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Eddi Din [679]
3 years ago
8

If you pick up and carry a piece of heavy furniture which type of friction do you have with the floor

Physics
1 answer:
Phoenix [80]3 years ago
7 0
Sliding Friction hope its right
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A). Archaea Hope it helps you
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3 years ago
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A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels
Aleksandr [31]

Answer:

<em>(a) The average velocity is 16 m/s</em>

<em>(b) The acceleration is 0.4 m/s^2</em>

<em>(c) The final velocity is 24 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:

v_f=v_o+at\qquad\qquad [1]

The distance traveled by the object is given by:

\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]

(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.

We know the distance is x=640 m and the time taken t= 40 s, thus:

\displaystyle \bar v=\frac{x}{t}=\frac{640}{40}=16

The average velocity is 16 m/s

Using the equation [1] we can solve for a:

\displaystyle a=\frac{v_f-v_o}{t}

(c) From [2] we can solve for a:

\displaystyle a= 2\frac{x-v_ot}{t^2}

Since vo=8 m/s, x=640 m, t=40 s:

\displaystyle a= 2\frac{640-8\cdot 40}{40^2}=0.4

The acceleration is 0.4 m/s^2

(b) The final velocity is calculated by [1]:

v_f=8+0.4\cdot 40

v_f=8+16=24

The final velocity is 24 m/s

3 0
3 years ago
What is the difference between the initial position and the final position of an object?
Grace [21]
The initial is where you are starting and the final postion is where the object ends up
5 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
This version of Einstein’s equation is often used directly to find what value?
stepladder [879]

A. THE ENERGY THAT IS RELEASED IN A NUCLEAR REACTION.

M =MASS

C^2= SPEED OF LIGHT SQUARED

8 0
3 years ago
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