If you were given distance & period of time, you would be able to calculate the speed.
Hope this helps!
Answer:
B. - 0.328
Explanation
Potential Energy:<em> This is the energy of a body due to position.</em>
<em>The S.I unit of potential energy is Joules (J).</em>
<em>It can be expressed mathematically as</em>
<em>Ep = mgh........................... Equation 1</em>
<em>Where Ep = potential energy, m = mass of the coin, h = height, g = acceleration due to gravity,</em>
<em>Given: m = 2.74 g = 0.00274 kg, h = 12.2 m, g = 9.8 m/s²</em>
Substituting these values into equation 1
Ep = 0.00274×12.2×9.8
Ep = 0.328 J.
Note: Since the potential energy at the surface is zero, the potential Energy with respect to the surface = -0.328 J
The right option is B. - 0.328
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Answer:
2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-
Explanation:
Here are the steps of how to arrive at the answer:
The volume of a cylinder = ((pi)D²/4) × H
Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m
H = Height of the reservoir = 87.32m
Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³
1ppm = 1g/m³
0.8ppm = 0.8 × 1g/m³
= 0.8g/m³
Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.
Thank you for reading.
For A 53 g ice cube at −30◦C is dropped into a container of water at 0◦C, the amount of water that freezes onto the ice? is mathematically given as
x = 9.93 g
<h3>What is the amount of water that freezes onto the ice?</h3>
Where
Energy received = energy given out
Generally, the amount of water is mathematically given as
(53)(0.5)(30) = (80)(x)
Therefore
x = (49)(0.5)(16)/(80)
x = 9.93 g
In conclusion, the mass of water
x = 9.93 g
Read more about mass
brainly.com/question/15959704
<span>We can answer this using
the rotational version of the kinematic equations:</span><span>
θ = θ₀ + ω₀<span>t + ½αt²
-----> 1</span></span>
ω² = ω₀² + 2αθ
-----> 2
Where:
θ = final angular
displacement = 70.4 rad
θ₀ = initial
angular displacement = 0
ω₀ = initial angular
speed
ω = final angular speed
t = time = 3.80 s
α = angular acceleration
= -5.20 rad/s^2
Substituting the values
into equation 1:<span>
70.4 = 0 + ω₀(3.80)
+ ½(-5.20)(3.80)² </span><span>
ω₀ = (70.4
+ 37.544) / 3.80 </span><span>
ω₀ = 28.406
rad/s </span><span>
Using equation 2:
ω² = (28.406)² + 2(-5.2)70.4
ω = 8.65 rad/s
</span>
