Answer:
A. 52 min
.A. 47 watts
Explanation:
Given that;
jim weighs 75 kg
and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.
Using the following relation to determine the amount of calories burned per minute while walking; we have:

here;
MET = energy cost of a physical activity for a period of time
Obtaining the data for walking with a speed of 3.3 mph From the standard chart for MET, At 3.3 mph; we have our desired value to be 4.3
However;
the calories burned in a minute = 
= 5.644
Therefore, for walking for 52 mins; Jim burns approximately 293.475 kcal which is nearest to 300 kcal.
4.
Given that:
mass m = 75 kg
intensity = 6 kcal/min
The eg ergometer work rate = ??
Applying the formula:

where ;



∴
Converting to watts;
Since; 6.118kg-m/min is = 1 watt
Then 291.66 kgm /min will be equal to 47.67 watts
≅ 47 watts
Answer:
a2 = 2.5 m/s2
Explanation:
F1 = m1 a1 We use the same force so F1 = F2
= 5kg × 15m/s2 F2 = m2 a2
= 75N a2 is required
a2 = F2 / m2
= 75N / 30 kg
= 2.5 m/s2
Answer with Explanation:
We are given that
Resistance,R=130 ohm
Potential difference, V=30 V
Capacitor,C=2.1pF=



a.Maximum flux



b.Maximum displacement current,
c.We have to find electric flux at t=0.5 ns






d.Displacement current at t=0.5ns


<h3>
Answer: The acceleration doubles</h3>
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Explanation:
Consider a mass of 10 kg, so m = 10
Let's say we apply a net force of 20 newtons, so F = 20
The acceleration 'a' is...
F = ma
20 = 10a
20/10 = a
2 = a
a = 2
The acceleration is 2 m/s^2. Every second, the velocity increases by 10 m/s.
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Now let's double the net force on the object
F = 20 goes to F = 40
m = 10 stays the same
F = ma
40 = 10a
10a = 40
a = 40/10
a = 4
The acceleration has also doubled since earlier it was a = 2, but now it's a = 4.
---------------
In summary, if you double the net force applied to the object, then the acceleration doubles as well.