Question 4 of 10

Physics

1 answer:

kiruha [24]3 years ago

5 0

B hey what do u know i took that test to

You might be interested in

How can you determine the amount of work done on an object? (joules, work, power...etc.)

Novay_Z [31]

Work done is equal to force by distance; so you take the force exerted, in newtons, and multiply that by the direction it's moved (from the starting point in a line, not along the path it's taken.)

7 0

3 years ago

Diffraction patterns are due to: interference refraction dispersion scattering

uranmaximum [27]

Diffraction patterns are due to interference<span>. Diffraction is a phenomena which occurs when a wave encounters an obstacle. It is the bending of light around the corners if the obstacle.</span>

3 0

3 years ago

Read 2 more answers

Please help fill in the blank Earth’s axis passes through the (blank) and south poles.

LuckyWell [14K]

Answer:

North

Explanation:

Have a nice day :)

7 0

3 years ago

A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the

Bingel [31]

Answer:

$E_T= 28J$

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2$

Where:

$\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block$

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

$E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}$

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since $\Delta x=0$ , so:

$E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}$

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2$