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2 years ago

Write an expression for elastic potential energy of a spring?

Physics

1 answer:

AveGali [126]2 years ago

6 0

$\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}$

$\leadsto$ The elastic potential energy of a spring can be said to be the energy possessed by a spring due to a force that compresses or stretches the spring. This can also be said to be the work which is done by the applied force against the restoring force of the spring, which is stored as the elastic potential energy.

The elastic potential of a spring can be expressed as:

▪ $\longrightarrow \sf{PE= \: \dfrac{1}{2} kx}$

$\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}$

$\bm{PE= \: \dfrac{1}{2} kx}$

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4. A driver takes 3.5 s to react to a complex situation. How far does the vehicle travel before the driver initiates a physical

katen-ka-za [31]

This question is incomplete, the complete question is;

A driver takes 3.5 s to react to a complex situation while driving at a speed of 60 mi/h.

How far does the vehicle travel before the driver initiates a physical response to the situation (i.e., putting his or her foot on the brake)

Answer:

the distance travelled by the vehicle, before the driver initiates a physical response is 308 ft

Explanation:

Given that;

Reaction time t = 3.5 sec

Speed S = 60 mi/h

we convert mile to foot and hour to seconds

Speed S = 60 mi/h = (60×5280)/(60×60) = 316800/3600 = 88 ft/sec

Now, The distance travelled by vehicle before the driver initiates physical response is equal to Lag distance. i.e

Lag distance = V × t

we substitute

Lag distance = 88 ft/sec 3.5 sec

Lag distance = 308 ft

Therefore, the distance travelled by the vehicle, before the driver initiates a physical response is 308 ft

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3 years ago

Explain why pumice floats and why ironwood sinks

almond37 [142]

Answer:

A quick way of describing density is to describe an object as heavy or light for its size. Pumice stone, unlike regular rock, does not sink in water because it has a low density. An ironwood branch is very dense and sinks in water.

Hope that helps. x

6 0

2 years ago

Read 2 more answers

What is the difference between displacement and distance travelled

IrinaK [193]

Displacement in Space
It is the length of a body's real route. It is the shortest distance between the body's final and beginning positions.
It's a number with a scalar value. It's a quantity with a vector.
It can't possibly be negative. It might be a negative number, a zero number, or a positive number.

8 0

3 years ago

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A car with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. if a car is init

Sladkaya [172]

The time required by the car to stop is 4.916 sec.

Since the car is moving with the constant deceleration we can apply the first equation of motion to calculate the time required by the car to stop.

The first equation of motion is given as

V=u+at

Here, V=final speed of the car=0 mi/h as the car stops

u =initial speed of the car=55 mi/hr=24.58 m/s

a= acceleartion =-5 m/s^2 (here negative sign indicates for deceleration)

Now applying the values in the first equation

V=u+at

0=24.58-5*t

t=4.916 sec

Therefore the car will stops in 4.916 sec.

8 0

3 years ago

A record is spinning at the rate of 25rpm. If a ladybug is sitting 10cm from the center of the record.

marin [14]

A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

C) Tangential speed: 26.4 cm/s

D) Distance travelled: 528 cm

Explanation:

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

$\omega = 25 rpm$

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

$\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s$

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

$\omega = 2 \pi f$ (1)

where

$\omega$ is the angular speed

f is the frequency of revolution

For the ladybug in this problem,

$\omega=0.42 rev/s$

Keeping in mind that $1 rev = 2\pi rad$ , the angular speed can be rewritten as

$\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42$

And re-arranginf eq.(1), we can find the frequency:

$f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz$

And the frequency is the number of complete revolutions made per second.

For an object in circular motion, the tangential speed is related to the angular speed by the equation

$v=\omega r$

where

$\omega$ is the angular speed

v is the tangential speed

r is the distance of the object from the axis of rotation

For the ladybug here,

$\omega = 2\pi \cdot 0.42 rad/s$ is the angular speed

r = 10 cm = 0.10 m is the distance from the center of the record

So, its tangential speed is

$v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s$

The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

$d=vt$