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Answered by vanishasaxena09
1

Answer:

A rain drop of radius= 2mm

falls from a height of = 500m

(Given)

2×10-3 m

Density of water , p= 10 3 kgm -3

The mass contained in the rain drop, m= pv

➡ 3 4 × 3.141× (2×10-3)

3×10 3 kg

Gravitational force experienced by the rain drop,= F mg ➡ 3 4 × 3.141 ×(2× 10-3)

3×10 3×9.8 N

The work done by the gravity on the drop is given by: W 1=Fs= 3 4×3.14×(2×10 3)

3×10 3× 9.8×250=0.082 J

The work done on the drop in the second half of the ✈journey will be same as that in the first half

.

. . W 2

= 0.082 J

The total energy of the drop remains conserved during it's motion.

The total energy at the top E T=Mgh+0

=4/3×3.14×(10-3)

3×10 3× 500×10-5

=0.164 J

Due to the presence of a resistive force the drop hits hits the ground with a velocity of m/s.

Total energy in the ground: E G= 2 1 mv 2+01/4×3×3.141 × (2×10-3)

3×10 3 × 9.8×10 2

= 1.675×10-3 J

So., the work done by the resistive force=E G -E T

=0.162 J

Explanation:

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Answered by marinettedupai57
0

Answer: A rain drop of radius= 2mm

falls from a height of = 500m

(Given)

2×10-3 m

Density of water , p= 10 3 kgm -3

The mass contained in the rain drop, m= pv

➡ 3 4 × 3.141× (2×10-3)

3×10 3 kg

Gravitational force experienced by the rain drop,= F mg ➡ 3 4 × 3.141 ×(2× 10-3)

3×10 3×9.8 N

The work done by the gravity on the drop is given by: W 1=Fs= 3 4×3.14×(2×10 3)

3×10 3× 9.8×250=0.082 J

The work done on the drop in the second half of the ✈journey will be same as that in the first half

.

. . W 2

= 0.082 J

The total energy of the drop remains conserved during it's motion.

The total energy at the top E T=Mgh+0

=4/3×3.14×(10-3)

3×10 3× 500×10-5

=0.164 J

Due to the presence of a resistive force the drop hits hits the ground with a velocity of m/s.

Total energy in the ground: E G= 2 1 mv 2+01/4×3×3.141 × (2×10-3)

3×10 3 × 9.8×10 2

= 1.675×10-3 J

So., the work done by the resistive force=E G -E T

=0.162 J

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