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2 years ago

Can an object still float when it's support beneath it has sunk?

2 answers:

5 0

It’s depends on the density of the object

if the density of the object is light than the fluid in which it’s floating than the object will float and if it’s density is more than it will sink

Alex777 [14]2 years ago

4 0

Answer:

yes it can

Explanation:

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Why do scientists not use US customary units when reporting their data?

Marianna [84]

From britaññica it said time and money. They didn’t have either to switch over from the industrial period and never did. Also from my own person reasoning i think most of the world uses not US customary, so to make stuff more accessible. hope this helps!

8 0

3 years ago

Which statement accurately describes quantum mechanics and nanotechnology?

IgorLugansk [536]

Answer:

Explanation:

edge

6 0

3 years ago

For a given Prandtl-Meyer expansion, the upstream Mach number is 3 and the pressure ratio across the wave is P2/P1 = 0.4. Calcul

loris [4]

Answer:

The angle for the forward Mach line is 19.47°

The angle for the rearward Mach line is 5.21°

Explanation:

From table A-1 (Modern Compressible Flow: with historical perspective):

$\frac{P_{o} }{P_{1} } =36.73$ (M₁ = 3)

If Po₁ = Po₂

$\frac{P_{o2} }{P_{2} } =\frac{P_{o1} }{P_{1} } *\frac{P_{1} }{P_{2} } =36.73*\frac{1}{4} =91.825$

Table A-1:

$\frac{P_{o2} }{P_{2} } =91.825,M_{2} =3.63$

Table A-5:

v₁ = 49.76°

μ₁ = 19.47°

v₂ = 60.55°

μ₂ = 16°

θ = 60.55 - 49.76 = 10.79°

The angle for the forward Mach line is:

μ₁ = 19.47°

The angle for the rearward Mach line is:

θr = μ₂ - θ = 16 - 10.79 = 5.21°

3 0

3 years ago

Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da

valentinak56 [21]

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

k = 3906250 N/m

Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

3 0

2 years ago

A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom o

Jet001 [13]

Answer:

C) 3,000 kg m/s

Explanation:

We can consider the horizontal velocity of the motorcycle to be zero, since it rolls off the edge of the cliff very slowly. So, we only need to find the vertical velocity at the time of the impact with the ground.

The vertical velocity of the motorcycle at time t is given by (free-fall motion):

$v(t)=v_0 -gt$