Answer all these and i'll mark brainliest.

Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another

Tanya [424]

Answer: $I_2=0.618\times 10^{-7} W/m^2$

Explanation:

Given

Distance between source and receiver $d_1=3 m$

Sound Intensity $I_1=1.1\times 10^{-7} W/m^2$

Distance of of second observer $d_2=4 m$

Intensity varies as

$I\propto \frac{1}{d^2}$

using this

$I=\frac{k}{d^2}$

$\frac{I_1}{I_2}=\frac{d_2^2}{d_1^2}$

$\frac{1.1\times 10^{-7}}{I_2}=\frac{4^2}{3^2}$

$I_2=0.75^2\times 1.1\times 10^{-7}$

$I_2=0.618\times 10^{-7} W/m^2$

6 0

3 years ago

A ships initial position is 25 km east of the origin. What is the ship’s final position if it sails at a constant velocity of 42

Evgen [1.6K]

Answer:

88km E

Explanation:

25 km is added already

then if we take the formula of speed(V)

V=d/t

and rearrange it to subject d (distance)

d= V x t

we get

d= 42km/h x 1.5hrs

we get

63 km

now if we add the already travelled 25 km to this additional distance

63km + 25km = 88km

Thus the final distance traversed by the ship from its origin is

88km E

Plz mark me brainliest

Hope this helps

7 0

3 years ago

How much time does it take for a car traveling 19.1 m/s to cover a distance of 1200 m?

nalin [4]

Answer:

63

Explanation:

it would take around 63 if ur asking for 1200/19.1

4 0

3 years ago

You have a suspect on the run in a dry, wooded area; they’ve ditched the gun but are hiding from you. how long do you have to fi

erma4kov [3.2K]

Answer:

C. a few days

Explanation:

The evidence that we will have to investigate is the gunshot residue on the body of the suspect. When a person fires a gun there will be a deposition of material from the components of the chemicals present in the gun. This deposit is called gunshot residue and consists of particles of the propellant or explosive primer.

Experimentation has shown that the period of time that is person will test possible for gunshot residue is 5.27 days.

7 0

3 years ago

In diagram B, prove that the tension in the middle string is 3.3 N.

zepelin [54]

Answer:

The tension in the middle string is 3.3 N

Explanation:

Two boxes move with same acceleration

The tension in the middle string is T

Lets start with the forces acting in the 2nd box

→ The tension T in the string in the direction of motion

Due to Newtons's Law

→ ∑ Force in direction of motion = mass × acceleration

→ T = M a

→ M = 1 kg

Substitute the value of M is the equation

→ T = a ⇒ (1)

The force acting on the 1st box

→ The force F in the direction of motion, and T in the string in opposite

direction of motion

→ F - T = 2M a

→ F = 10 N , M = 1 kg

Substitute These values in the equation

→ 10 - T = 2(1) a

→ 10 - T = 2 a ⇒ (2)

Substitute equation (1) in (2)

→ 10 - T = 2 T

Add T to both sides

→ 10 = 3 T

Divide both sides by 3

→ T = 3.3 N

<em>The tension in the middle string is 3.3 N</em>

7 0

3 years ago