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2 years ago

1. Sam walked 4 meters north, 5 meters east, and 4 meters south. What is his

Physics

1 answer:

tatyana61 [14]2 years ago

4 0

Answer:

His distance would be 80 or 13 something. I dont really understand this question. Just be sure to look into other answers pls.

Explanation:

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If the intensity of a loud car horn is 0.005 W/m^2 when you are 2 meters away from the source. Calculate the sound intensity lev

nekit [7.7K]

Answer:

Explanation:

We have given the intensity of the loud car horn $I=0.005w/m^2$

We know that $I_O=10^{-12}w/m^2$

Now the sound intensity level is given by $\beta =10log\frac{I}{I_0}=10log\frac{0.005}{10^{-12}}=96.98dB$ , which is nearly equal to 97

So the sound intensity level will be 97 dB

So option (c) will be the correct option

4 0

3 years ago

A block is being pulled upward along an inclined surface at a constant speed. Which of the following statements is correct? Grou

Alex17521 [72]

Answer:The net force on the block is zero.

Explanation:

Given

Block is being pulled upward along an inclined surface at a constant speed

As speed is constant and moved in a straight line along the plane therefore its velocity is also constant .

and change in velocity is equal to acceleration therefore acceleration is zero here i.e. net force is zero acting on the body.

5 0

3 years ago

A standing wave pattern is created on a string with mass density μ = 3.4 × 10-4 kg/m. A wave generator with frequency f = 61 Hz

uranmaximum [27]

Answer:

1) λ = 0.413 m , 2)v = 25,213 m / s , 3) T = 0.216 N , 4) m = 22.04 10-3 kg

Explanation:

1) The resonance occurs when the traveling wave bounces at the ends and the two waves are added, the ends as they are fixed have a node, the wavelength and the length of the string are related

λ = 2L / n n = 1, 2, 3 ...

In this case L = 0.62 m and n = 3

Let's calculate

λ = 2 0.62 / 3

λ = 0.413 m

2) the velocity related to wavelength and frequency

v = λ f

v = 0.413 61

v = 25,213 m / s

3) let's use the equation

v = √T /μ

T = v² μ

T = 25,213² 3.4 10⁻⁴

T = 0.216 N

4) the rope tension is proportional to the hanging weight

T-W = 0

T = W

W = m g

m = W / g

m = 0.216 / 9.8

m = 22.04 10-3 kg

5) n = 2

λ = 2 0.62 / 2

λ = 0.62 m

6) v = λ f

v = 0.62 61

v = 37.82 m / s

7) T = v² μ

T = 37.82² 3.4 10⁻⁴

T = 0.486 N

8) m = W / g

m = 0.486 / 9.8

m = 49.62 10⁻³ kg

9) n = 1

λ = 2 0.62

λ = 1.24 m

v = 1.24 61

v = 75.64 m / s

T = v² miu

T = 75.64² 3.4 10⁻⁴

T = 2.572 10⁻² N

m = 2.572 10⁻² / 9.8

m = 262.4 10⁻³ kg

5 0

3 years ago

1) Consider a situation where immediately after birth, twins are separated. One continues to live on earth, while the other is w

Hatshy [7]

No they will not look the same.

4 0

3 years ago

Read 2 more answers

How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s

poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

$W = \Delta KE$

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

Here,

m = mass

$v_{f,i}$ = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2$

$W_1 = 150(m) J$

Second case) When the particle goes from 20m/s to 30m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2$

$W_1 = 250(m) J$

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0

3 years ago