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Rufina [12.5K]
2 years ago
11

ASAP PLS HELP What is chemical potential energy?

Physics
1 answer:
mash [69]2 years ago
5 0

Answer:

chemical potential of a species is energy that can be absorbed or released due to a change of the particle number of the given species, e.g. in a chemical reaction or phase transition.

Explanation:

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Two 2.0kg bodies A and B collide The velocities before the collision are U1=15i+30j and U2=-10j+5.0j After the collision V1=-5.0
Svetllana [295]

Use the law of conservation of momentum. Since the momentum is a linear measure, we can treat each of the dimension separately:

i-direction:

m_1u_{1i}+m_2u_{2i}=m_1v_{1i}+m_2v_{2i}\\v_{2i} = \frac{m_1u_{1i}+m_2u_{2i}-m_1v_{1i}}{m_2}=\frac{(2\cdot 15-2\cdot10+2\cdot5)kg\frac{m}{s}}{2kg}=10\frac{m}{s}

j-direction:

m_1u_{1j}+m_2u_{2j}=m_1v_{1j}+m_2v_{2j}\\v_{2j} = \frac{m_1u_{1j}+m_2u_{2j}-m_1v_{1j}}{m_2}=\frac{(2\cdot 30+2\cdot5-2\cdot20)kg\frac{m}{s}}{2kg}=15\frac{m}{s}

Answer: Final velocity is: (10i + 15j) m/s

Change in the kinetic energy:

\Delta E_k = E_{ku}-E_{kv} = \frac{1}{2}m(u_1^2+u_2^2-v_1^2-v_2^2)=\\=\frac{1}{2}2kg(1125+125-425-325)\frac{m^2}{s^2}=500J

Answer: The system lost 500J worth of kinetic energy in the collision

4 0
3 years ago
You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.How muc
ahrayia [7]

Work is equal to the force applied times the displacement. Since you pull the wagon at constant speed this means that there is no acceleration on the wagon as it does not change speed. F=ma. Since a=0, F=0. Therefore no work has been done in this situation

3 0
3 years ago
You are standing next to a table and looking down on a record player sitting on the table. Take the spindle (axis of rotation) t
VLD [36.1K]

Answer:

a) the rotational speed of the clay is  3.45 rad/s

b) the value of A in the equation of motion is 0.15 m

c) the value of ϕi is 90° or π/2 rad.

Explanation:

 Given that;

Revolution per minute rpm = 33( 1/3) =  100/3

The frequency f = 100 / 3(60) = 0.55 Hz

a)

Rotational speed W = 2πf

we substitute

W = 2π × 0.55

W = 3.45 rad/s

Therefore, the rotational speed of the clay is  3.45 rad/s

b)

given equation; y(t)=Asin(ωt+ϕi)

given that radius = 0.15 m

y(t)=(0.2)sin(ωt+ϕi)

Therefore, the value of A in the equation of motion is 0.15 m

c)

since y(t) has the maximum value at t =0

so at t=0

y(0) = (0.15)sin(ω(0)+ϕi)

= 0.15sin(ϕi)

this will give maximum value when ϕi = 90°

so

y(0) = (0.15)sin(ω(0)+ϕi)

= 0.15sin(90°)

= 0.15

hence, the value of ϕi is 90° or π/2 rad.

7 0
2 years ago
A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s
lisabon 2012 [21]

Answer:

Magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Given

Contact Time = t = 0.05 seconds

Mass (of ball) = 0.80kg

Initial Velocity = u = 25m/s

Final Velocity = 25m/s

Magnitude of the average force exerted on the wall by the ball is given by;

F = ma

Where m = 0.8kg

a = Average Acceleration

a = (u + v)/t

a = (25 + 25)/0.05

a = 50/0.05

a = 1000m/s²

Average Force = Mass * Average Acceleration

Average Force = 0.8kg * 1000m/s²

Average Force = 800kgm/s²

Average Force = 800N

Hence, the magnitude of the average force exerted on the wall by the ball is 800N

3 0
3 years ago
Assuming that Albertine's mass is 60.0 kg , for what value of μk, the coefficient of kinetic friction between the chair and the
makvit [3.9K]

Answer:

\mu_k=0.101

Explanation:

It is given that,

Mass of Albertine, m = 60 kg

It can be assumed, the spring constant of the spring, k = 95 N/m

Compression in the spring, x = 5 m

A glass sits 19.8 m from her outstretched foot, h = 19.8 m

When she just reach the glass without knocking it over, a force of friction will also act on it. Using the conservation of energy for the spring mass system such that,

\dfrac{1}{2}kx^2=\mu_k mgh

\mu_k=\dfrac{kx^2}{2mgh}

\mu_k=\dfrac{95\times (5)^2}{2\times 60\times 9.8\times 19.8}

\mu_k=0.101

So, the coefficient of kinetic friction between the chair and the waxed floor is 0.101. Hence, this is the required solution.

3 0
3 years ago
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