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3 years ago

Please help!!

Physics

2 answers:

Luba_88 [7]3 years ago

8 0

Answer:

Bath CD jshchdhdhfhfhhfhd jpg de f for frr for gi Jhong GO by be jr jpg be

erastova [34]3 years ago

7 0

34 bc of the velocity of the arrow will come down with less dense

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¿Cuál es el parámetro que indica la cantidad de energía liberada en un movimiento sísmico?

artcher [175]

Answer:

Explanation:

en un movimiento sísmico?

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2 years ago

According to the picture, how would the gravitational potential energy of the car change if its weight was doubled? A) The gravi

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Samuriguy125 on YT

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3 years ago

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A mosquito can fly with a speed of 1.10 m/s with respect to the air. Suppose a mosquito flies east at this speed across a swamp.

sineoko [7]

Answer:

The velocity of Mosquito with respect to earth will be 0.302m/s

Explanation:

V(ma) = 1.10 m/s, east Velocity of mosquito with respect to air

V(ae) = 1.4 m/s at 35° Velocity of air with respect to Earth in west of south direction.

Velocity of Mosquito with respect to earth will be

V(me) = V(ma) + V(ae)

We need to find the mosquito’s speed with respect to Earth in the x direction.

V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )

Angle (ae) = –90.0° − 35°=−125°

V(x, me) = 1.10 + (1.4)Cos(-125)

= 1.10 + 1.4(-0.57)

= 1.10 -0.798

= 0.302

So the velocity of Mosquito with respect to earth will be 0.302m/s

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3 years ago

(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i

ddd [48]

Answer:

a) $P = 370.993\,kPa$ , b) $F = 25.948\,kN$

Explanation:

a) The absolute pressure at a depth of 27.5 meters is:

$P = P_{atm} + P_{man}$

$P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)$

$P = 370.993\,kPa$

b) The force exerted by the water is:

$F = (P - P_{atm})\cdot A$

$F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}$

$F = 25.948\,kN$

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3 years ago

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A torsion-bar spring consists of a prismatic bar, usually of round cross section, that is twisted at one end and held fast at th

ra1l [238]

Answer:

d₁ = 0.29 in

d₂ = 0.505 in

Explanation:

Given:

T = 1500 lbf in

L = 10 in

x = 0.5 L = 5 in

$T_{1} =\frac{T(L-x)}{L} =\frac{1500*(10-5)}{10} =750lbfin$

First case: T = T₁ + T₂

T₂ = T - T₁ = 1500 - 750 = 750 lbf in

If the shafts are in series:

θ = θ₁ + θ₂

θ = ((T₁ * L₁)/GJ) + ((T₂ * L₂)/GJ)

Second case: If d₁ ≠ d₂

θ = ((T₁ * L₁)/GJ₁) + ((T₂ * L₂)/GJ₂) = 0 (eq. 1)

t₁ = t₂

$\frac{16T_{1} }{\pi d_{1}^{3} } =\frac{16T_{2} }{\pi d_{2}^{3} }$ (eq. 2)

T₁ + T₂ = 1500 (eq. 3)

θ₁ first case = θ₁ second case

Replacing:

$\frac{750*5}{G(\frac{\pi }{32})*0.5^{4} } =\frac{T_{1}*3.7 }{G(\frac{\pi }{32})*d_{1} ^{4} }\\T_{1} =16216d_{1} ^{4}$

The same way to θ₂:

$\frac{750*5}{G(\frac{\pi }{32})*0.5^{4} } =\frac{T_{2}*6.3 }{G(\frac{\pi }{32})*d_{2} ^{4} } \\T_{2} =9523.8d_{2} ^{4}$

From equation 2, we have:

d₁ = 0.587 * d₂

From equation 3, we have:

d₂ = 0.505 in

d₁ = 0.29 in

7 0

3 years ago