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Bess [88]
3 years ago
12

Which statement best describes an atom? (1 point) protons and neutrons grouped in a specific pattern protons and electrons sprea

d around randomly a group of protons and neutrons that are surrounded by electrons a ball of electrons and neutrons surrounded by protons
Physics
1 answer:
sashaice [31]3 years ago
7 0

A group of protons and neutrons surrounded by electrons

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A 250 g object is hung onto a spring. It stretches 18 cm. Find the spring's spring constant
Juliette [100K]
<h2>Answer: 13.61 N/m</h2>

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force F applied to it, <u>as long as the spring is not permanently deformed</u>:

F=k (x-x_{o})    (1)

Where:

k is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.

x_{o} is the length of the spring without applying force.

x is the length of the spring with the force applied.

According to this, we have a spring where only the force due gravity is applied.

In other words, the force applied is the weigth W of the block:

W=m.g   (2)

Where m=250g=0.25kg is the mass of the block and g=9.8\frac{m}{s^{2}}  is the gravity acceleration.

W=(0.25kg)(9.8\frac{m}{s^{2}})   (3)

W=2.45N   (4)

Knowing the force applied W and x=18cm=0.18m and x_{o}=0, we can substitute the values in equation (1) and find k:

W=k (x-x_{o})    (5)

2.45N=k (0.18m-0m)    (6)

<u>Finally:</u>

k=13.61\frac{N}{m}  

4 0
3 years ago
Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of
WINSTONCH [101]

a₀).  You know ...
         -- the object is dropped from 5 meters
             above the pavement;
         --  it falls for 0.83 second.

a₁).  Without being told, you assume ...
         -- there is no air anyplace where the marshmallow travels,
             so it free-falls, with no air resistance;
         -- the event is happening on Earth,
            where the acceleration of gravity is  9.81 m/s² .

b).  You need to find how much LESS than 5 meters
       the marshmallow falls in 0.83 second.
    
c).  You can use whatever equations you like.
       I'm going to use the equation for the distance an object falls in
       ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d).  To see how this all goes together for the solution, keep reading:


The distance that an object falls in ' T ' seconds
when it's dropped from rest is

                                 (1/2 G) x (T²) .

On Earth, ' G ' is roughly  9.81 m/s², so in 0.83 seconds,
such an object would fall

                               (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

                         (5.00 - 3.38)  =  1.62 meters

above the pavement.  That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0
3 years ago
What is the definition of reflection of light?​
Savatey [412]

Answer: Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. Common examples include the reflection of light, sound and water waves.

Explanation:

8 0
3 years ago
How many milligrams are equal to 8 grams?
Andre45 [30]

Answer:

it should be 8,000

Explanation:

because if you multiply the mass value by 1000, you will get 8,000  

:):):):)

6 0
3 years ago
A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1
nikklg [1K]

Answer:

Part a)

F_n = 306 N

Part B)

v = 12.1 m/s

Explanation:

Part A)

At the top of the hump the force on the rider is

1) Normal force

2) weight

so here we know that

mg - F_n = \frac{mv^2}{R}

F_n = mg - \frac{mv^2}{R}

F_n = (100)(9.81) - \frac{100(9^2)}{12}

F_n = 306 N

Part B)

At the top of the loop we will have

F_n + mg = \frac{mv^2}{R}

in order to remain in contact the normal force must be just greater than zero

so we will have

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = \sqrt{15\times 9.81}

v = 12.1 m/s

5 0
3 years ago
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