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Maksim231197 [3]
3 years ago
7

Please help fast due next period!!!

Physics
1 answer:
OLga [1]3 years ago
3 0

Answer:

The chemical formula for zinc (III) phosphide is Zn3P2

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What is the period of a simple pendulum 47 cm long (a) on the Earth, and ( b) when it is in a freely falling elevator?
Liula [17]

Answer:

a)1.37 s

b)∞ ( Infinite)

Explanation:

Given that

L= 47 cm              ( 1 m =100 cm)

L= 0.47 m

a)

On the earth :

Acceleration due to gravity = g

We know that time period of the simple pendulum given as

T=2\pi\sqrt{ \dfrac{L}{g_{{eff}}}

Here

g_{eff}= g

Now by putting the values

T=2\pi \times\sqrt{ \dfrac{0.47}{9.81}}

T=1.37 s

b)

Free falling elevator :

When elevator is falling freely then

g_{eff}= 0            ( This is case of weightless motion)

Therefore

T=2\pi\sqrt{ \dfrac{L}{0}

T=∞  (Infinite)

6 0
4 years ago
Does air pressure increase or decrease with an increase in altitude?
Yakvenalex [24]

Answer:

increase

Explanation:

think of climbing a mountain. the higher you go the harder it is to breathe. its because air pressure is increasing

5 0
3 years ago
Read 2 more answers
Please help, <br> What does a Transverse and longitude wave combine to form?
Wewaii [24]
Electromagnetic waves<span> transfer energy without going through a medium. ... Sometimes, a </span>transverse wave<span> and a </span>longitudinal wave can combine to form<span>another </span>kind<span> of </span>wave<span> called a surface </span>wave<span>. </span>Transverse Waves<span>. </span>Waves<span> in which the particles vibrate in an up-and-down motion

</span>
4 0
3 years ago
Read 2 more answers
If a string vibrates at the fundamental frequency of 528 Hz and also produces an overtime with a frequency of 1,056 Hz this over
Romashka-Z-Leto [24]
I have a hunch that you're not talking about overtime
OR overtune.  I think you're going for overtone .

For a fundamental frequency of 528 Hz,
1,056 Hz is the second harmonic.
5 0
3 years ago
A straight wire of length 0.53 m carries a conventional current of 0.2 amperes. What is the magnitude of the magnetic field made
olga55 [171]

Explanation:

It is given that,

Length of wire, l = 0.53 m

Current, I = 0.2 A

(1.) Approximate formula:

We need to find the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire, r = 2 cm = 0.02 m

The formula for magnetic field at some distance from the wire is given by :

B=\dfrac{\mu_oI}{2\pi r}

B=\dfrac{4\pi \times 10^{-7}\times 0.2\ A}{2\pi \times 0.02\ m}

B = 0.000002 T

B=10^{-5}\ T

(2) Exact formula:

B=\dfrac{\mu_oI}{2\pi r}\dfrac{l}{\sqrt{l^2+4r^2} }

B=\dfrac{\mu_o\times 0.2\ A}{2\pi \times 0.02\ m}\times \dfrac{0.53\ m}{\sqrt{(0.53\ m)^2+4(0.02\ m)^2} }

B = 0.00000199 T

or

B = 0.000002 T

Hence, this is the required solution.

4 0
3 years ago
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