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2 years ago

PLEASE HELP ASAPPPP! I ONLY HAVE 10 MINUTES!

Physics

1 answer:

alexgriva [62]2 years ago

6 0

Answer:

4 gamma closest thing to this V

Explanation:

Technetium. Tc is a very versatile radioisotope, and is the most commonly used radioisotope tracer in medicine.

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Part A What is the resistance of a 4.4 m length of copper wire 1.3 mm n diameter? The resistivity of copper is 1.68x 10-8 Ω-m Ex

Natalija [7]

Explanation:

It is given that,

Length of the copper wire, l = 4.4 m

Diameter of copper wire, d = 1.3 mm = 0.0013 m

Radius of copper wire, r = 0.00065 m

The resistivity of the copper wire, $\rho=1.68\times 10^{-8}\ \Omega-m$

We need to find the resistance of the copper wire. It is given by :

$R=\rho\dfrac{l}{A}$

$R=1.68\times 10^{-8}\ \times \dfrac{4.4\ m}{\pi (0.00065)^2}$

R =0.055 ohms

So, the resistance of the copper wire is 0.055 ohms. Hence, this is the required solution.

7 0

2 years ago

A boy pushes his baby sister in a stroller that has a mass of 45,000 g. What is the acceleration on the stroller if the force is

dlinn [17]

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Force.

Force = Mass * Acceleration.

here, Mass = 45 kg and Force = 65 N

hence, Acceleration = 65/45

===> Acc. = 1.44 m/s^2

hence the acceleration is 1.44 m/s^2

8 0

2 years ago

Una manguera de agua de 1.3 cm de diametro es utilizada para llenar una cubeta de 24 Litros. Si la cubeta se llena en 48 s. A) ¿

Viefleur [7K]

Answer:

We must translate this:

a 1.3 cm diameter water hose is used to fill a 24-liter bucket. If the bucket is filled in 48 s.

A) What is the speed with which the water leaves the hose?

B) if the diameter of the hose is reduced to 0.63 cm and assuming the same flow, what will be the speed of the water leaving the hose?

A) If the velocity of the water is Xcm/s

and the radius of the hose is equal to half its diameter, so it is 1.3cm/2

Then in one second we can considerate that a cylinder of:

V = pi*(1.3cm/2)^2*X cm^3 of water.

So we have that quantity in one second of flow.

where pi = 3.14

then in 48 seconds, the amount of water in the bucket is:

V = 48*pi*(1.3/2)^2*X = 24 L = 24,000 cm^3

Now we need to solve this for X.

48*3.14*(1.3/2)^2*X = 24000

63.679*x = 24000

x = 24000/63.679 = 376.89

So the velocity of the water is 376 cm per second.

B) if the diameter is 0.64cm, we have the equation:

48*3.14*(0.63/2)^2*x = 24000

14.955*X = 24000

X = 24000/14.955 = 1604.814 cm/s

6 0

3 years ago

Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea

Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

Where,

L = Vibrating length string

T = Tension in the string

$\mu =$ Linear mass density

At the same time we have the expression for the number of beats described as

$n = |f_1-f_2|$

Where

$f_1$ = First frequency

$f_2$ = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

$f \propto \sqrt{T}$

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

$n = f_2-f_1$