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Elena-2011 [213]
3 years ago
6

Explain why your PE and KE are usually not both high at the same time (If PE is high then usually KE is low)

Physics
1 answer:
AfilCa [17]3 years ago
4 0
Might help:

an object can have both kinetic and potential energy at the same time. for example, an object which is falling, but has not reached the ground has kinetic energy because it is moving downwards, and potential energy because it is able to move downwards even further than it already has. as an object falls its potential energy decreases, while its kinetic energy increases. the decrease in potential energy is exactly equal to the increase in kinetic energy.
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Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te
seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy  5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant \gamma =1.4

specific heat is given as =\frac{\gamma R}{\gamma -1}

gas constant =287 J⋅kg−1⋅K−1

Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}

specific heat at constant volume

Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}

change in internal energy = Cv(T_2 -T_1)

                            \Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^{-1}

change in enthalapy \Delta H = Cp(T_2 -T_1)

                                 \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}

change in entropy

\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})

\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})

\Delta S = 382.79 J kg^{-1} K^{-1}

7 0
3 years ago
A toy dart gun generates a dart with a momentum of 140 kg*m/s and a
irinina [24]

Answer:

35 kg

Explanation:

From the question,

Momentum (I) = mass (m) × velocity (v)

I = m×v................... Equation 1

Where m = mass, v = velocity

make m the subject of the equation

m = I/v.................... Equation 2

Given: I = 140 kgm/s, v = 4 m/s

Substitute these values into equation 2

m = 140/4

m = 35 kg

Hence the mass of the dart is 35 kg

6 0
3 years ago
Read 2 more answers
How much equal charge should be placed on the earth and the moon so that the electrical repulsion balances the gravitational for
kumpel [21]

As we know that electrostatic force between two charges is given as

F = \frac{kq_1q_2}{r^2}

here we know that electrostatic repulsion force is balanced by the gravitational force between them

so here force of attraction due to gravitation is given as

F_g = 1.98 \times 10^{20} N

here we can assume that both will have equal charge of magnitude "q"

now we have

1.98 \times 10^{20} = \frac{kq^2}{r^2}

1.98 \times 10^{20} = \frac{(9\times 10^9)(q^2)}{(3.84 \times 10^8)^2}

1.98 \times 10^{20} = (6.10 \times 10^{-8}) q^2

now we have

q = 5.7 \times 10^{13} C

6 0
3 years ago
How do atoms become more chemically stable?
Marat540 [252]

Answer:

For an atom to become totally stable, it needs to have a full outer shell. To do this, two or more atoms will share or give away electrons to each other in a process called bonding.

Explanation:

When an atom loses or gains an electron, it becomes an ion. If it gains an electron, it's a cation, and if it loses one, it's an anion. This happens most commonly in chemical reactions, in which atoms share electrons to form a stable outer shell of 8. For example, the water molecule consists of two hydrogen atoms and an oxygen atom.

3 0
3 years ago
Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav
umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

v_i = 4.0 \ m/s

then,

v_f=\frac{4.0}{2}

    =2.0 \ m/s

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ a=\frac{v_f^2-v_i^2}{2s}

On substituting the given values, we get

⇒    =\frac{(2.0)^2-(4.0)^2}{2\times 18.2}

⇒    =\frac{4-8}{37}

⇒    =\frac{-4}{37}

⇒    =0.108 \ m/s^2

As we know,

⇒  f=ma

and,

⇒  \mu_rmg=ma

⇒       \mu_r=\frac{a}{g}

On substituting the values, we get

⇒       =\frac{0.108}{9.8}

⇒       =0.011

7 0
3 years ago
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