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3 years ago

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A 12-kg dog jumps up in the air to catch a ball. The dog's center of mass is normally 0.20 m above the ground, and he is 0.50 m

long. The lowest he can get his center of mass is 0.10 m above the ground, and the highest he can get it before he can no longer push against the ground is 0.30m . If the maximum force the dog can exert on the ground in pushing off is 2.1 times the gravitational force Earth exerts on him, how high can he jump?

1 answer:

ad-work [718]3 years ago

5 0

Explanation:

Given Data:

mass of dog = 12 Kg

dog's center of mass = 0.20m

length of dog = 0.50m

height of dog's jump = ?

Solution:

Work done of gravitational force = Gain in Potential energy

2.1 × mgΔh = mg (h - 0.1)

2.1 × (0.3 - 0.1) = (h - 0.1)

h = 0.52 m

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A solid disc with a radius of 5.00 m and a mass of 20.0 kg is initially at rests and lies on the plane of the paper. A smaller s

drek231 [11]

Answer:

Explanation:

This problem is based on conservation of angular momentum.

moment of inertia of larger disc I₁ = 1/2 m r² , m is mass and r is radius of disc . I

I₁ = .5 x 20 x 5²

= 250 kgm²

moment of inertia of smaller disc I₂ = 1/2 m r² , m is mass and r is radius of disc . I

I₂ = .5 x 10 x 2.5²

= 31.25 kgm²

3500 rmp = 3500 / 60 rps

n = 58.33 rps

angular velocity of smaller disc ω₂ = 2πn

= 2π x 58.33

= 366.3124 rad /s

applying conservation of angular momentum

I₂ω₂ = ( I₁ +I₂) ω , ω is the common angular velocity

31.25 x 366.3124 = ( 250 +31.25) ω

ω = 40.7 rad / s .

4 0

2 years ago

What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a w

Tatiana [17]

Answer:

$a. \ f_1=7.9057Hz\\\\b. \ f_2=15.8114Hz\\\\c. \ f_3=23.7171Hz$

Explanation:

a. The wire's length is 10m long and has a mass 100g and a tension of 250N.

Frequency is given by the equation:

$f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu}$ #where t=250N*10=2500N, $\mu=0.1kg$

#substitute for actual values for the lowest frequency.

$F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{1}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=7.9057Hz$ #n=1, lowest frequency

Hence, the lowest frequency for standing waves is 7.9057Hz

b.The wire's length is 10m long and has a mass 100g and a tension of 250N.

Frequency is given by the equation:

$f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu}$ #where t=250N*10=2500N, $\mu=0.1kg$

#The second lowest frequency happens at $n=2$ :

$F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{2}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=15.8114Hz$

Hence, the second lowest frequency is 15.8114Hz

c.Given that the wire's length is 10m long and has a mass 100g and a tension of 250N.

Frequency is given by the equation:

$f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu}$ #where t=250N*10=2500N, $\mu=0.1kg$

The third lowest frequency happens at $n=3$

$F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{3}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=23.7171Hz$

Hence, the third lowest frequency is 23.7171Hz

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2 years ago

A 1.78-m3 rigid tank contains steam at 220°C. One-third of the volume is in the liquid phase and the rest is in the vapor form.

mafiozo [28]

Answer:

a) P = 2319.6[kPa]; b) 2.6%

Explanation:

Since the problem data is not complete, the following information is entered:

A 1.78-m3 rigid tank contains steam at 220°C. One-third of the volume is in the liquid phase and the rest is in the vapor form. Determine (a) the pressure of the steam, and (b) the quality of the saturated mixture.

From the information provided in the problem we can say that you have a mixture of liquid and steam.

a) Using the steam tables we can see (attached image) that the saturation pressure at 220 °C is equal to:

$P_{sat} =2319.6[kPa]$

$v_{f}=0.001190[m^{3}/hr]\\v_{g}=0.08609[m^{3}/hr]\\$

b) Since the specific volume of the gas and liquid is known, we can find the mass of each phase using the following equation:

$m_{f}=\frac{V_{f} }{v_{f} } \\m_{g}=\frac{V_{g} }{v_{g} } \\where:\\V_{f}=volume of the fluid[m^3]\\v_{f}=specific volume of the fluid [m^3/kg]\\$

We know that the volume of the fluid is equal to:

$V_{f}=1/3*V_{total} \\V_{total}=1.78[m^3]\\$

Now we can find the mass of the gas and the liquid.

$m_{f}=\frac{1/3*1.78}{0.001190} \\m_{f}=498.6[kg]\\m_{g}=\frac{2/3*1.78}{0.08609}\\m_{g}=\ 13.78[kg]$

The total mass is the sum of both

$m_{total} =m_{g} + m_{fluid} \\m_{total} = 498.6 + 13.78\\m_{total} = 512.38[kg]$

The quality will be equal to:

$x = \frac{m_{g} }{m_{T} }\\ x= \frac{13.78}{512.38} \\x = 0.026 = 2.6%$

5 0

3 years ago

It takes a bus driver 30 minutes to pick up students from four stops. The last stop is at the corner of Green Street and Route 7

deff fn [24]

Answer:

No, the distance from the last stop to the school and the time it takes to travel that distance are required.

7 0

3 years ago

For your final exam in electronics, you’re asked to build an LC circuit that oscillates at 10 kHz. In addition, the maximum curr

Marina CMI [18]

Answer:

L= 2 mH

$C=1.26\times 10^{-7}\ F$

Explanation:

Given that

Frequency , f= 10 kHz

Maximum current ,I = 0.1 A

Maximum energy stored ,E= 1 x 10⁻⁵ J

The maximum energy stored in the inductor is given as follows

$E=\dfrac{1}{2}LI^2$

Where ,L= Inductance

I=Current

E=Energy

Now by putting the values in the above equation

$10^{-5}=\dfrac{1}{2}\times L\times 0.1^2$

$L=\dfrac{2\times 10^{-5}}{0.1^2}\ H$

L=0.002 H

L= 2 mH

We know that frequency f is given as

$2\pi f=\dfrac{1}{\sqrt{LC}}$

C=Capacitance , f=frequency ,L=Inductance

Now by putting the values

$2\pi \times 10\times 10^3=\dfrac{1}{\sqrt{0.002\times C} }$

$62831.85=\dfrac{1}{\sqrt{0.002\times C}}$

$\sqrt{0.002\times C=\dfrac{1}{62831.85}$

$0.002\times C=0.0000159^2$

$C=\dfrac{0.0000159^2}{0.002}\ F$

$C=1.26\times 10^{-7}\ F$

Therefore the inductance and capacitance will be 2 mH and 1.26 x 10⁻⁷ F respectively.

6 0

2 years ago