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3 years ago

BRAINLEST HURRY PLZZ HELP the options for number 1 2 and 3 are A B C D

Physics

2 answers:

SCORPION-xisa [38]3 years ago

6 0

1. B
2.C
3.D
............

coldgirl [10]3 years ago

3 0

1.B
2.D
3.C
read at ur own risk

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3 years ago

Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground = 3.13 m/s

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3 years ago

19. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to

Aleks [24]

Recall this gas law:

$\frac{P₁V₁}{T₁}$ = $\frac{P₂V}{T₂}$

P₁ and P₂ are the initial and final pressures.

V₁ and V₂ are the initial and final volumes.

T₁ and T₂ are the initial and final temperatures.

Given values:

P₁ = 475kPa

V₁ = 4m³, V₂ = 6.5m³

T₁ = 290K, T₂ = 277K

Substitute the terms in the equation with the given values and solve for Pf:

$\frac{475*4}{290} = \frac{P₂*6.5}{277}$

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3 years ago

The spring of a spring balance is 6.0 in. long when there is no weight on the balance, and it is 8.4 in. long with 4.0 lb hung f

Sergeeva-Olga [200]

Answer:

W = 55.12 J

Explanation:

Given,

Natural length = 6 in

Force = 4 lb, stretched length = 8.4 in

We know,

F = k x

k is spring constant

4 = k (8.4-6)

k = 1.67 lb/in

Work done to stretch the spring to 10.1 in.

$W =k\int_{6}^{10.1} x$

$W = \dfrac{k}{2}[x^2]_6^{10.1}$

$W = \dfrac{1}{2}\times 1.67\times (10.1^2-6.0^2)$

W = 55.12 J

Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.

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