Data:
Centripetal Force = ? (Newton)
m (mass) = 68 Kg
s (speed) = 3.9 m/s
R (radius) = 6.5 m
Formula:
Solving:
Answer:
<span>
B.159 N</span>
<span>The formulas are,
v1d1² = v2d2² ........ (1)
h = (v2²-v1²)/2g ...... (2)
Given that,
v1 = 1.71 m/s
we assume that the stream has decreased by a factor
d2 =0.805d1
then,
v1d1² = v2 (0.805d1)²
cancelled both side d1² then we get,
v1 = v2 (0.805)²
v1 = v2 (0.648025)
Sub v1 = 1.71,
1.71 = v2 (0.648025)
v2 = 1.71/0.648025
v2 = 2.638787083831642
v2 = 2.64 m/s
The vertical distance formula,
h = (v2²-v1²)/2g
We know that value of gravity constant is 9.8 m/s²
h = {(2.64)² - (1.71)²)/2(9.8)
h = {(6.9696) - (2.9241)}/19.6
h = (4.0455)/19.6
h = 0.2064030612244898
h = 0.21 cm
Therefore, the vertical distance h = 0.21 cm.</span>
Answer:
v = 1/3 m / s = 0.333 m / s
in the direction of the truck
Explanation:
The average speed is defined by the variation of the position between the time spent
v = Δx / Δt
since the position is a vector we must add using vectors, we will assume that the displacement to the right is positive, the total displacement is
Δx = 20 - 15 +20
Δx = 25 m
therefore we calculate
v = 25/75
v = 1/3 m / s = 0.333 m / s
in the direction of the truck
We are missing an important piece of information needed to answer this question: the number of kcal Charles losses per day. However, we can come up with a general equation in which kcal/day is the only independent variable.
We know that it takes 3500 kcal to lose one pound. To lose 5 pounds, Charles needs to lose 5 x 3500 kcal = 17,500 kcal.
To find how many days it takes Charles to lose 17,500 kcal (5 pounds), we must divide that amount by the number of kcal Charles loses per day.
Here is the equation to calculate that number
Number of days= 17500 / (kcal per day)
If given calories, remember that 1000 calories = 1 kcal, and .001 kcal = 1 cal
Answer:
Few millimeter thick aluminium, water, wood, acrylic glass or plastic.
Explanation:
The materials that are best for protection against beta particles are few millimeter thickness of aluminium, but for the high energy beta-particles radiations the low atomic mass materials such as plastic, wood, water and acrylic glass can be used.
These materials can also be used in personal protective equipment which includes all the clothing that can be worn to prevent any injury or illness due to the exposure to radiation.
