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BlackZzzverrR [31]

3 years ago

14

Two tuning forks are played at the same time. One has a frequency of 490 Hz and the other is 488 Hz. How many beats per second a

re heard?

2 answers:

Gnoma [55]3 years ago

7 0

Hope this helps you.

Setler79 [48]3 years ago

7 0

A.490
B.488
C.2
D.4

C is the answer!

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Identifying the factors contributing to and acting as determinant factors of health disparities during the program theory and de

Sav [38]

Answer:

a) True

Explanation:

A program-specific message provided to an individual or group with the intention of raising awareness of a health condition, motivating behavior change, removing perceived barriers to participating in a health habit, or something else relating to the program's aims and objectives. The most effective intervention messages are usually theory-based and culturally adapted.

5 0

2 years ago

If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resol

Mrrafil [7]

Answer:

The maximum wavelength is 492 nm.

Explanation:

Given that,

Angular separation $\theta=3.0\times10^{-5}\ rad$

Suppose a telescope with a small circular aperture of diameter 2.0 cm.

We need to calculate the maximum wavelength

Using formula of angular separation

$\sin\theta=\dfrac{1.22\lambda}{d}$

$\lambda=\dfrac{d\sin\theta}{1.22}$

Put the value into the formula

$\lambda=\dfrac{2.0\times\sin(3\times10^{-5})}{1.22}$

For small angle $\sin\theta\approx\theta$

$\lambda=\dfrac{0.02\times3\times10^{-5}}{1.22}$

$\lambda=4.92\times10^{-7}\ m$

$\lambda=492\ nm$

Hence, The maximum wavelength is 492 nm.

5 0

3 years ago

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

NemiM [27]

Answer:

their final velocity is 0.091 m/s

Explanation:

Given;

mass of the first train, m₁ = 138,000 kg

mass of the second train, m₂ = 123,000 kg

initial velocity of the first train, u₁ = 0.288 m/s

initial velocity of the seocnd train, u₂ = -0.131 m/s (opposite direction to the first)

Let their common final velocity after been coupled = v

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(138,000 x 0.288) + (-0.131 x 123,000) = v(138,000 + 123,000)

39,744 - 16,113 = v(261,000)

23,631 = v(261,000)

v = 23,631 / 261,000

v = 0.091 m/s

Therefore, their final velocity is 0.091 m/s

5 0

3 years ago

Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the

IRISSAK [1]

Answer:

a₁ = 0.63 m/s² (East)

a₂ = -1.18 m/s² (West)

Explanation:

m₁ = 95 Kg

m₂ = 51 Kg

F = 60 N

a₁ = ?

a₂ = ?

To get the acceleration (magnitude and direction) of the man we apply

∑Fx = m*a (⇒)

F = m₁*a₁ ⇒ 60 N = 95 Kg*a₁

⇒ a₁ = (60N / 95Kg) = 0.63 m/s² (⇒) East

To get the acceleration (magnitude and direction) of the woman we apply

∑Fx = m*a (⇒)

F = -m₂*a₂ ⇒ 60 N = -51 Kg*a₂

⇒ a₂ = (60N / 51Kg) = -1.18 m/s² (West)

For every case we apply Newton’s 3 d Law

8 0

3 years ago

Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side

pishuonlain [190]

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

8 0

3 years ago