This type of a problem can be solved by considering energy transformations. Initially, the spring is compressed, thus having stored something called an elastic potential energy. This energy is proportional to the square of the spring displacement d from its normal (neutral position) and the spring constant k:

$E_p=\frac{1}{2}kd^2= \frac{1}{2}175\frac{N}{m}\cdot 0.37^2m^2=11.98J$

So, this spring is storing almost 12 Joules of potential energy. This energy is ready to be transformed into the kinetic energy when the masses are released. There are two 0.2kg masses that will be moving away from each other, their total kinetic energy after the release equaling the elastic energy prior to the release (no losses, since there is no friction to be reckoned with).

The kinetic energy of a mass m moving with a velocity v is given by:

$E_k = \frac{1}{2}mv^2$

And we know that the energies are conserved, so the two kinetic energies will equal the elastic potential one:

$E_p = 2E_k=mv^2$

From this we can determine the speed of the mass:

$E_p =mv^2\implies v=\pm \sqrt{\frac{E_p}{m}}=\pm\sqrt{\frac{11.98J}{0.2kg}}=\pm 7.74\frac{m}{s}$

The speed will be 7.74m/s in in one direction (+), and same magnitude in the opposite direction (-).

4 0

3 years ago

A charged particle moves at 2.5 × 104 m/s at an angle of 25° to a magnetic field that has a field strength of 8.1 × 10–2 T. If t

nekit [7.7K]

the magnitude of charge=q=8.76 x 10⁻⁵C

Explanation:

the magnetic force Fm is given by

Fm= q V B sinθ

q= charge

v= velocity= 2.5 x 10⁴ m/s

B= magnetic field strength= 8.1 x 10⁻²T

Fm= magnetic force= 7.5 x 10⁻² N

θ=25°

so 7.5 x 10⁻² =q (2.5 x 10⁴ ) (8.1 x 10⁻²) sin25

q=8.76 x 10⁻⁵C

4 0

3 years ago

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Briefly describe the characteristics of each soil horizon from the top layer to the bottom layer

inna [77]

<span>There is six horizen. 1. O Horizon - The top, organic layer of soil, 2. A Horizon - The layer called topsoil; 3. E Horizon - This layer is beneath the A Horizon and above the B Horizon. It is made up mostly of sand. 4. B Horizon - Also called the subsoil - this layer is beneath the E Horizon and above the C Horizon. 5. C Horizon - it's called regolith: the layer beneath the B Horizon and above the R Horizon. 6 R Horizon - this is last and the unweathered rock layer that is beneath all the other layers.</span>

7 0

3 years ago