capacity and $\Delta T$ the increase in temperature

d. The energy is being absorbed by the silver, since its temperature increases, this means that its molecules move faster so energy should be provided to the silver by the surroundings

e. The amount of energy being transferred is

$Q=m C_s \Delta T=(39.2 g)(0.240 J/gC)(412.9^{\circ}C-25.9^{\circ}C)=3641=3.64 kJ$

4. 89 kJ

a. Both the phase of the substance (from solid to liquid) and then the temperature

b. The latent heat of fusion of ice: $\lambda=334 J/g$ and the specific heat capacity of water: $C_s=4.186 J/gC$

c. The equation to use will be $Q=m\lambda + m C_s \Delta T$ , where m is the mass of ice, $\lambda$ the latent heat of fusion of ice, Cs is the specific heat capacity of water and $\Delta T$ the increase in temperature

d. The energy is being absorbed by the ice, at first to break the bonds between the molecules of ice and to cause the melting of ice, and then to increase the temperature of the water

e. The amount of energy being transferred is

$Q=m\lambda +m C_s \Delta T=(156.3 g)(334 J/g)+(156.3 g)(4.186 J/gC)(56.232^{\circ}C-0^{\circ}C)=8.9\cdot 10^4 J=89 kJ$

6 0

3 years ago

What is the velocity of 360?

9966 [12]

Answer:

3/360 is 120. your velocity is 120

Explanation:

8 0

3 years ago

1. Draw magnetic field (B field) line diagrams for the following. Make sure the diagrams are large enough so the direction and s

larisa [96]

I attached the requested diagram.

<em>In the case of the magnetic field in a bar</em> by convention, the direction of the field is taken out of the north pole and towards the south pole of the magnet. These types of images are commonly made of some ferrous material.

<em>In the case of the horseshoe </em>magnet, the highly concentrated magnetic field is distinguished between its legs. In the figure it is shown in a contribution from North to South, again by agreement, however outside the two poles, the magnetic field falls rapidly. A horseshoe magnet is basically a bent bar magnet.

4 0

3 years ago

A box of spherical jawbreaker candies is 23 g and its volume is 32.3 cm3. If the average mass of a single jawbreaker is 0.94 g,

Alex777 [14]

The radius of each jawbreaker is approximately 0.68 cm.

<h3>Volume of a sphere;</h3>

v = 4 /3 πr³

where

r = radius

Therefore,

23 g = 32.3 cm³

0.94 g = ?

cross multiply

volume of a single jawbreaker = 32.3 × 0.94 / 23 = 30.362 / 23 = 1.32 cm³

Therefore,

volume of each jawbreaker = 4 /3 πr³

1.32 = 4 / 3 × 3.14 × r³

r³ = 1.32 /4.18666666667

r³ = 0.31533683707

r = ∛0.31533683707

r = 0.680651651 = 0.68

Therefore, the radius of each jawbreaker is approximately 0.68 cm.

learn more on radius here: brainly.com/question/19172427

5 0

3 years ago

A player strikes a hockey puck giving it a velocity of 30.252 m/s. The puck slides across the ice for 0.267 s after which time i

erica [24]

Answer:

The average drag force is 1.206 (-i) N

Explanation:

You have to apply the equations of<em> Impulse</em>:

I=FmedΔt

Where I and Fmed (the average force) are vectors.

The Impulse can also be expressed as the change in the <em>quantity of motion</em> (vector P)

I=P2-P1

P=mV (m is the mass and v is the velocity)

You can calculate the quantity of motion at the beggining and at the end of the given time:

Replace the mass in kg, dividing the mass by 1000 to convert it from g to kg.

P1=(0.179kg)(30.252m/s) i= 5.414 i kg.m/s

P2=0.179kg)(28.452m/s) i = 5.092 i kg. m/s

Where i is the unit vector in the x-direction.

Therefore:

I= 5.092 i - 5.414 i = -0.322 i

The average drag force is:

Fmed= I/Δt = -0.322 i/ 0.267s = -1.206 i N

3 0

3 years ago