The height difference is found by

Then the change in potential energy is
Answer:
1.t=-1.96sec
2.H=4.8m
3.T=1.96sec
4.R=19.2m
Explanation:
u=9.8,t=?,sin theta=1
using formula t=2usintheta/g
t=2x9.8x1=19.6/10
t=1.96seconds
using formula H=u(squared)sin(squared)theta/2g
H=9.8(squared)x1(squared)/2x10
H=96x1/20
H=96/20
H=4.8m
using formula T=2usintheta/g
T=2x9.8x1/10
T=19.6/10
T=1.96sec
using the formula R=u(squared)sin2theta/g
R=9.8(squared)x2/10
R=96x2/10
R=192.08/10
R=19.2
The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m
<h3>What is volume?:</h3>
This is the product of the height of a solid object and its crossectional area.
The Volume of the rod is can be calculated using the formula below.
Note: A rod has the shape of a cylinder.
Formula:
- V = πr²h............... Equation 1
Where:
- V = Volume of the rod
- r = radius of the rod
- h = height of the rod.
From the question,
Given:
- r = 4mm = 0.004 m
- h = 25 cm = 0.25 m
- π = 3.14
Substitute these values into equation 1
- V = 3.14(0.004²)(0.25)
- V = 1.26×10⁻⁵ m³
<h3>What is linear charge density:</h3>
This is the ratio of the charge on an object to the length of the object.
The linear charge density of the rod can be calculated using the formula below.
- D = Q/h.................... Equation 2
Where:
- D = Linear charge density of the rod
- Q = Charge on the rod.
- h = height or length of the rod
From the question
Given:
- Q = 0.91 C
- h = 25 cm = 0.25 m
Substitute these values into equation 2
- D = 0.91/0.25
- D = 3.64 C/m
Hence, The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m
Learn more about charge density here: brainly.com/question/14568868
Explanation:
Given that,
Current of clothes dryer, I = 16 A
Voltage, V = 240 V
Time, t = 45 min = 2700 s
Current of personal computer, I' = 2.7 A
Voltage, V' = 120 C
Energy used by clothes dryer is given by :

Let t' is the time for this computer to "surf" the internet. Again using formula of energy used as :

So, for 8.83 hours you could use this computer to "surf" the internet.
Answer:
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