Explanation:
What is the weight of a 2.00-kilogram object on the surface of Earth?
2.00 N
4.91 N
9.81 N
19.6 N
Given parameters:
Mass of the object = 2kg
Unknown:
Weight of the object = ?
Solution:
The weight of an object is the force of gravity acting on the object;
Weight = mass x acceleration due to gravity
Acceleration due to gravity = 9.8m/s²
Now insert the parameters and solve;
Weight = 2 x 9.8 = 19.6N
A person weighing 785 Newtons on the surface of the Earth would weigh 47 Newtons on the surface of Pluto. What is the magnitude of the gravitational acceleration on the surface of Pluto?
1.7 m/s²
0.59 m/s²
0.29 m/s²
9.8 m/s²
Given parameters:
Weight on Earth = 785N
Weight on Pluto = 47N
Unknown:
Acceleration due to gravity on Pluto = ?
Solution
The mass of the body both on Earth and Pluto is the same.
Weight = mass x acceleration due to gravity
Now find the mass on Earth;
Acceleration due to gravity on Earth = 9.8m/s²
785 = mass x 9.8
mass = 
= 80.1kg
So;
Acceleration due to gravity on Pluto = 
Acceleration due to gravity = 
= 0.59m/s²
There are two different ways to understand time and these are:
A. What time is it?
B. How much time?
The examples of these two different ways are:
A. What time is it? The best example that would help us understand and know what time are the clock and the calendar. This gives us the exact hour, minutes and seconds. The calendar tells us the exact day, month and year.
B. How much time? This makes us understand how much time did it take from the starting time. An example for this would be a stopwatch.
Answer: Before the jump, the snowboarder would carry potential energy.
During the jump he will carry kinetic energy.
And after the jump, assuming hes at a full stop, he will carry potential energy once again.
This would be B
Hope this helped
Answer:
388.97 nm
Explanation:
The computation of the wavelength of this light in benzene is shown below:
As we know that
n (water) = 1.333
n (benzene) = 1.501
And, the wavelength of water is 438 nm
![\lambda (benzene) = \lambda (water) [\frac{n(water)}{n(benzene}]](https://tex.z-dn.net/?f=%5Clambda%20%28benzene%29%20%3D%20%5Clambda%20%28water%29%20%5B%5Cfrac%7Bn%28water%29%7D%7Bn%28benzene%7D%5D)
Now placing these values to the above formula
So,
= 388.97 nm
We simply applied the above formula so that we can easily determine the wavelength of this light in benzene could come
