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Law Incorporation [45]

3 years ago

13

A spring with a rest length of 0.7 m has a spring constant of 70 N/m. It is stretched and now has a length of 2.5 m. What is the

potential energy stored in the spring?
____ J

1 answer:

pentagon [3]3 years ago

7 0

Answer:

<em>113.4 J</em>

Explanation:

<u>Elastic Potential Energy</u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

The spring has a natural length of 0.7 m and a spring constant of k=70 N/m. When the spring is stretched to a length of 2.5 m, the change of length is

Δx = 2.5 m - 0.7 m = 1.8 m

The energy stored in the spring is:

$\displaystyle PE = \frac{1}{2}70(1.8)^2$

PE = 113.4 J

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Shkiper50 [21]

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3 years ago

Please help.. urgent Which statement is equivalent to Newton's first law? a. 15,300 N b. 1.20*10^3 N c. 2,030 N d. 1,560 N

taurus [48]

According to Newton laws of motion,
F = m*a
Here, m = 1,560 Kg
a = 1.30 m/s²

Substitute their values,
F = 1,560 * 1.30
F = 2028 N ~ 2030 N [ Closest value ]

In short, Your Answer would be Option C

Hope this helps!

6 0

3 years ago

A 3.5 kg object moving at 8.1 m/s in the positive direction of an x axis has a one-dimensional elastic collision with an object

Tamiku [17]

Answer:

So the mass of the second object M will be 1.951 kg

Explanation:

We have given mass of the first object $m_1=3.5kg$ and its velocity $v_1=8.1m/sec$

Mass of the second object $m_2=M$ it is at rest so its velocity $v_2=0$

From conservation of momentum we know that

Initial momentum = final momentum

So $m_1v_1+m_2v_2=(m_1+m_2)v$

$3.5\times 8.1+M\times 0=(3.5+M)5.2$

$28.35=18.2+5.2M$

M = 1.951 kg

8 0

3 years ago

on takeoff, the propellers on a uav (unmanned aerial vehicle) increase their angular velocity from rest at a rate of ω = (25.0t)

Rudik [331]

Hi there!

a)
We can find the angular velocity at t = 2.0 s by plugging in this value into the equation.

$\omega (t) = 25.0t \\\\\omega (2) = 25.0(2) = \boxed{50.0 \frac{rad}{s}}$

b)

The angular acceleration is the derivative of the angular velocity, so:
$\alpha (t) = \frac{d\omega}{dt} (25.0t) = 25.0$

Thus, the angular acceleration is a <u>constant 25 rad/s².</u>

7 0

2 years ago

Two sinusoidal waves traveling in opposite directions interfere to produce a standing wave with the wave function y = (2.00) sin

olga2289 [7]

Answer:

Part a)

$\lambda = 4\pi$

Part b)

$f = 47.7 Hz$

Part c)

$v = 600 m/s$

Explanation:

Part a)

As we know that angular wave number is given as

$k = \frac{2\pi}{\lambda}$

$k = 0.500$

$0.500 = \frac{2\pi}{\lambda}$

$\lambda = \frac{2\pi}{0.500}$

$\lambda = 4\pi$

Part b)

As we know that angular frequency is given as

$\omega = 300 rad/s$

$\omega = 2\pi f$

$300 = 2\pi f$

$f = \frac{300}{2\pi}$

$f = 47.7 Hz$

Part c)

Speed of the wave is given as

$v = \lambda \times frequency$

so we have

$v = 4\pi \times 47.7$

$v = 600 m/s$

8 0

3 years ago