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1 year ago

For for each situation say how the ideas of force pressure and area can be applied :

Physics

1 answer:

fomenos1 year ago

4 0

<h3 /><h3><u><em>Solution-:</em></u></h3><h3><u><em>more force as expansion is much</em></u></h3>

<u><em>also, less force area is much</em></u>

<h3> mark me Brainliest</h3>

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A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

jeyben [28]

Answer:

$V_0$

Explanation:

Given that, the range covered by the sphere, $M$ , when released by the robot from the height, $H$ , with the horizontal speed $V_0$ is $D$ as shown in the figure.

The initial velocity in the vertical direction is $0$ .

Let $g$ be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. $V_0$ will remain constant throughout the projectile motion.

So, if the time of flight is $t$ , then

$D=V_0t\; \cdots (i)$

Now, from the equation of motion

$s=ut+\frac 1 2 at^2\;\cdots (ii)$

Where $s$ is the displacement is the direction of force, $u$ is the initial velocity, $a$ is the constant acceleration and $t$ is time.

Here, $s= -H, u=0,$ and $a=-g$ (negative sign is for taking the sigh convention positive in $+y$ direction as shown in the figure.)

So, from equation (ii),

$-H=0\times t +\frac 1 2 (-g)t^2$

$\Rightarrow H=\frac 1 2 gt^2$

$\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)$

Similarly, for the launched height $2H$ , the new time of flight, $t'$ , is

$t'=\sqrt {\frac {4H}{g}}$

From equation (iii), we have

$\Rightarrow t'=\sqrt 2 t\;\cdots (iv)$

Now, the spheres may be launched at speed $V_0$ or $2V_0$ .

Let, the distance covered in the $x-$ direction be $D_1$ for $V_0$ and $D_2$ for $2V_0$ , we have

$D_1=V_0t'$

$D_1=V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_1=\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_1=1.41 D$ (approximately)

This is in the $3$ points range as given in the figure.

Similarly, $D_2=2V_0t'$

$D_2=2V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_2=2\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_2=2.82 D$ (approximately)

This is out of range, so there is no point for $2V_0$ .

Hence, students must choose the speed $V_0$ to launch the sphere to get the maximum number of points.

7 0

3 years ago

A train travelling at 50km/h approaches another train moving towards the first at 90km/h. If they are 35km apart (on a straight

Ede4ka [16]

it will take 36 seconds

8 0

2 years ago

Which of these forces determines the chemical properties of an atom? friction force nuclear force gravitational force electrical

zysi [14]

The answers is an electrical force.
Under normal conditions, atoms interact with each other via electrons that are furthest away from the nucleus. These electrons from the what is called the outer shell of the atom, electrons from the outer shell that can participate in chemical reactions are called valence electrons.

3 0

3 years ago

I really need help! Please no one has helped me! Will give BRAINLIEST, THANKS, and RATE!! Please help if you are good at science

adell [148]

I took this test and the correct answer is c
I hope this helps

3 0

2 years ago

Read 2 more answers

A block is resting on a platform that is rotating at an angular speed of 2.4 rad/s. The coefficient of static friction between t

Sloan [31]

Answer:

r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.

Explanation:

From a sum of forces:

$Ff = m*a$ where Ff = μ * N and $a = \frac{V^2}{r}=\omega^2*r$

N - m*g = 0 So, N = m*g. Replacing everything on the original equation:

$\mu*m*g = m*\omega^2*r$ (eq2)

Solving for r:

$r = \frac{\mu*g}{\omega^2}=1.41m$

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.

4 0

2 years ago