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Vika [28.1K]
3 years ago
14

Identical forces act for the same length of time on two different masses. The change in momentum of the smaller mass is

Physics
1 answer:
xenn [34]3 years ago
7 0

Answer:

Equal to change in momentum of larger mass.

Explanation:

We are given that

Two difference  masses .

Force act on both masses  for the same length of time.

We have to find the change in momentum of the smaller mass.

Let M and m are two masses

M>m

We know that

Change in momentum for large mass=F\Delta t

Change in momentum for small mass=F\Delta t

Because Force and length of time are same for both masses .

Hence, the change in momentum of smaller mass is equal to change in momentum of larger mass.

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A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
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Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

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8 0
3 years ago
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Hey!
pav-90 [236]

Answer:

Hydraulic pressure exerted on glass slab, ρ=10 atm

Bulk modulus of glass, B=37×10^9 Nm^−2

Bulk modulus, B=P/(ΔV/V)

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ΔV/V= Fractional change in volume

ΔV/V=P/B

=10×1.013×10^5 /(37×10 ^9)

=2.73×10^-5

Therefore, the fractional change in the volume of the glass slab is 2.73×10^-5

Hope it helps

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Explanation:

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