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Salsk061 [2.6K]

3 years ago

10

When a ball strikes a surface , it changes its shape temporarily, what type of energy is stored with the ball as it is compresse

d?

1 answer:

adoni [48]3 years ago

8 0

Since ball is elastic i.e it regains its shape, elastic potential energy is stored in the ball.

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What is the total resistance in the circuit? (include unit in answer - ohms)

solmaris [256]

Answer:60 ohms

Explanation:

R1=30 ohms

R2=15 ohms

R3=15 ohms

Let the total resistance be R

R=R1 + R2 + R3

R=30 + 15 +15

R=60

Total resistance is 60 ohms

4 0

3 years ago

A horizontal spring with spring constant 85 N/m extends outward from a wall just above floor level. A 5.5 kg box sliding across

Aloiza [94]

Answer: The box was moving with a velocity of 0.256m/s when it hit the spring

Explanation: Please see the attachments below

6 0

3 years ago

A head-on, elastic collision between two particles with equal initial speed v leaves the more massive particle (mass m1) at rest

ZanzabumX [31]

<span>1/3 The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r" The equation for kinetic energy is E = 1/2MV^2. So the energy for the system prior to collision is 0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5 The energy after the collision is 0.5rv^2 Setting the two equations equal to each other 0.5r + 0.5 = 0.5rv^2 r + 1 = rv^2 (r + 1)/r = v^2 sqrt((r + 1)/r) = v The momentum prior to collision is -1r + 1 Momentum after collision is rv Setting the equations equal to each other rv = -1r + 1 rv +1r = 1 r(v+1) = 1 Now we have 2 equations with 2 unknowns. sqrt((r + 1)/r) = v r(v+1) = 1 Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r. r(sqrt((r + 1)/r)+1) = 1 r*sqrt((r + 1)/r) + r = 1 r*sqrt(1+1/r) + r = 1 r*sqrt(1+1/r) = 1 - r r^2*(1+1/r) = 1 - 2r + r^2 r^2 + r = 1 - 2r + r^2 r = 1 - 2r 3r = 1 r = 1/3 So the less massive particle is 1/3 the mass of the more massive particle.</span>

8 0

3 years ago

Read 2 more answers

The position-time equation for a cheetah chasing an antelope is:

pochemuha

Answer:

x = 1.6 + 1.7 t^2 omitting signs

a) at t = 0 x = 1.6 m

b) V = d x / d t = 3.4 t

at t = 0 V = 0

c) A = d^2 x / d t^2 = 3.4 (at t = 0 A = 3.4 m/s^2)

d) x = 1.6 + 1.7 * (4.4)^2 = 34.5 (position at 4.4 sec = 34.5 m)

4 0

2 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

3 years ago