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3 years ago

14

What describes how the spring constant affects the potential energy of an object for a given displacement from an equilibrium po

sition?

1 answer:

Anit [1.1K]3 years ago

8 0

Answer:

The higher the spring constant, the greater the elastic potential energy.

You might be interested in

The mass of the Earth is 5.972 × 10^27 g. Convert this mass to kg.

sineoko [7]

Answer:

5.972x10^27 x 10^-3 = 5.972x10^24

Explanation:

1g = 10^-3 kg => So the mass of Earth in kg is 5.972x10^24

3 0

2 years ago

What is Newton's third law of motion? Give an example.

Vikentia [17]

Answer: Examples of Newton's third law of motion are ubiquitous in everyday life. For example, when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air. Engineers apply Newton's third law when designing rockets and other projectile devices.

Explanation: hope this helps :)

7 0

2 years ago

Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the co

ladessa [460]

Answer:

$\dot W_{in} = 49.386\,kW$

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

$\dot W_{in} + \dot m \cdot c_{p}\cdot (T_{1}-T_{2}) = 0$

The power consumed by the compressor can be calculated by the following expression:

$\dot W_{in} = \dot m \cdot c_{v}\cdot (T_{2}-T_{1})$

Let consider that air behaves ideally. The density of air at inlet is:

$P\cdot V = n\cdot R_{u}\cdot T$

$P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

$\rho = \frac{P\cdot M}{R_{u}\cdot T}$

$\rho = \frac{(104\,kPa)\cdot (28.02\,\frac{kg}{kmol})}{(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (292\,K)}$

$\rho = 1.2\,\frac{kg}{m^{3}}$

The mass flow through compressor is:

$\dot m = \rho \cdot \dot V$

$\dot m = (1.2\,\frac{kg}{m^{3}})\cdot (0.15\,\frac{m^{3}}{s} )$

$\dot m = 0.18\,\frac{kg}{s}$

The work input is:

$\dot W_{in} = (0.18\,\frac{kg}{s} )\cdot (1.005\,\frac{kJ}{kg\cdot K})\cdot (565\,K-292\,K)$

$\dot W_{in} = 49.386\,kW$

5 0

2 years ago

If the cross-sectional area of a wire is decreased while all other factors remain the same, how will the resistance change?

Nataly [62]

Answer:

A. It will increase

Explanation:

Area of a wire is:

R = ρL/A

where ρ is the resistivity of the wire,

L is the length of the wire,

and A is the cross-sectional area.

If A decreases, while ρ and L stay the same, then R will increase.

4 0

2 years ago

A force of 30 N is exerted on an object on a frictionless surface for a distance of 6.0 meters. If the object has a mass of 10 k

tangare [24]

Answer:The velocity of the object will be 5 $\sqrt7$ m/s or 13.23m/s

Explanation:

force exerted by the object= 30N

distance displayed by the object by the action of force=6.0m

mass of object=10kg

velocity gained by the object=?

$\frac{1}{2}mv^{2}= forcexdisplacement\\\frac{1}{2}10v^{2} = 30x6\\ 5v^{2}=180\\ v^{2}= 180-5\\ v^{2} =175\\v=\sqrt{175} \\v=5\sqrt{7} or 13.23$

6 0

1 year ago