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3 years ago

The height of a transverse waves is known as its?

Physics

1 answer:

Greeley [361]3 years ago

7 0

Answer:

amplitude

Explanation:

amplitude of a wave is the maximum displacement of a point from its rest position.

hope you get it,

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In what direction is the weight vector always drawn?

Varvara68 [4.7K]

Yo sup??

The weight vector is usually drawn vertically downwards from the centre of the body.

It can be respectively resolved as well.

Hope this helps

3 0

3 years ago

What is the derivative of x^2 - x + 3 at the point x = 5?

Hunter-Best [27]

Answer:

Explanation:

d/dx (x² - x + 3)

= 2x - 1

when x = 5,

2x-1

= 2(5) - 1

= 10 - 1

= 9

5 0

3 years ago

Phoebe's insulated foam cup is filled with 0.15 kg of coffee (mostly water) that is too hot to drink, so she adds

Nesterboy [21]

Answer:

65 Celsius.

Explanation:

8 0

3 years ago

How much energy is used when the incandescent bulb is left on for 12 hours

Reil [10]

You haven't told us the "wattage" rating of the bulb. We'll just have to call it ' W ' .

The bulb uses energy at the rate of W watts, or 0.001W kilowatts.

In 12 hours, it uses <em>0.012W kilowatt-hours </em>of energy.

= = = = =

W watts = W Joules/second

1 hour = 3600 seconds

12 hours = (12 x 3600) seconds

Energy = (W Joule/sec) x (12 x 3600 sec)

<em>Energy = 43,200W Joules</em>

3 0

3 years ago

A marble runs off the edge of a table that is 1.5 m high and the marble lands 0.50 m from the base of the table. a. How much tim

Free_Kalibri [48]

Answer:

t = 0.55[sg]; v = 0.9[m/s]

Explanation:

In order to solve this problem we must establish the initial conditions with which we can work.

y = initial elevation = - 1.5 [m]

x = landing distance = 0.5 [m]

We set "y" with a negative value, as this height is below the table level.

in the following equation (vy)o is equal to zero because there is no velocity in the y component.

therefore:

$y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\ where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]$

Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.

$(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]$

3 0

3 years ago