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german
3 years ago
8

For each picture below identify the material and closing the sandwich as transparent translucent or opaque

Physics
1 answer:
astra-53 [7]3 years ago
5 0
I need a picture to help answer
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An inclinded plane makes work easier by increasing the distance and blank the force
iragen [17]

Hi! I believe your answer is decreasing. <u>An inclined plane makes work easier by decreasing the amount of effort force needed, but increases the distance</u>. I hope this helps you! Good luck and have a great day. ❤️✨

7 0
3 years ago
Which of these is an example of a mechanical wave
anastassius [24]
Some of the most common examples of mechanical waves are water waves, sound waves, and seismic waves. There are three types of mechanical waves: transverse waves, longitudinal waves, and surface waves.
6 0
3 years ago
A roller coaster speeds up with constant acceleration for 2.3 s until it reaches a velocity of 35 m/s. During this time, the rol
Arada [10]

The initial velocity is 0.65 m/s

Explanation:

The motion of the roller coaster is a uniformly accelerated motion, so we can solve the problem by using the following suvat equation:

s=(\frac{u+v}{2})t

where

s is the displacement

u is the initial velocity

v is the final velocity

t is the time

For the roller coaster in this problem:

v = 35 m/s

t = 2.3 s

s = 41 m

Solving the equation for u, we find the initial velocity:

u=\frac{2s}{t}-v=\frac{2(41)}{2.3}-35=0.65 m/s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

5 0
3 years ago
A bus moving along a level road increases its speed from 20 m/s to 35 m/s in 15.0s. What is the car's acceleration?​
Ksju [112]

Explanation:

initial velocity U = 20m/s

Final velocity V = 35m/s

time = 15.0 secs

change in velocity = 35 - 15

= 20m/s

acceleration a = change in velocity/time V/t

a = (35-20)/15

a= 15/15

Hence, your acceleration is 1m/s^2

5 0
2 years ago
What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

4 0
3 years ago
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