An athlete swims the length of 50m pool in 20sec, and

Two floors in a building are separated by 4.1 m. People move between the two floors on a set of stairs. (a) Determine the change

Ket [755]

Answer:

a) The change in potential energy of a 3.0 kilograms backpack carried up the stairs.

b) The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.

Explanation:

Let consider the bottom of the first floor in a building as the zero reference ( $z = 0\,m$ ). The change in potential energy experimented by a particle ( $\Delta U_{g}$ ), measured in joules, is:

$\Delta U_{g} = m\cdot g\cdot (z_{f}-z_{o})$ (1)

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$z_{o}$ , $z_{f}$ - Initial and final height with respect to zero reference, measured in meters.

Please notice that $m\cdot g$ is the weight of the particle, measured in newtons.

a) If we know that $m = 3\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $z_{o} = 0\,m$ and $z_{f} = 4.1\,m$ , then the change in potential energy is:

$\Delta U_{g} = (3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.1\,m-0\,m)$

$\Delta U_{g} = 120.626\,J$

The change in potential energy of a 3.0 kilograms backpack carried up the stairs.

b) If we know that $m\cdot g = 650\,N$ , $z_{o} = 4.1\,m$ and $z_{f} = 0\,m$ , then the change in potential energy is:

$\Delta U_{g} = (650\,N)\cdot (0\,m-4.1\,m)$

$\Delta U_{g} = -2665\,J$

The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.

5 0

3 years ago

Express the surface area of a cube as a function of its volume v.

vredina [299]

The surface area of a cube is:

S = 6l², making l the subject:
l = √(S/6)

While the volume is:

V = l³

If we substitute l to get:

V = (√(S/6))³

√(S/6) = ³√V

S/6 = (³√V)²

S = 6(³√V)² ⇒ 6V^(2/3)

5 0

3 years ago

When a 4.25 kg object is placed on top of a vertical spring, the spring compresses a distance of 2.62 cm. what is the force cons

ANTONII [103]

The force applied to the spring is the weight of the object that compresses it, so it is equal to:
$W=mg=(4.25 kg)(9.81 m/s^2)=41.7 N$

Because of this force, the spring compresses by $x=2.62 cm=0.0262 m$ . Using Hook's law,
$F=kx$ ,
since we know the intensity of the force (the weight W) and the compression of the spring, x, we can find k, the spring constant:
$k= \frac{F}{x}= \frac{W}{x}= \frac{41.7 N}{0.0262 m} =1591.6 N/m$

6 0

3 years ago

An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su

s2008m [1.1K]

Answer: A. 23.59 minutes.

B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

Latent heat of fusion is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

Specific heat capacity of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, given that:

mass of ice, m= 0.555 kg

temperature of ice, T= -16.6°C

rate of heat transfer, $q=820 J.min^{-1}$

specific heat of ice, $c_{i}= 2100 J.kg^{-1}.K^{-1}$

latent heat of fusion of ice, $L_{i}=334\times10^{3}J.kg^{-1}$

Asked:

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

We have the formula:

$Q=mc\Delta T$

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

$Q= 0.555\times2100\times16.6$

$Q= 19347.3 J$

Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

$t=\frac{Q}{q}$

$=\frac{19347.3}{820}$

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

We have the equation: latent heat, $Q_{L}= mL_{i}$

$Q_{L}= 0.555\times334\times10^{3}= 185370 J$

Now the time required for supply of 185370 J:

$t=\frac{Q_{L}}{q}$

$t=\frac{185370}{820}$

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0

3 years ago

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If

mart [117]

Answer: 29.50 m

Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x= vo*t- (a/2)* t^2

Vf=0= vo-a*t then t=vo/a

Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m

5 0

3 years ago