Explain why not everyone can become a bodybuilder, even if they train hard enough.

Bob is threatening Tom’s life with a giant laser with wavelength (650 nm), a distance (D = 10 m) from the wall James is shackled

Fittoniya [83]

Answer:

He should stand from the center of laser pointed on the wall at 1.3 m.

Explanation:

Given that,

Wave length = 650 nm

Distance =10 m

Double slit separation d = 5 μm

We need to find the position of fringe

Using formula of distance

$d\sin\theta=n\lambda$

$d\dfrac{y}{D}=n\lambda$

$y=\dfrac{\lambda D}{d}$

Put the value into the formula

$y=\dfrac{650\times10^{-9}\times10}{5\times10^{-6}}$

$y=1.3\ m$

Hence, He should stand from the center of laser pointed on the wall at 1.3 m.

8 0

2 years ago

An atom that has fewer neutrons than protons and more electrons than protons is a ion

riadik2000 [5.3K]

i think it is an ion

6 0

3 years ago

Please help me on this

QveST [7]

Answer:

i think its bulb 2

Explanation:

4 0

1 year ago

A block of mass 22 kg is sliding along the ice at constant speed 5.0 m/s just ahead of it is q block of mass 29 kg sliding in th

mars1129 [50]

Answer:

New speed of the 22-kg block is 1.57 m/s

Explanation:

Mass of block

Mass of another block

Initial speed of the block

Initial speed of the another block

For conservation of momentum, we have

Substitute all the values and solving for final speed of the 22kg block is

new speed of the 22-kg block is 1.57 m/s

Couldnt write the answer so check picture

8 0

2 years ago

What is the greatest distance an image can be located behind a convex spherical mirror?

Afina-wow [57]

Answer:

Maximum distance of image from mirror is equal to focal length of the mirror

Explanation:

As we know by the equation of mirror we have

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

here we know for convex mirror

object position is always negative as it will be placed behind the mirror always

while the focal length of the convex mirror is always taken positive

So here we have

$\frac{1}{d_i} + \frac{1}{-d_o} = \frac{1}{f}$

$\frac{1}{d_i} = \frac{1}{d_o} + \frac{1}{f}$

so here maximum value of image distance is equal to focal length of the mirror

6 0

2 years ago