Ask question

Ask question

All categories

2 years ago

6

Think of a way you could demonstrate elastic force to a younger student. Describe the procedure you would follow and the materia

ls you would use.

1 answer:

Soloha48 [4]2 years ago

7 0

Answer:

I feel like to demonstrate you would use an elastic band as the material. You obviously have to put force in order to see how far it stretches. From this you can also find about its resistance and durability

Also you have to make sure the distance between the two hands are equal as you want an accurate result.

You might be interested in

A hiker walks 15km due east then heads due north for 8km . What is the direction of the resultant vector?

4vir4ik [10]

his displacements are given as

$d_1 = 15 km$ East

$d_2 = 8 km$ North

now we can find the direction of displacement by using the concept

$\theta = tan^{-1}\frac{\deta y}{\delta x}$

$\theta = tan^{-1}\frac{8}{15}$

$\theta = 28.1 degree$

magnitude of displacement is given as

$d = \sqrt{x^2 + y^2}$

$d = \sqrt{15^2 + 8^2}$

$d= 17 km$

<em>so the displacement is 17 km at angle 28.1 degree North of East</em>

8 0

2 years ago

Transforma la siguiente unidad al sistema internacional:<br> 37 ft3

tigry1 [53]

F up HC Chi UC Chi UC CI UC CJ

3 0

3 years ago

Can someone explain it with steps?

Anton [14]

Answer:

Option C is the correct answer

Explanation:

Distance travelled by car during reaction time

$=15\times0.4\\\\=6m$

The car stopped before hitting the animal by $1 m$

Distance travelled during deceleration is $21-6-1=14m$

Hence by $v^2=u^2+2as$

We have

$0^2=15^2+2 \cdot a \cdot 14\\\\a=\frac{-225}{28} \\\\=-8.03m/s^2$

Option C is the correct answer

5 0

2 years ago

Read 2 more answers

A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?

rosijanka [135]

Answer:

a

$l_o =52 \ m$

b

$l = 37.13 \ LY$

Explanation:

From the question we are told that

The speed of the spaceship is $v = 0.800c$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

The length is $l = 31.2 \ m$

The distance of the star for earth is $d = 145 \ light \ years$

The speed is $v_s = 2.90 *10^{8}$

Generally the from the length contraction equation we have that

$l = l_o \sqrt{1 -[\frac{v}{c } ]}$

Now the when at rest the length is $l_o$

So

$l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }$

$l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }$

$l_o=52 \ m$

Considering b

Applying above equation

$l =l_o \sqrt{1 - [\frac{v}{c } ]}$

Here $l_o =145 \ LY(light \ years )$

So

$l=145 * \sqrt{1 - \frac{v_s^2}{c^2 } }$

$l =145 * \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }$

$l = 37.13 \ LY$

4 0

2 years ago

a 1.2x10^3 kilogram car is accelerated uniformly from 10. meters per second to 20 meters per second in 5.0 seconds. what is the

irinina [24]

Force , F = ma

F = m(v - u)/t

Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
t = time, Force is in Newton.

m= 1.2*10³ kg, u = 10 m/s, v = 20 m/s, t = 5s

F = 1.2*10³(20 - 10)/5

F = 2.4*10³ N = 2400 N

7 0

2 years ago