A wave has a wavelength of 0.98 m and travels 2.0x10^2 m in 0.78 s. What is the frequency of the wave?

sergiy2304 [10]

There's a nasty wrinkle here that's kind of sneaky, and makes the work harder than it should be.

Look at the first question. There's a number there that's dropped in so quietly that you're almost sure to miss it, but it changes the whole landscape of both of these problems. That's where it says

" ... 20 cm mark (30 cm from the fulcrum) ... " .

That tells us that the yellow bar resting on the pivot is actually a meter stick, but the pictures don't show the centimeter marks on the stick. The left end of the stick is "0 cm", the right end of the stick is "100 cm", and the pivot is under the "50 cm" mark.

When the question talks about hanging a weight, it tells the centimeter mark on the stick where the weight is tied. To solve the problem, we have to first figure out how far that is from the pivot, then calculate how far from the pivot to put the weight on the other side, and finally what centimeter mark that is on the stick.

How to solve the problems:

-- The "moment" of a weight is (the weight) x (its distance from the pivot) .

-- To balance the stick, (the sum of the moments on one side) = (the sum of the moments on the other side).

= = = = = = = = = =

#1). Only one moment on the left side.

(160 gm) x (30 cm from pivot) = 4,800 gm-cm

To balance, we need 4,800 gm-cm of moment on the right side.

(500 gm) x (distance from pivot) = 4,800 gm-cm

Distance from pivot = (4,800 gm-cm) / (500 gm) = 9.6 cm

The 500 gm has to hang 9.6 cm to the right of the pivot. But that's not the answer to the problem. They want to know what mark on the stick to hang it from. The pivot is at the 50cm mark. The 500gm has to hang 9.6 cm to the right of the pivot. That's the 59.6 cm mark on the stick.

= = = = =

#2). There are 2 weights hanging from the left side. We have to find the moment of each weight, add them up, then create the same amount of moment on the right side.

one weight: 120gm, hanging from the 25cm mark.

That's 25cm from the pivot. Moment = (120gm) (25cm) = 3,000 gm-cm

the other weight: 20gm, hanging from the 10cm mark;

That's 40cm from the pivot. Moment = (20gm) (40cm) = 800 gm-cm

Add up the moments on the left side:

(3,000 gm-cm) + (800 gm-cm) = 3,800 gm-cm.

To balance, we need 3,800 gm-cm of moment on the right side.

(500 gm) x (its distance from the pivot) = 3,800 gm-cm

Distance from the pivot = (3,800 gm-cm) / (500 gm) = 7.6 cm

The pivot is at the 50cm mark on the stick. You have to hang the 500gm from 7.6cm to the right of that. The mark at that spot on the stick is (50cm + 7.6cm) = 57.6 cm .

3 0

3 years ago

Read 2 more answers

A car with mass 1500 kg moves with constant velocity of 36 m/s. The driver sees a group of cows in front and he immediately step

Crazy boy [7]

From laws of motion:

$S = ( \frac{v + u}{2} ) \times t$
Where S is the distance/displacement (as you would call it) which is unknown
v = final velocity which is 0m/s (this is because the car stops)
u = initial velocity which is 36m/s (from the data given)
t = time taken for the distance to be covered and it is 6s

Substitute the values, hence:
$S = ( \frac{0 + 36}{2} ) \times 6$
$S = (18) \times 6 \\ \\ S = 108m$

But this is merely the distance he travelled in the 6 seconds he was trying to stop the car.

Therefore, the distance between the car and the cows = 160-108
Distance = 52m

6 0

3 years ago

At what distance from the centre of the earth its acceleration due to gravity becomes one third only? [Mass and radius of earth

Kipish [7]

Answer:

235

Explanation:

3 0

3 years ago

Read 2 more answers

A container with volume 1.64 L is initially evacuated. Then it is filled with 0.226 g of N2N

vaieri [72.5K]

Answer:

0.015 atm

Explanation:

The pressure of the gas can be calculated using Ideal Gas Law:

$p = \frac{nRT}{V}$

Where:

n: is the number of moles of the gas

R: is the gas constant = 0.082 L*atm/(K*mol)

V: is the volume of the container = 1.64 L

T: is the temperature

We need to find the number of moles and the temperature. The number of moles is:

$n = \frac{m}{M}$

Where:

M: is the molar mass of the N₂ = 14.007 g/mol*2 = 28.014 g/mol

m: is the mass of the gas = 0.226 g

$n = \frac{0.226 g}{28.014 g/mol} = 8.07 \cdot 10^{-3} moles$

Now, the temperature can be found using the following equation:

$v_{rms} = \sqrt{\frac{3RT}{M}}$

Where:

R: is the gas constant = 0.082 L*atm/K*mol = 8.314 J/K*mol

$v_{rms}$ : is the root-mean-square speed of the gas = 182 m/s

By solving the above equation for T, we have:

$T = \frac{v_{rms}^{2}*M}{3R} = \frac{(182 m/s)^{2}*28.014 \cdot 10^{-3} Kg/mol}{3*8.314 J K^{-1}mol^{-1}} = 37.20 K$

Finally, we can find the pressure of the gas:

$p = \frac{nRT}{V} = \frac{8.07 \cdot 10^{-3} mol*0.082 L*atm* K^{-1}*mol^{-1}*37.20 K}{1.64 L} = 0.015 atm$

Therefore, the pressure of the gas is 0.015 atm.

I hope it helps you!

8 0

3 years ago

On isolated ground receptacles, the metal yoke ____ integrally bonded to the equipment grounding terminal of the receptacle.

Ede4ka [16]

On isolated ground receptacles, the metal yoke is not allowed to be integrally bonded to the equipment grounding terminal of the receptacle.

Any device with two distinct switches or receptacles is a duplex device. It can be shaped to fit a Decora opening or a typical duplex plate opening. It should be noted that they can be combination devices with a switch/outlet, switch/pilot light, etc.

Because of grounding connection removal and receptacle, it is utterly undesirable to connect the two bare equipment grounding conductors together directly.

The equipment grounding conductor associated with those circuits must be connected to the box when circuit conductors are terminated on equipment inside a metal box to prevent unneeded current discharge.

Learn more about grounding conductors here brainly.com/question/14886979

#SPJ4.

4 0

2 years ago