Ask question

Ask question

All categories

3 years ago

11

you walk from the park to your friend's house, then back to your house. (a) What is the distance traveled? (b) What is your disp

lacement? If you walk from your house to the library, then back to your house, repeat (a) and (b).

1 answer:

777dan777 [17]3 years ago

3 0

Explanation:

The total distance in a path is called distance.

The shortest distance between two points is called displacement.

a) Here, the distance travelled between the park to your friend's house and back is

Distance between park to friends house + Distance from friend's house to your house.

b) Displacement would be the shortest distance between the park and your house.

a) Distance walked between your house to library and back is

Distance between your house and library + Distance between your house and library

b) Displacement would be zero (0) as the distance between you initial point and final point is zero. Here, the initial and final points are the same

You might be interested in

A force of 35 N acts on a ball for 0.2 s. If the ball is initially at rest:

olya-2409 [2.1K]

To solve this problem we will apply the concepts related to momentum and momentum on a body. Both are equivalent values but can be found through different expressions. The impulse is the product of the Force for time while the momentum is the product between the mass and the velocity. The result of these operations yields equivalent units.

PART A ) The Impulse can be calculcated as follows

$L= F\Delta t$

Where,

F = Force

$\Delta t =$ Change in time

Replacing,

$L = (35N)(0.2s)$

$L= 7N\cdot s$

PART B) At the same time the momentum follows the conservation of momentum where:

Initial momentum= Final momentum

And the change in momentum is equal to the Impulse, then

$\Delta p = L$

And

$\Delta p = p_f - p_i$

There is not initial momentum then

$\Delta p = p_f$

$L = p_f$

$p_f = 7N\cdot s = 7kg\cdot m/s$

8 0

3 years ago

A bystander observes the musicians heading toward each other. When musician #1 is 100 m away, the intensity is 1.24 x 10-8 W/m^2

777dan777 [17]

Explanation:

Given that,

Distance 1, r = 100 m

Intensity, $I_1=1.24\times 10^{-8}\ W/m^2$

If distance 2, r' = 25 m

We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :

$I=\dfrac{P}{4\pi r^2}$ .........(1)

Let I' is the intensity at r'. So,

$I'=\dfrac{P}{4\pi r'^2}$ ............(2)

From equation (1) and (2) :

$I'=\dfrac{Ir}{r'^2}$

$I'=\dfrac{1.24\times 10^{-8}\times 100}{25^2}$

$I'=1.98\times 10^{-9}\ W/m^2$

Intensity level is given by :

$dB=10\ log(\dfrac{I'}{I_o})$ , $I_o=10^{-12}\ W/m^2$

$dB=10\ log(\dfrac{1.98\times 10^{-9}}{10^{-12}})$

dB = 32.96 dB

Hence, this is the required solution.

7 0

3 years ago

a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f

34kurt

Answer:

$\frac{h_{liquid} }{ h_{body} }$ = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

B = ρ_liquid g V_liquid

let's write the translational equilibrium condition

B - W = 0

let's use the definition of density

ρ_body = m / V_body

m = ρ_body V_body

W = ρ_body V_body g

we substitute

ρ_liquid g V_liquid = ρ_body g V_body

$\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }$

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

V = A h_bogy

Thus

$\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }$

we substitute

5/9 = $\frac{h_{liquid} }{ h_{body} }$

8 0

3 years ago

Words that describe direction east, up, and left is it true or false

m_a_m_a [10]

Answer:

True

Explanation:

East, up, and left all define as a direction.

7 0

3 years ago

The picture below shows a wheelbarrow. Using a wheelbarrow can make it easier to lift a heavy object by

timofeeve [1]

Answer:

B IS CORRECT

Explanation:

8 0

3 years ago