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3 years ago

If the kinetic and potential energy in a system are equal, then the potential energy increases. What happens as a resul

1 answer:

7 0

-- If the system is 'closed', then nothing ... including energy ... can get in or out, and the total energy inside has to be constant.

If half of the energy in the system starts out as potential energy and the rest starts out as kinetic, and then the potential energy increases, there's only one place the increase could have come from ... it could only have been converted from kinetic energy. So the kinetic energy in the system must decrease.

In fact, this isn't even a "result". The kinetic energy has to decrease before the potential energy can increase, because that's where the increase has to come from.

If the system is 'open', then energy can come in and go out. If the potential energy inside suddenly increases, we don't know where it came from, so we can't say anything about what happens to the system.

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A 200g particle oscillating in SHM travels 66cm between the two extreme points in its motion with an average speed of 110cm/s. F

IgorC [24]

For a:v = d / Δt
110 = 0.66 / Δt
Δt = 0.66 / 110
Δt = 0.006 s
the period is:
T = 2Δt
T = 2*0.006
T = 0.012 s
the frequency is the inverse of the period. so: f = 1 / T
f = 83.3333333 Hz (about; Hz = 1/s)
b. T = 2π√(m/k)
being the mass m = 200g = 0.2 kg = 2*10^-1 kg, π = 3.14 (about) and T = 0.012, k is equal to:
0.012 = 6.28√(2*10^-1 / k)
0.012 / 6.28 = √(2*10^-1 / k)
0.00191082803 = √(2*10^-1 / k)
2*10^-1/ k = 0.000003
2*10^-1 / k = 3*10^-6
k = 2*10^-1 / 3*10^-6
k = 6.67*10^-5

now using hooke's law:
F = -kx
F = - 6.67*10^-5* 3.3*10^-1
F = -2.20x10^-5m
F = -0.22 *10^4 N

4 0

4 years ago

A steel bar that is at 10 ° c is 5 meters long, a bar for heated to 120 ° c, how long is that bar? Α = 1.2.10- ° c

svet-max [94.6K]

Answer:

1) 5.0066 m

2A) β = 3×10⁻⁷ / °C

2B) 2500.045 cm²

3A) γ = 8.1×10⁻⁵ / °C

3B) 1618.144 cm³

Explanation:

1) Linear thermal expansion is:

ΔL = α L₀ ΔT

where ΔL is the change in length,

α is the linear thermal expansion coefficient,

L₀ is the original length,

and ΔT is the change in temperature.

Given L₀ = 5 m, ΔT = 110°C, and α = 1.2×10⁻⁵ / °C:

ΔL = (1.2×10⁻⁵ / °C) (5 m) (110°C)

ΔL = 0.0066 m

The length increases by , so the new length is:

L = L₀ + ΔL

L = 5 m + 0.0066 m

L = 5.0066 m

2A) The surface expansion coefficient is:

β = 2α

β = 2 (1.5×10⁻⁷ / °C)

β = 3×10⁻⁷ / °C

2B) The change in area is:

ΔA = β A₀ ΔT

ΔA = (3×10⁻⁷ / °C) (50 cm × 50 cm) (60°C)

ΔA = 0.045 cm²

So the new area is:

A = A + ΔA

A = 2500 cm² + 0.045 cm²

A = 2500.045 cm²

3A) The volumetric expansion coefficient is:

γ = 3α

γ = 3 (2.7×10⁻⁵ / °C)

γ = 8.1×10⁻⁵ / °C

3B) The change in volume is:

ΔV = γ V₀ ΔT

ΔV = (8.1×10⁻⁵ / °C) (1600 cm³) (140°C)

ΔV = 18.144 cm³

So the new area is:

V = V + ΔV

V = 1600 cm³ + 18.144 cm³

V = 1618.144 cm³

6 0

3 years ago

Which of the following is an example of bad sportsmanship?

Minchanka [31]

I believe that B is the answer.

3 0

4 years ago

HELPPpPpPpPpPp!!!! ASAP !!!

DENIUS [597]

Answer:

50 N

Explanation:

8 0

2 years ago

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$