Strong nuclear forces act over _____ distances within atoms.

A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.0 km, the je

jenyasd209 [6]

<h2>The answer got is reasonable.</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 300 m/s

Acceleration, a = ?

Final velocity, v = 400 m/s

Displacement,s = 4 km = 4000 m

Substituting

v² = u² + 2as

400² = 300² + 2 x a x 4000

a = 8.75 m/s² = 8.8 m/s²

The acceleration is 8.8 m/s²

The answer got is reasonable.

7 0

3 years ago

Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is

attashe74 [19]

Answer:

0.125 volts

Explanation:

The induced emf can be sufficient to stimulate neuronal activity.

One such device generates a magnetic field within the brain that rises from zero to 1.5 T in 120 ms.

We need to find the induced emf within a circle of tissue of radius 1.6 mm and that is perpendicular to the direction of the field. The formula for the induced emf is given by :

$\epsilon=-\dfrac{d\phi}{dt}$

Where

$\phi$ is magnetic flux

So,

$\epsilon=-\dfrac{d(BA)}{dt}\\\\=2\pi r\times \dfrac{dB}{dt}\\\\=2\pi \times 1.6\times 10^{-3}\times \dfrac{1.5-0}{120\times 10^{-3}}\\\\=0.125\ V$

So, the induced emf is equal to 0.125 volts.

7 0

3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

An 80.0 kg skier slides down a hill shaped as shown. Assume

umka21 [38]

The height above the ground from where the skier start is 11.5 m.

<h3>Conservation of energy</h3>

The height above the ground from where the skier start is determined by applying the principle of conservation of energy as shown below;

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

$h = \frac{v^2}{2g} \\\\h = \frac{15^2}{2 \times 9.8} \\\\h = 11.5 \ m$

Thus, the height above the ground from where the skier start is 11.5 m.

Learn more about conservation of energy here: brainly.com/question/166559

8 0

3 years ago

Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m

Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

$m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s$