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3 years ago

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1 answer:

blagie [28]3 years ago

3 0

Answer:

A

Explanation:

hope its right

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______________ are low energy waves found in the electromagnetic spectrum and ______________ are high energy waves found in the

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Radios are and infrareds are high e

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4 years ago

Read 2 more answers

A 25.0 kg bag of peat moss sits in the back of a flatbed truck, driving up a hill. The bag experiences a 225N normal force. The

malfutka [58]

Answer:

a

$\theta = 23.32^o$

b

$\mu_s = 0.27$

c

$s = 0.948 \ m$

Explanation:

From the question we are told that

The mass of the bag is $m_b = 25.0 \ kg$

The normal force experienced is $F_n = 225 \ N$

The maximum acceleration of the bag is $a = 2.40 \ m/s^2$

Generally this normal force experience by the bag is mathematically represented as

$F_n = mg cos \theta$

=> $225 = (25 * 9.8) cos \theta$

=> $0.9183 = cos \theta$

=> $\theta = cos^{-1}[0.9183]$

=> $\theta = 23.32^o$

Generally for the bag not to slip , it means that the frictional force is equal to the sliding force

$F_f = F_s$

Hence $F_f$ is mathematically represented as

$F_f = \mu_s * F_n$

While $F_s$ is mathematically represented as

$F_s = m * a$

So

$\mu_s * F_n = m * a$

=> $\mu_s * 225 = 25 * 2.40$

=> $\mu_s = 0.27$

Generally from the workdone equation we have that

$KE_f - KE_i = W_f$

Here $W_f$ is the work done by friction which is mathematically represented as

$W_f = m * g * \mu_k * s$

Here s is the distance covered by the bag

$KE_f$ is zero given that velocity at rest is zero

and

$KE_i = \frac{1}{2} * m* v_i^2$

so

$\frac{1}{2} * m* v_i^2 = m * g * \mu_k * s$

=> $\frac{1}{2} * v_i^2 = g * \mu_k * s$

substituting 2.55 m/s for v_i and 0.350 for \mu_k we have that

$\frac{1}{2} * 2.55^2 = 9.8 * 0.350 * s$

=> $s = 0.948 \ m$

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4 years ago

A water wave has a wavelength of 204 m and a frequency of 0.5 Hz. How far does it travel in 1 s?

Zigmanuir [339]

Answer:

c = 204 x 5 = 1020 m/s so it travels 1020 meters in 1 second.

Explanation:

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3 years ago

Electronic flash unit cameras contain a capacitor for storing the energy used to produce the flash. In one such unit the flash l

Stolb23 [73]

Answer:

(A) 421 J energy stored in the capacitor for one flash.

(B) The value of capacitance is 0.0537 F

Explanation:

Given :

(A)

Time $t = \frac{1}{675}$

Average power $P = 2.7 \times 10^{5}$ W

From power equation,

$P= \frac{E}{t}$

So energy in one light is given by,

$E = Pt$

$E = 2.7 \times 10^{5} \times \frac{1}{675} = 400$ J

Since efficiency is 95 % so we can write, energy stored in one flash,

$E_{tot} = \frac{400}{0.95} = 421$ J

(B)

From the formula of energy stored in capacitor,

$E = \frac{1}{2}C V^{2}$

Where $E = E_{tot}$ and $V = 125$ V

$C = \frac{2E}{V^{2} }$

$C = \frac{2 \times 421}{15625}$

$C = 0.0537 F$

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3 years ago

The emission spectrum of iodine is shown below.<br> Which is the absorption spectrum?

garik1379 [7]

Sorry Sorry Sorry Sorry Sorry Sorry Sorry Sorry

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3 years ago