What is a creative name of a roller coaster about the solar system

Two uniform bars of the same dimensions are constructed from the same material. One bar has five evenly spaced holes through it

Elan Coil [88]

Solution :

The angular acceleration, $\alpha$ is obtained from the equation of the $\text{Newton's second law}$ of rotational motion,

Thus,

$\tau = F \times d$

or $\tau = I \times \alpha$

where $\tau$ is torque, F is force, d is moment arm distance, I is the moment of inertia

Thus, $\alpha=\frac{(F\times d)}{I}$

Now if the force and the moment arm distance are constant, then the $\text{angular acceleration is inversely proportional to the moment of inertia.}$

That is when, F = d = constant, then $\alpha \propto \frac{1}{I}$ .

Thus, moment of inertia, I is proportional to mass of the bar.

The mass is less for the bar in case (1) in comparison with that with the bar in case (2) due to the holes that is made in the bar.

Therefore, the bar in case (1), has less moment of inertia and a greater angular acceleration.

6 0

3 years ago

A navy diver hears the underwater sound wave from an exploding ship across the harbor. They immediately lift their head out of t

tamaranim1 [39]

Answer:

s = 1800 m = 1.8 km

Explanation:

The distance, the speed, and the time of reach of the sound are related by the following formula:

$s = vt$

where,

s = distance

v = speed

t = time

FOR WATER:

$s = v_wt$ ---------------------- eq (1)

where,

s = distance between ship and diver = ?

$v_w$ = speed of sound in water = 1440 m/s

t = time taken by sound in water

FOR AIR:

$s = v_a(t+4\ s)$ ---------------------- eq (2)

where,

s = distance between ship and diver = ?

$v_a$ = speed of sound in water = 344 m/s

t + 4 s = time taken by sound in water

Comparing eq (1) and eq (2),because distance remains constant:

$v_wt=v_a(t+4\ s)\\\\(1440\ m/s)t = (344\ m/s)(t+4\ s)\\(1440\ m/s - 344\ m/s)t=1376\ m\\t = \frac{1376\ m}{1096\ m/s}$

t = 1.25 s

Now using this value in eq (1):

$s = (1440\ m/s)(1.25\ s)$

s = 1800 m = 1.8 km

4 0

3 years ago

In building a particle accelerator, you manage to produce a uniform electric field of magnitude 6.03 × 10 5 N/C in one 35.5 cm s

mamaluj [8]

Answer:

V = 2.14×10⁵ V.

W = 3.424×10⁻¹⁴ J.

Explanation:

Electric Potential: This can be defined as the work done in bringing a unit positive charge from infinity to that point, against the action of a field.

The S.I unit is V.

The expression containing electric potential, distance and electric field is given as,

V = E×r .............. Equation 1

Where V = Electric potential difference across the length of the accelerator's section, E = Electric Field, r = Length of the section.

Given: E = 6.03×10⁵ N/C, r = 35.5 cm = 0.355 m.

Substitute into equation 1

V = 6.03×10⁵×0.355

V = 2.14065×10⁵ V.

V ≈ 2.14×10⁵ V.

amount of Work required to move a proton through the section is given as,

W = qV ............... Equation 2

Where W = work required to move a proton through the section, q = charge on a proton V = Electric potential.

Given: V = 2.14×10⁵ V, q = 1.60 x 10⁻¹⁹ C.

Substitute into equation 2

W = (2.14×10⁵)(1.60 x 10⁻¹⁹)

W = 3.424×10⁻¹⁴ J.

6 0

3 years ago

You live on a planet far from ours. "Based on extensive communication with a physicist on earth", you have determined that all l

Ugo [173]

Answer:

8.56 m/s2

Explanation:

Using law of energy conservation while taking into account of the rotational and translation kinetic energy, when the solid cylinder rolls down the incline we have the potential energy converted to kinetic energy:

$E_p = E_{kv} + E_{k\omega}$

$mgh = mv^2/2 + I\omega^2/2$

where m is the mass, $I = mr^2/2$ is the moments of inertia of the solid cylinder $\omega = v / r$ is the angular speed of the cylinder

$mgh = mv^2/2 + \frac{mr^2}{2}\frac{(v/r)^2}{2}$

$mgh = mv^2/2 + mv^2/4 = 3mv^2/4$

$h = 3v^2/(4g)$

So if you plot a liner chart of h vs $v^2$ and get a slope of 6.42 then that means 3/(4g) = 6.42 so $g = 6.42*4/3 = 8.56 m/s^2$

The gravitational acceleration on this planet is 8.56 m/s2

3 0

4 years ago

Nuclear chain reactions within power plants do NOT produce bomb-like explosions primarily because the ________.

Gelneren [198K]

Answer: D. Density of uranium within nuclear fuel rods is insufficient to become explosive

Explanation: Nuclear power plants use the same fuel as nuclear bombs, i.e. radioactive Uranium-235 isotope. However, in a nuclear power plant, the energy is released more slowly unlike in a nuclear bomb. The energy released is through nuclear fission, and radioactive decay occurs at the same rate as in nuclear bombs. therefore, option A, B and C are incorrect.

The primary reason why nuclear chain reactions within power plants do NOT produce bomb-like explosions is because the uranium fuel rods used in electricity generation is not sufficiently enriched in Uranium-235 to produce a nuclear detonation. This is the same idea in option D which is the correct option.

6 0

3 years ago