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SOVA2 [1]
3 years ago
14

What is the speed of a wave that has a frequency of 2,400 Hz and a wavelength of 0.75

Physics
1 answer:
likoan [24]3 years ago
6 0

Answer:

1800 m/s

Explanation:

The equation is v = fλ

λ= 0.75

f = 2400 Hz

V = 2400 × 0.75

V = 1800 m/s

[ you did not give units for wavelength, I assumed it would be m/s]

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A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875
mario62 [17]

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

So v= 1.25mm x 5498.5

= 6.82m/s

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5 0
3 years ago
Read 2 more answers
3. How is using a model to study cells helpful?
4vir4ik [10]

Answer:

Yes, it is very helpful.

Explanation:

It's helpful since in a cell, plant or animal, there are a lot of different things. It's hard to memorize everything and know what they look like. Using a model can help you memorize everything better and even understand it better. If someone asked me where or what something was in a cell I think I would be able to recognize it better.

I hope this helps!

6 0
3 years ago
the total positive charge is QQQ = 1.62×10−6 CC , what is the magnitude of the electric field caused by this charge at point P,
balu736 [363]

Answer:

6.1 × 10^9 Nm-1

Explanation:

The electric field is given by

E= Kq/d^2

Where;

K= Coulombs constant = 9.0 × 10^9 C

q = magnitude of charge = 1.62×10−6 C

d = distance of separation = 1.53 mm = 1.55 × 10^-3 m

E= 9.0 × 10^9 × 1.62×10−6/(1.55 × 10^-3 )^2

E= 14.58 × 10^3/2.4 × 10^-6

E= 6.1 × 10^9 Nm-1

8 0
3 years ago
Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from
qwelly [4]
Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N

4 0
3 years ago
Colonel John P. Stapp, USAF, participated in studying whether a jet pilot could survive emergency ejection. On March 19, 1954, h
borishaifa [10]

we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi ​ =632mi/h=632mi/h( 1mi 1609m ​ )( 3600s 1h ​ )=282m/s (a) taking v xf ​ =v xi ​ +a x ​ t with v xf ​ =0 a x ​ = t v xf ​ −v xf ​ ​ = 1.40s 0−282m/s ​ =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f ​ −x i ​ = 2 1 ​ (v xi ​ +v xf ​ )t= 2 1 ​ (282m/s+0)(1.40s)=198m

7 0
3 years ago
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