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GalinKa [24]
3 years ago
8

Identify the statment below that could be an example of properly written long-term goal

Physics
2 answers:
Alex Ar [27]3 years ago
8 0

<u>Option (d)</u>

<u>Explanation:</u>

The statement that could be an example of properly written long-term goal is  Each week I am going to add .25 miles onto my run so i can reach a total of 5 miles by the end of this month.

It is corresponding to long term portrayed by the statement is, every week if the student is targeting for a goal and aims to complete 0.25 miles on his long run, by the end of the 4th week of the month, he would have achieved 5 miles of his long term target.

Inessa05 [86]3 years ago
4 0

''Each week I am going to add .25 miles onto my run so i can reach a total of 5 miles by the end of this month.''

Answer: Option(d)

<u>Explanation:</u>

Long term goal is a set of pre-preparation activities of something that can be done in the future. This goal helps one to achieve a successful career in life.

Effective planning and proper time allotments are the two basic things to be followed while framing a long-term goal.

In Option D, the statement explains the pre-planned activity that can be done in the future, and it explains that a person is adding 0.25 miles of run everyday so that he can reach 5 miles of run at the end of this month.

This indicates long term objective of that person.

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In the figure, if Q = 52 µC q =10 µC and d = 55 cm, what is the magnitude of the electrostatic force on q?​
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Answer:

F = 15.47 N

Explanation:

Given that,

Q = 52 µC

q = 10 µC

d = 55 cm = 0.55 m

We need to find the magnitude of the electrostatic force on q. The formula for the electrostatic force is given by :

F=k\dfrac{q_1q_2}{d^2}\\\\F=9\times 10^9\times \dfrac{52\times 10^{-6}\times 10\times 10^{-6}}{(0.55)^2}\\\\F=15.47\ N

So, the magnitude of the electrostatic force is 15.47 N.

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A 4.2-g bullet is fired at a speed of 370 m/s into a stationary lead block, which moves a distance of 1.2 m before coming to res
Natali5045456 [20]

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(a) F = 239.575 N (b) t = 0.00649s or 6.49 ms

Explanation:

(a) By law of energy conservation, the bullet kinetic energy will be transferred to work done on stopping it from moving.

Formula for Kinetic Energy E_k = \frac{mv^2}{2} where m is bullet mass, v is the velocity

Formula for work W = FS where F is the average force and S is the distance travelled.

E_k = W

\frac{mv^2}{2} = FS

F = \frac{mv^2}{2S}

Substitute m = 4.2 g = 0.0042 kg, v = 370 m/s and S = 1.2 (m)

F = \frac{0.0042*370^2}{2*1.2} = 239.575 N

(b) If the force is constant, since the mass is constant and F = ma according to Newton's 2nd law, the acceleration on bullet is also constant

a = \frac{F}{m} = \frac{239.575}{0.0042} = 57041.67 (m/s^2)

We also have v(t) = v_0 + at

At the time the bullet is coming to rest, v(t) = 0, a = -57041.67 m/s^2

Therefore, 0 = 370 - 57041.67t

t = \frac{370}{57041.67} = 0.00649 s = 6.49 ms

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