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2 years ago

15

The product nuclei may or may not be _______

1 answer:

Vsevolod [243]2 years ago

8 0

Answer:

i have no clue

Explanation:

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Please someone help me with this!!!

exis [7]

Answer: Gravitational potential energy changes.

Explanation: This is because depending on the amount of mass in an object that’s the amount of gravity pulling you down to the center of the earth

5 0

2 years ago

I need help with science due tommorow

Andrews [41]

The first one is the light bends sheikh is known as refraction

8 0

3 years ago

O be useful in science, a hypothesis must be *

shepuryov [24]

Answer:

the answer is c testable

3 0

3 years ago

Read 2 more answers

You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength

Ray Of Light [21]

Answer:

wavelength $\lambda$ = 437.27 nm

Explanation:

given data

first bright fringe = 2.96 mm

slit separation = 0.325 mm

distance D = 2.20 m

solution

we know that this is double slit experiment

so we apply here Fringe width formula that is

β = $\frac{D\lambda}{d}$ ....................1

$\lambda$ is Wavelength of light and D is Distance between screen and slit and d is slit width

so put here value and we get

$\lambda$ = $\frac{2.96*0.325*10^{-6}}{2.20}$

$\lambda$ = 437.27 × $10^{-9}$ m

wavelength $\lambda$ = 437.27 nm

4 0

2 years ago

dybincka [34]

Answer:

1. 218.55 N

2. $30.96^{o}$

3. $2.1 m/s^{2}$

Explanation:

Part 1;

Net force $F=mg sin \theta$ where m is mass, g is gravitational force and $\theta$ is the angle of inclination

$F= 46*9.8*sin 29^{o}= 218.55N$

Frictional force, $F_{r}$ is given by

$F_{r} = \mu_{s}mg cos \theta$ where $\mu_{s}$ is the coefficient of static friction

$F_{r} = 0.6*46*9.8 cos 29$

$F_{r}=236.57N$

Since $F_{r}>F$ , therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

$\mu_{s}= \frac {sin \theta}{cos \theta}$

$\mu_{s}= tan \theta$

The maximum angle of inclination $\theta = tan^{-1} \mu_{s}$

$\theta = tan^{-1} (0.6)$

$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

$a = g ( sin 38 - \mu_{k} cos 38 )$

= 9.8 ( sin 38 - (0.51) cos 38 )

= $2.1m/s^{2}$

7 0

3 years ago