Which of the following prefixes represents the largest value

Ivan

Helium lithium and calcium

3 0

3 years ago

You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves down a frictionless track to a height o

irina [24]

Answer: 20.765 m/s

Explanation:

This problem can be solved by the conservation of energy principle, this means the initial energy $E_{o}$ must be equal to the final energy $E_{f}$ :

$E_{o}=E_{f}$ (1)

Where each energy is the sum of kinetic energy $K$ and potential energy $U$ :

$K_{o}+U_{o}=K_{f}+U_{f}$ (2)

Where:

$K_{o}=\frac{1}{2}mV_{o}^{2}$

Being $m$ your mass and $V_{o}=0 m/s$ your initial velocity, since the roller coaster sterted from rest.

$U_{o}=mgh_{o}$

Being $g=9.8 m/s^{2}$ the acceleration due gravity and $h_{o}=25 m$ your initial height

$K_{f}=\frac{1}{2}mV_{f}^{2}$

Being $V_{f}$ your final velocity

$U_{f}=mgh_{f}$

Being $h_{f}=3 m$ your final height

Rewritting (2):

$\frac{1}{2}mV_{o}^{2}+mgh_{o}=\frac{1}{2}mV_{f}^{2}+mgh_{f}$ (3)

$mgh_{o}=m(\frac{1}{2}V_{f}^{2}+gh_{f})$ (4)

Isolating $V_{f}$ :

$V_{f}=\sqrt{2g(h_{o}-h_{f})}$ (5)

$V_{f}=\sqrt{2(9.8 m/s^{2})(25 m-3 m)}$ (6)

Finally:

$V_{f}=20.765 m/s$ This is your spedd when you arrive at 3 m height

7 0

4 years ago

In deep space (no gravity), the bolt (arrow)

salantis [7]

t = 0.527 s

It accelerates for 0.527 s.

Explanation:

We use the formula:

v = u+at

Given:

v = 106 m/s

u = 0 (since no gravity)

$a=201 \mathrm{m} / \mathrm{s}^{2}$

So applying the formula,

v = u+at

106 = 0 + 201t

t = 106/201

t = 0.527 s

4 0

3 years ago

The force of attraction that a -40.0 μC point charge exerts on a +108 μC point charge has magnitude 4.00 N. How far apart are th

Karolina [17]

Answer:

F = k q1 q2 / r^2

r^2 = k q1 q2 / F = 9E9 * 4E-5 * 10.8E-5 / 4

r^2 = 9 * 4 * 10.8 / 4 * E-1 = 9.72 m^2

r = 3.12 m

8 0

3 years ago

1 mole of air undergoes a Carnot cycle. The hot reservoir is at 800 oC and the cold reservoir is at 25 oC. The pressure ranges b

BabaBlast [244]

Answer:

The net work produced is 30.37 KJ

The efficiency of cycle is 72.3%

Explanation:

For the net work produced, we have the formula:

Work = (Th - Tc)(Sh - Sc)(M)

Where,

Th = higher temperature = 800° C + 273 = 1073 k

Tc = lower temperature = 25° C + 273 = 298 k

Sh = specific entropy at higher temperature

Sc = specific entropy at lower temperature

M = molar mass of air

Using, ideal gas table to find entropy. The table is attached.

therefore,

Work = (1073 k - 298 k)(3.0485235 KJ/kg.k - 1.69528 KJ/kg.k)(0.02896 kg)

Work = (775 k)(1.3532435 KJ/kg.k)(0.02896 kg)

Work = 30.37 KJ

Now, for the efficiency (n), we have a formula:

n = 1 - Tc/Th

n = 1 - (298 k)/(1073 k)

n = 0.723 = 72.3 %

5 0

3 years ago