If I double the amount I stretch a spring, what happens to its EPE

1. James drives 400 km in 5 hours to his grandmothers. What are the units for speed going to be?

Airida [17]

Answer:

See the answer below

Explanation:

1. Speed is calculated as the ratio of distance and time. Hence, Jame's speed can be calculated as:

400/5 km/hr = 80 km/hr

The unit for the speed would be km/hr. This can also be converted to m/s:

80 km = 80,000 m

1 hr = 3,600 s

80 km/hr = 80,000/3600 m/s = 22.22 m/s

2. Since James drove 400 km in 5 hours, the distance he drove is 400 km.

3. The time it took for James to get there is 5 hours.

6 0

2 years ago

Describe the water cycle process starting from an afternoon thunderstorm. There are many different variations that could happen.

spin [16.1K]

Explanation:

The water cycle basically involves five steps:

evaporation and transpiration ⇄
condensation, ⇄
precipitation, ⇄
runoff, ⇄
infiltration ⇄

So when a thunderstorm occurs it helps in completing the precipitation process by enabling the release of water vapor stored up in the atmosphere to fall on the ground as rain.

After this, the water runoffs to the surface of the ground, on plants, into rocks, rivers, and lakes.

Next, the Infiltration process enables the water on the ground surface to enter the soil some of which becomes groundwater.

The cycle begins again as the evaporation and transpiration process begins, where the groundwater as a result of heat from the sun is taken back into the atmosphere, while water in plants by means of transpiration goes back into the atmosphere.

It then condenses and falls back as precipitation again.

3 0

2 years ago

A balloon filled with helium gas has an average density of Q,-0.41 kg/m'. The density of the air is Qa-1.23 kg/m3. The volume of

Citrus2011 [14]

Answer:

a) $(Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]$

b) a = 19.61[m/s^2]

Explanation:

The total mass of the balloon is:

$massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\$

The buoyancy force acting on the balloon is:

$Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]$

Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.

In the attached image we can see the free body diagram and the equation deducted by Newton's second law

6 0

2 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

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

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is: