In a free body diagram for an object projected upwards;
- the acceleration due to gravity on the object is always directed downwards.
- the velocity of the object is always in the direction of the object's motion.
An object projected upwards is subjected to influence of acceleration due to gravity.
As the object accelerates upwards, its velocity decreases until the object reaches maximum height where its velocity becomes zero and as the object descends its velocity increases, which eventually becomes maximum before the object hits the ground.
To construct a free body diagram for this motion, we consider the following;
- the acceleration due to gravity on the object is always directed downwards
- the velocity of the object is always in the direction of the object's motion.
<u>For instance:</u>
upward motion for velocity ↑ downward motion for velocity ↓
↑ ↓
↑ ↓
acceleration due to gravity ↓
↓
↓
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Answer:
The system's potential energy is -147 J.
Explanation:
Given that,
Energy = 147 J
We know that,
System is isolated and it is free from external forces.
So, the work done by the external forces on the system should be equal to zero.
We need to calculate the system's potential energy
Using thermodynamics first equation
Put the value into the formula
Hence, The system's potential energy is -147 J.
Answer:
Explanation:
Flux is given by
A = Area
E = Electric field = 76.7 N/C
Angle is given by
The flux through the sheet is
Answer:
20 seconds.
Explanation:
The following data were obtained from the question:
Distance = 10 m
Speed = 0.5 m/s
Time =...?
The speed of an object is simply defined as the distance travelled by the object per unit time. Mathematically, it is expressed as:
Speed = Distance /time
With the above formula, we can obtain the time taken for the ball to travel a distance of 10 m as shown below:
Distance = 10 m
Speed = 0.5 m/s
Time =...?
Speed = Distance /time
0.5 = 10/time
Cross multiply
0.5 × time = 10
Divide both side by 0.5
Time = 10/0.5
Time = 20 secs.
Therefore, it will take 20 seconds for the ball to travel a distance of 10 m.
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<span>Cl^- 1s^2 2s^2p^6 3s^2 3p^6 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 17- 10 =7 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6; 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 19- 10 = 9
</span>
S = 2 + 6.8 + 2.45 = 11.25
<span>Zeff(Cl^-) = 17 – 11.25 = 5.75 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6 same S as for Cl^- but Z increases by 2 hence </span>
<span>Zeff(K^+) = 19 - 11.25 = 7.75</span>