Hydrogen-2 and Hydrogen-3 fuse to form Helium-4 and a neutron. How much energy is released in this nuclear reaction?

At which latitudes shown in the image of Earth do people experience the greatest tangential speed?

PilotLPTM [1.2K]

There's actually only one latitude implied by anything attached the question . . . Zero, since there are no choices listed.

That's brilliant ! Because zero latitude (the equator) IS the place on Earth where tangential speed is greatest.

Very clever !

5 0

4 years ago

Who is good on civics i have a C in my class please help!!!!!!

Alex73 [517]

Answers: PHYSICS??

Explanation:

3 0

4 years ago

Two objects collide and stick together. Kinetic energy

rodikova [14]

Answer:

its inertia.

Inertia is the tendency of an object to continue in the state of rest or of uniform motion

4 0

3 years ago

Light of wavelength 578.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 62.5 cm from the slit.

yaroslaw [1]

Answer:

$80.6\mu m$

Explanation:

When light passes through a narrow slit, it produces a diffraction pattern on a distant screen, consisting of several bright fringes (constructive interference) alternated with dark fringes (destructive interference).

The formula to calculate the position of the m-th maximum in the diffraction pattern produced in the screen is:

$y=\frac{m\lambda D}{d}$

where

y is the distance of the m-th maximum from the central maximum (m = 0)

$\lambda$ is the wavelength of light used

D is the distance of the screen from the slit

d is the width of the slit

In this problem, we have:

$\lambda=578.0 nm = 578\cdot 10^{-9} m$ is the wavelength

D = 62.5 cm = 0.625 m is the distance of the screen

We know that the distance between the third order maximum (m=3) and the central maximum is 1.35 cm (0.0135 m), which means that

$y_3 = 0.0135 m$

For

m = 3

Therefore, rearranging the equation for d, we find the width of the slit:

$d=\frac{m\lambda D}{y_3}=\frac{(3)(578\cdot 10^{-9})(0.625)}{0.0135}=80.3\cdot 10^{-6} m=80.6\mu m$

8 0

4 years ago

A gun is fired vertically into a 1.40 kg block of wood at rest directly above it. If the bullet has a mass of 21.0 g and a speed

ahrayia [7]

To solve this problem we will use the theorems related to the conservation of momentum to determine the final speed of the system, once the bullet hits it. From there, we will calculate with the kinematic equations of linear motion, the height traveled with that speed. The conservation of momentum say us that,

$m_1u_1+m_2u_2 = (m_1+m_2)v_f$