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Goryan [66]
4 years ago
15

Please answer these!

Physics
1 answer:
solong [7]4 years ago
8 0

6. f = 128.62 s^{-1} , T= 0.0077775 s

7. f  = 2.2 * 10^{4} s^{-1}, T = 4.545 * 10^{-5} s

8. 32.64 s^{-1}

9. 3.29 * 10^{14} s^{-1}

Explanation:

Step 1:

6.

For light and sound v = fλ

where v represents the velocity

f represents the frequency

λ represents the wavelength

λ = 2.69 m

v = 346 m/s

f = v/λ = 346/2.69 = 128.62 s^{-1}

Time period is the reciprocal of frequency

T = 1/128.62 = 0.0077775 s

Step 2:

7.

λ = 110 cm = 1.1 m

v = 2.42*10^{4} m/s

f = 2.42*10^{4}/1.1 = 2.2 * 10^{4} s^{-1}

T = 1/(2.2*10^{4}) = 4.545 * 10^{-5} s

Step 3:

8.

λ = 10.6 m

v = 346 m/s

f = v/λ = 346/10.6 = 32.64 s^{-1}

Step 4:

9.

λ = 5.89 * 10^{-7} m

v = 1.94 * 10^{8} m/s

f = v/λ = 3.29 * 10^{14} s^{-1}

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Answer:

- Waves with higher amplitude transfer HIGHER energy.

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3 0
3 years ago
Cuestionario:
Sati [7]

Explanation:

1.El objetivo principal de realizar la llamada entrada en calor o calentamiento, es preparar el cuerpo para la actividad física o deportiva. Numerosas lesiones y ciertos problemas cardíacos como algunas arritmias, están asociados a la ejercitación violenta sin mediar un adecuado calentamiento.

3 0
3 years ago
Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of
Alchen [17]

Answer:

\theta=10.20^{\circ}  

\Delta l=0.10 ft    

Explanation:

First of all, we analyze the system of blocks before starting to move.

\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0  

\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0

11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0

11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0

11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0  

16sin(\theta)-2.91cos(\theta)=0  

tan(\theta)=0.18  

\theta=arctan(0.18)  

\theta=10.20^{\circ}  

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.  

P_{A}sin(\theta)-F_{fA}-F_{spring}=0

Where:

F_{spring} = k\Delta l=2.1\Delta l

P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0

\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}

\Delta l=0.10 ft    

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

4 0
4 years ago
I NEED HELP ASAP!! I NEED IT BEFORE TOMORROW!! D: 10pts+5pts!
Wewaii [24]
9. Compounds can form from two nonmetals by sharing their electrons in a
C) covalent

11. An atom that has an excess positive or negative electrical charge caused by the loss or addition of an electron is called a(n) ______.
B) ion

5 is either A or C
4 0
3 years ago
If a scuba diver fills his lungs to full capacity of 5.5 L when 10 m below the surface, to what volume would his lungs expand if
umka21 [38]

Answer:

The volume at the surface is 10.97 L.

Explanation:

Given that,

Volume = 5.5 L

Height = 10 m

Density of sea water= 1025 kg/m³

We need to calculate the pressure at that point

Using formula of pressure

P'=P+\rho gh

Put the value into the formula

P'=1.01\times10^{5}+1025\times9.8\times10

P'=201450\ Pa

We need to calculate the volume at the surface

Using equation of ideal gas

PV= RT

So, for both condition

PV=P'V'

Put the value into the formula

V=\dfrac{201450\times5.5}{1.01\times10^{5}}

V=10.97\ L

Hence, The volume at the surface is 10.97 L.

3 0
3 years ago
Read 2 more answers
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