The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.
<h3>
Force required to pull one end at a constant speed</h3>
The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is determined by applying Newton's second law of motion as shown below;
F = ma
where;
- m is mass
- a is acceleration
At a constant speed, the acceleration of the object will be zero.
F = m x 0
F = 0
Thus, the force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.
Learn more about constant speed here: brainly.com/question/2681210
Answer:

Explanation:
As per thermal radiation we know that rate is heat radiation is given as

here we know that
T = 34 degree C = 307 K

e = 0.557


now we have




Answer:
t = 2.2 s
Explanation:
Given that,
A person observes a firework display for A safe distance of 0.750 km.
d = 750 m
The speed of sound in air, v = 340 m/s
We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.

So, the required time is 2.2 seconds.
Answer:
red I think
Explanation:
it's on red so I googled some of it and the closest was red
Answer: 3.33 m/s
Explanation:
Assuming the questions is to convert 12 km/h to meter per second (m/s), let's begin:
In order to make the conversion, we have to know the following:

And:

Keeping this in mind, we can make the conversion:

Then:
