Calculate the pressure in SI unit when a force of 100 dyne acts on area of 0.02cm²

Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400

hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

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

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

$v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s$

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

$v=u+at$

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

$v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms$

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

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

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

$v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m$

The beetle can jump to a height of <u>2.1 m</u>

7 0

2 years ago

A 0.4-kg toy train car moving forward at 3 m/s collides with and sticks to a 0.8–kg toy car that is traveling in the opposite di

Gemiola [76]

Hey there!

Seems like you're looking for the size and direction to the final velocity of the two cars. To find it, you must solve it like this.

0.4 kg(3 m/s) + 0.8kg(–2 m/s) = 1.2 kg m/s -1.6 kg m/s = –0.4 kg m/s

–0.4 kg m/s = 1.2 kg(v) = (–0.4 kg m/s)/(1.2 kg) = v = –0.33 m/s

So, the cars are traveling at -0.33 m/s in the direction of the second car.

Hope this helps

<em>Tobey</em>

4 0

2 years ago

Question is down below

rosijanka [135]

The vertical components of velocity is 10.35 m/s and the horizontal component of velocity is 38.6 m/s

<h3>What are the components of velocity?</h3>

We know that velocity is a vector quantity, a vector often can be resolved into its components. The vertical components is V sinθ while the horizontal component is vcosθ.

Hence;

Vertical component = 40 m/s sin 15 degrees = 10.35 m/s

Horizontal component = 40 cos 15 degrees = 38.6 m/s

Learn more about components of velocity:brainly.com/question/14478315

#SPJ1

7 0

1 year ago

A 5 kgkg sphere having a charge of ++ 8 μCμC is placed on a scale, which measures its weight in newtons. A second sphere having

Mrac [35]

Answer:

F_Balance = 46.6 N ,m' = 4,755 kg

Explanation:

In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in

∑ F = 0

Fe –W + F_Balance = 0

F_Balance = - Fe + W

The electric force is given by Coulomb's law

Fe = k q₁ q₂ / r₂

The weight is

W = mg

Let's replace

F_Balance = mg - k q₁q₂ / r₂

Let's reduce the magnitudes to the SI system

q₁ = + 8 μC = +8 10⁻⁶ C

q₂ = - 3 μC = - 3 10⁻⁶ C

r = 0.3 m = 0.3 m

Let's calculate

F_Balance = 5 9.8 - 8.99 10⁹ 8 10⁻⁶ 3 10⁻⁶ / (0.3)²

F_Balance = 49 - 2,397

F_Balance = 46.6 N

This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.

Mass reading is

m' = F_Balance / g

m' = 46.6 /9.8

m' = 4,755 kg

6 0

3 years ago

What is kepler's law??

Elan Coil [88]

<h2>QUESTION:- </h2>

➜what is kepler's law??

$\huge\red{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W}\orange{ER}}}}$

Kepler gave the three laws or theorems of motion of the orbitals bodies

${\huge {\bold{ \red{ \star}}}}{ \blue{ \bold{FIRST \: \: \: LAW}}}$

This law state that the celestial bodies revolves around the stars in elliptical orbit and star as a single focus.

Example :- Earth revolves around the Sun as assuming it as single focus

This also shows that earth revolves around the sun in elliptical orbit.

${\huge {\bold{ \blue{ \star}}}}{ \green{ \bold{SECOND \: \: \: LAW}}}$

Area covered by the planet is equal in equal duration of time irrespective of the position of the planet.

It also states that Angular momentum is constant

As Angular momentum is constant it means areal velocity is also constant.

$\frac{ \triangle \: A}{ \triangle \: T} = \frac{L}{2m}△T△A=2mL$

where:-

A is the area.

T is the time.

L is the angular momentum.

M is the mass of the body.

${\huge {\bold{ \green{ \star}}}}{ \purple{ \bold{THIRD \: \: \: LAW}}}$

square of the time of the revolution is directly proportional to the cube of the distance between the planet and star in Astronomical unit.

${T}^{2} = {a}^{3}T2=a3$

where:-

T = time of revolution

a is the distance between the planet and star.

$\purple\star \: {Thanks \: And \: Brainlist} \blue \star \\ {\orange{ \star}}{if \: U \: Like d \: My \: Ans} {\green{ \star }}$

8 0

3 years ago

Calculate the pressure in SI unit when a force of 100 dyne acts on area of 0.02cm²​

Calculate the pressure in SI unit when a force of 100 dyne acts on area of 0.02cm²