We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"
- it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
From the question we are told
the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?
Generally the equation for the Force is mathematically given as
F=\frac{F}{dx}
Therefore
F=-kdx
k=600Nm^{-1}
now
K.E=0.5x ds^2
K.E=600*(-0.1^2)
K.E=3J
Therefore
the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
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The height to which the weight-watcher must climb to work off the equivalent 991 (food) Calories is 0.59 Km
<h3>How to determine the energy. </h3>
1 food calorie = 103 calories
Therefore,
991 food calories = 991 × 103
991 food calories = 102073 calories
Multiply by 4.2 to express in joule (J)
991 food calories = 102073 × 4.2
991 food calories = 428706.6 J
<h3>How to determine the height </h3>
- Energy (E) = 428706.6 J
- Mass (m) = 73.9 kg
- Acceleration due to gravity (g) = 9.8 m/s²
E = mgh
Divide both side by mg
h = E / mg
h = 428706.6 / (73.9 × 9.8)
h = 591.95 m
Divide by 1000 to express in km
h = 591.95 / 1000
h = 0.59 Km
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The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s
Answer:
Part 1 
Part 2 
Part 3 
Explanation:
Given
Number of protons 
Radius of nucleus 
Distance of the electrons 
Part 1
Electric field produced by just outside its surface
Part 2
Electric field produced by just outside its surface
Part 3
The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other
hence, the solution is
Part 1
Part 2
Part 3
