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3 years ago

15

Suppose you lift a 20 kg box by a height of 1.0 m.

2 answers:

zhannawk [14.2K]3 years ago

8 0

Answer: The work done in lifting something is W=mgh where m is the mass of the objet, g is the gravity acceleration and h is the height, this work is done because you are "fighting" against the force of gravity.

so in this case we have W = 20kg*9.8m/s*1m= 196N

9966 [12]3 years ago

6 0

W=mgh W=(20)(9.8)(1) w=196J

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A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit

likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density $\mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m$

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 120=753.6rad/sec$

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by $P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt$

From the relation we can see that power $P\ \propto\ A^2$

So if amplitude is halved then power will be $\frac{1}{4}$ times

So power will be equal to $\frac{0.2023}{2}=0.0505watt$

4 0

3 years ago

An object of irregular shape has a characteristic length of L = 0.5 m and is maintained at a uniform surface temperature of Ts =

goblinko [34]

Answer:

The value of the average convection coefficient is 20 W/Km².

Explanation:

Given that,

For first object,

Characteristic length = 0.5 m

Surface temperature = 400 K

Atmospheric temperature = 300 K

Velocity = 25 m/s

Air velocity = 5 m/s

Characteristic length of second object = 2.5 m

We have same shape and density of both objects so the reynold number will be same,

We need to calculate the value of the average convection coefficient

Using formula of reynold number for both objects

$R_{1}=R_{2}$

$\dfrac{u_{1}L_{1}}{\eta_{1}}=\dfrac{u_{2}L_{2}}{\eta_{2}}$

$\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}$

Here, $k_{1}=k_{2}$

$h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}$

$h_{2}=\dfrac{q}{T_{2}-T_{1}}\times\dfrac{L_{1}}{L_{2}}$

Put the value into the formula

$h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}$

$h_{2}=20\ W/Km^2$

Hence, The value of the average convection coefficient is 20 W/Km².

7 0

3 years ago

if thanos lost in End Game" what would happend to the gountlet and who would take over the world with it?????

AleksandrR [38]

Have you seen Endgame?

3 0

3 years ago

Read 2 more answers

In establishing standing waves with a Slinky®, your wrist oscillates with f frequency of 4.0 Hz, and the length between any two

vladimir1956 [14]

Your answer would be <span>20.</span>

4 0

3 years ago

A series of optical telescopes produced an image that has a resolution of about 0.00350 arc second.

Mila [183]

Answer:

The smallest diameter is $D =122 \ m$

Explanation:

From the question we are told that

The resolution of the telescope is $\theta = 0.00350 \ arc \ second$

The wavelength is $\lambda = 1.70 \mu m = 1.70 *10^{-6} \ m$

From the question we are told that

$1 arc \ sec = \frac{1}{3600^o}$

So $0.00350 \ arc \ second = x$

Therefore

$x = 0.00350 * \frac{1}{3600 }$

$x = ( 9.722*10^{-7} )^o$

Now $1^o = \frac{\pi}{180}$

So $(9.722*10^{-7})^o = \theta$

=> $\theta = (9.722*10^{-7}) * \frac{\pi}{180}$

$\theta = 1.69*10^{-8} rad$

The smallest diameter is mathematically represented as

$D = \frac{1.22 \lambda }{\theta }$

substituting values

$D = \frac{1.22 * 1.7 *10^{-6}} {1.69 *10^{-8} }$

$D =122 \ m$

6 0

3 years ago