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2 years ago

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

Physics

1 answer:

fomenos2 years 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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Look at this picture of a frog.

fredd [130]

Answer:

Biosphere

Explanation:

The biosphere consist of all living organisms

6 0

3 years ago

Read 2 more answers

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the sp

laila [671]

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

$a = \frac{v^2}{r}$

At constant speed, we will have;

$v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1$

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

7 0

3 years ago

A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming

Monica [59]

Answer:

The deceleration is $a = - 76.27 m/s^2$

Explanation:

From the question we are told that

The height above firefighter safety net is $H = 14 \ m$

The length by which the net is stretched is $s = 1.8 \ m$

From the law of energy conservation

$KE_T + PE_T = KE_B + PE_B$

Where $KE_T$ is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

and $PE_T$ is the potential energy of the before jumping which is mathematically represented at

$PE_T = mg H$

and $KE_B$ is the kinetic energy of the person just before landing on the safety net which is mathematically represented at

$KE_B = \frac{1}{2} m v^2$

and $PE_B$ is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

So the above equation becomes

$mgH = \frac{1}{2} m v^2$

=> $v = \sqrt{2 gH }$

substituting values

$v = 16.57 m/s$

Applying the equation o motion

$v_f = v + 2 a s$

Now the final velocity is zero because the person comes to rest

$0 = 16.57 + 2 * a * 1.8$

$a = - \frac{16.57^2 }{2 * 1.8}$

$a = - 76.27 m/s^2$

4 0

3 years ago

A tennis player smashes a ball of mass m horizontally at a vertical wall. The ball rebounds at the same speed v with which it st

Morgarella [4.7K]

Answer:

Explanation:

Change in momentum: This is defined as the product of mass and change in velocity of a body. or it can be defined as the product of force and time of a body. The fundamental unit of change in momentum is kg.m/s

Change in momentum = M(V-U)......................... Equation 1

where M = mass of the ball, V = final velocity of the ball, U = initial velocity of the ball.

Let: M = m kg and V = U = v m/s

Substituting these values into equation 1

Change in momentum = m(v-v)

Change in momentum = m(0)

Change in momentum = 0 kg.m/s

<em>Therefore the momentum of the ball has not changed.</em>

<em>The right option is B.0</em>

5 0

3 years ago

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C

miskamm [114]

<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

$\boxed{q_{B}=+2q}$

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential $V$ , which is <em>the energy per unit charge, </em>so the potential $V$ at any point in an electric field with a test charge $q_{0}$ at that point is:

$V=\frac{U}{q_{0}}$

The potential $V$ due to a single point charge q is:

$V=k\frac{q}{r}$

Where $k$ is an electric constant, $q$ is value of point charge and $r$ is the distance from point charge to where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius $r$ . Before the sphere A and B touches we have:

$V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r$

When they touches each other the potential is the same, so:

$V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}$

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

$q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}$

Therefore:

$(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}$

So after A and B touch and are separated the charge on sphere B is:

$\boxed{q_{B}=+2q}$

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

$\boxed{q_{C}=+1.5q}$

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

$q_{A}=q_{B}=+2q$

Second: C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here $q_{A}=+2q$ and C carries no net charge or $q_{C}=0$ . Also, $r_{A}=r_{C}=r$

$V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

Applying the same concept as the previous problem when sphere touches we have:

$k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}$

For the principle of conservation of charge:

$q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q$

Finally, from the third step:

Here $q_{B}=+2q \ and \ q_{C}=+q$ . Also, $r_{B}=r_{C}=r$

$V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

When sphere touches we have:

$k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}$

For the principle of conservation of charge:

$q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q$

So the charge that ends up on sphere C is:

$q_{C}=+1.5q$

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

$+4q$

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

$q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0$

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

$q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q$

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

$+4q$

8 0

3 years ago