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3 years ago

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Initially sliding with a speed of 1.9 m/s, a 1.8 kg block collides with a spring and compresses it 0.35 m before coming to rest.

what is the force constant of the spring?

1 answer:

Alika [10]3 years ago

3 0

Let k = the force constant of the spring (N/m).

The strain energy (SE) stored in the spring when it is compressed by a distance x=0.35 m is
SE = (1/2)*k*x²
= 0.5*(k N/m)*(0.35 m)²
= 0.06125k J

The KE (kinetic energy) of the sliding block is
KE = (1/2)*mass*velocity²
= 0.5*(1.8 kg)*(1.9 m/s)²
= 3.249 J

Assume that negligible energy is lost when KE is converted into SE.
Therefore
0.06125k = 3.249
k = 53.04 N/m

Answer: 53 N/m (nearest integer)

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Blake and Sandra are having a rummage sale. Blake drags 3 boxes a distance of 10 meters each. He exerts a force of 20 newtons on

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Read 2 more answers

A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its

Pavel [41]

Answer:

a

$v_2 = 5.6 \ m/s$

b

$P_2 = 80600 \ Pa$

Explanation:

From the question we are told that

The pressure of the water in the pipe is $P_1= 110 \ kPa = 110 *10^{3 } \ Pa$

The speed of the water is $v_1 = 1.4 \ m/s$

The original area of the pipe is $A_1 = \pi \frac{d^2 }{4}$

The new area of the pipe is $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

Generally the continuity equation is mathematically represented as

$A_1 * v_1 = A_2 * v_2$

Here $v_2$ is the new velocity

So

$\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=> $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

=> $d^2 * 1.4 = \frac{d^2}{4} * v_2$

=> $1.4 = 0.25 * v_2$

=> $v_2 = 5.6 \ m/s$

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

$P_1 + \frac{1}{2} * \rho * v_1 ^2 = P_2 + \frac{1}{2} * \rho * v_2 ^2$

Here $\rho$ is the density of water with value $\rho = 1000 \ kg /m^3$

$P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]$

=> $P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]$

=> $P_2 = 80600 \ Pa$

4 0

3 years ago

If a photon has frequency = 2.00 x 1014s-1 and the speed of light = 3.00 x 108ms-1, then what is its wavelength?

butalik [34]

Answer:

The photon has a wavelength of $1.5x10^{-6}m$

Explanation:

The speed of a wave can be defined as:

$v = \nu \cdot \lambda$ (1)

Where v is the speed, $\nu$ is the frequency and $\lambda$ is the wavelength.

Equation 1 can be expressed in the following way for the case of an electromagnetic wave:

$c = \nu \cdot \lambda$ (2)

Where c is the speed of light.

Therefore, $\lamba$ $\lambda$ can be isolated from equation 2 to get the wavelength of the photon.

$\lambda = \frac{c}{\nu}$ (3)

$\lambda = \frac{3.00x10^{8}m/s}{2.00x10^{14}s^{-1}}$

$\lambda = 1.5x10^{-6}m$

Hence, the photon has a wavelength of $1.5x10^{-6}m$

<em>Summary: </em>

Photons are the particles that constitutes light.

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3 years ago