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3 years ago

If you attach a 50.0 g mass to the spring whose data are shown in the graph, what will be the period of its oscillations?

Physics

1 answer:

telo118 [61]3 years ago

8 0

I just took the quiz an the answer is .28

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Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven t

Bond [772]

Answer:

$\frac{D_{jet}}{D_{prop}}=2.865$

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

$D=\frac{1}{2}CpAv^2\\$

The ratio between the drag force on the jet to the drag force on prop-driven transport is then given by:

$\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\ \frac{D_{jet}}{D_{prop}}=2.865$

4 0

2 years ago

2.when sugar or another substance is dissolved in water, it disappears from view and forms a homogeneous mixture with the water,

bekas [8.4K]

You cant really tell when it is dissolved

3 0

3 years ago

With what speed must m1 rotate in a circle of radius r if m2 is to remain hanging at rest?

11Alexandr11 [23.1K]

Velocity =2 pie*r/t
distance = 2 (pie) r
accelaretion =distance/t2
f=m*v2/r
v=square root of Fr/m

5 0

3 years ago

In Trial II, the same spring is used as in Trial I. Let us use this information to find the suspended mass in Trial II. Use 0.51

Masteriza [31]

Answer:

$M_2=0.79kg$

Explanation:

From the question we are told that:

Period $T=0.517s$

Trial 1

Spring constant $\mu=117N/m$

Period $T_1=0.37$

Mass $m=0.400kg$

Trial 2

Period $T_2=0.52$

Generally the equation for Spring Constant is mathematically given by

\mu=\frac{4 \pi^2 M}{T^2}

Since

$\mu _1=\mu_2$

Therefore

$\frac{4 \pi^2 M_1}{T_1^2}=\frac{4 \pi^2 M_2}{T_2^2}$

$M_2=M_1*(\frac{T_2}{T_1})^2$

$M_2=0.400*(\frac{0.52}{0.37}})^2$

$M_2=0.79kg$

5 0

3 years ago

Consider an NACA 2412 airfoil with a 2-m chord in an airstream with a velocity of 50 m/s at standard sea level conditions. If th

Akimi4 [234]

Answer:

4°

Explanation:

Given,

chord in the airstream, c = 2 m

speed, v = 50 m/s

lift per unit span = 1353 N/m

density of air, ρ = 1.225 kg/m³

Lift, $L = c_L \times \dfrac{1}{2} \rho v^2(c \times s)$

$\dfrac{L}{s}=c_L \times \dfrac{1}{2} \rho v^2\times c$

$50=c_L \times 0.5\times 1.225\times 50^2\times 2$

$c_L = 0.44$

From the plot of c_L and angle of attack

Angle of attack is equal to 4°

8 0

2 years ago