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frosja888 [35]
2 years ago
10

The school bag of four students A,B,C,D measures 9kg, 2800gm, 2kg and 8000gm respectively. Whose bag is the lightest

Physics
1 answer:
inna [77]2 years ago
5 0

Answer:

Student C

Explanation:

order from heaviest to lightest is...

9 kg (A) , 8000g (8 kg) (D) , 2800g (2.8kg) (B), 2 kg (C)

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D correct answer for que
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3 years ago
A supercapacitor is an electrical energy storage device. A supercapacitor, initially charged to 2.1 thousand millivolts, supplie
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Answer:

the time taken t is 9.25 minutes

Explanation:

Given the data in the question;

The initial charge on the supercapacitor = 2.1 × 10³ mV = 2.1 V

now, every minute, the charge lost is 9.9 %  

so we need to find the time for which the charge drops below 800 mV or 0.8 V

to get the time, we can use the formula for compound interest in basic mathematics;

A = P × ( (1 - r/100 )ⁿ

where A IS 0.8, P is 2.1, r is 9.9

so we substitute

0.8 = 2.1 × ( 1 - 0.099 )ⁿ

0.8/2.1 = 0.901ⁿ

0.901ⁿ = 0.381

n = 9.25 minutes

Therefore, the time taken t is 9.25 minutes

6 0
3 years ago
The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with
nikdorinn [45]

Answer:

A) v = 6.93 m/s

B) v = 4.9 m/s

C) x_m = 0.015m

D) v_max = 5.2 m/s

Explanation:

We are given;

x = 6 cm = 0.06 m

k = 400 N

m = 0.03 kg

F = 6N

A) from work energy law, work dome by the spring on ball which now became a kinetic energy is;

Ws = K.E = ½kx²

Similarly, kinetic energy of ball is;

K.E = ½mv²

So, equating both equations, we have;

½kx² = ½mv²

Making v the subject gives;

v = √(kx²/m)

Plugging in the relevant values to give;

v = √((400 × 0.06²)/0.03)

v = √48

v = 6.93 m/s

B) If there is friction, the total work is;

Ws = ½kx² - - - (1)

Work of the ball is;

Wb = KE + Wf

So, Wb = ½mv² + fx - - - (2)

Combining both equations, we have;

½mv² + fx = ½kx²

Plugging in the relevant values, we have;

(½ × 0.03 × v²) + (6 × 0.06) = ½ × 400 × 0.06²

0.015v² + 0.36 = 0.72

0.015v² = 0.72 - 0.36

v² = 0.36/0.015

v = √24

v = 4.9 m/s

C) The speed is greatest where the acceleration stops i.e. where the net force on the ball is zero. (ie spring force matches 6.0N friction)

So, from F = Kx;

(x is measured into barrel from end where F = 0)

Thus; 6.0 = 400x

x_m = 6/400

x_m = 0.015m from the end after traveling 0.045m

D) Initial force on ball = (Kx - F) =

[(400 x 0.06) - 6.0] = 18N

Final force on ball = 0N

Mean Net force on ball = ½(18 + 0)

Mean met force, F_m = 9N

Net Work Done on ball = KE = 9N x 0.045m = 0.405 J

Thus;

½m(v_max)² = 0.405J

(v_max)² = 2 x 0.405/0.03

(v_max)² = 27

v(max) = √27

v_max = 5.2 m/s

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Answer:

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Explanation:

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What is meant by critical angle of indence?​
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