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3 years ago

Timmy bikes to school 3 kilometers east and 2 kilometers south. What is Timmy's final displacement?

Physics

1 answer:

vovikov84 [41]3 years ago

8 0

Answer:

3.6km South East

Explanation:

Displacement is the shortest distance between the starting point and the ending point and the direction it is displaced in. To calculate the displacement we can use the pythagoras theorem because the 3km East and the 2km south form the two shorter sides of a right angled triangle between the starting and ending points. So, the displacement is the length C of the triangle which we can calculate as follows:

Pythagoras Theorem:

a^2+b^2=c^2

(2)^2+(3)^2=c^(2)

4+9=c^2

Square root 13 = c

c=3.6km (1dp)

The total displacement is 3.6km and is in the approximate direction of South East (because he travelled east and south).

Hope this helped!

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Joanna wants to determine the speed of sound in xenon. When she plays a tone with a frequency of 440 Hz, the resulting sound wav

iVinArrow [24]

The speed of the sound in the xenon is 178 m/s. And the right option is b 178 m/s

<h3 /><h3>What is speed?</h3>

Speed can be defined as the ratio of the total distance traveled by a body to the total time taken.

To calculate the speed of the sound in the xenon, we use the formula below.

Formula:

v = λf............. Equation 1

Where:

v = Speed of the sound in xenon
f = Frequency
λ = Wavelength.

From the question,

Given:

f = 440 Hz
λ = 40.4 cm = 0.404 m

Substitute the values above into equation 1

v = 440(0.404)
v = 177.76 m/s.
v ≈ 178 m/s

Hence, The speed of the sound in the xenon is 178 m/s. And the right option is b 178 m/s

Learn more about speed here: brainly.com/question/4931057

7 0

2 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$