Answer:
hypothesis , hope it helps
Explanation:
Your diagram should include four forces:
• the box's weight, pointing down (magnitude <em>w</em> = 43.2 N)
• the normal force, pointing up (mag. <em>n</em>)
• the applied force, pointing the direction in which the box is sliding (mag. <em>p</em> = 6.30 N, with <em>p</em> for "pull")
• the frictional force, pointing oppoiste the applied force (mag. <em>f</em> )
The box is moving at a constant speed, so it is inequilibrium and the net forces in both the vertical and horizontal directions sum to 0. By Newton's second law, we have
<em>n</em> + (-<em>w</em>) = 0
and
<em>p</em> + (-<em>f</em> ) = 0
So then the forces have magnitudes
<em>w</em> = 43.2 N
<em>n</em> = <em>w</em> = 43.2 N
<em>p</em> = 6.30 N
<em>f</em> = <em>p</em> = 6.30 N
Answer:
The workdone is 
Explanation:
From the question we are told that
The height of the cylinder is 
The face Area is 
The density of the cylinder is 
Where 
is the density of freshwater which has a constant value
Now
Let the final height of the device under the water be 
Let the initial volume underwater be 
Let the initial height under water be 
Let the final volume under water be 
According to the rule of floatation
The weight of the cylinder = Upward thrust
This is mathematically represented as
So 
=> 
Now the work done is mathematically represented as
![= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.](https://tex.z-dn.net/?f=%3D%20%20%20%5Crho_w%20g%20A%20%5B%5Cfrac%7Bh%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20h_f%7D%20%5Catop%20%7Bh%7D%7D%20%5Cright.)
![= \frac{g A \rho}{2} [h^2 - h_f^2]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%20%5Bh%5E2%20-%20h_f%5E2%5D)
![= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%28h%5E2%29%20%20%5B1%20%20-%20%5Cfrac%7Bh_f%5E2%7D%7Bh%5E2%7D%20%5D)
Substituting values
1). Take a sample of the substance. The sample should be the largest
possible that will allow it to be be easily handled and the following steps
to be performed with it.
(The density doesn't depend on the size of the sample, and every sample
of the same substance has the same density. But using a larger sample
can improve the accuracy of the measurements you make, and therefore
improve the accuracy of the density you derive for the substance.)
2). Ask or measure the mass of the sample.
3). Ask or measure the volume of the sample.
4). Divide the mass by the volume. Their quotient is the density
of the substance.
Answer:
10.88 km
Explanation:
We shall represent displacement in terms of i , j unit vectors in the direction of east and north .
4.5 km due west
D₁ = - 4.5 i
6.7 km at an angle of 27° south of west
D₂ = - 6.7 cos27 i - 6.7 sin27j
= - 6.7 x .89 i - 6.7 x .45 j
= - 5.96i - 3 j
Total displacement
= D₁ + D₂
= - 4.5 i - 5.96i - 3 j
= -10.46 i - 3j
Magnitude = √ ( 10.46² + 3²)
= √ ( 109.41 + 9)
= √ 118.41
= 10.88 km .
