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2 years ago

Difference between upthrust force and normal reaction force.

Physics

1 answer:

irakobra [83]2 years ago

7 0

Answer:

Explanation:

Both are contact forces arising at the interface between two bodies. In the fluid this interface might be irregular, and it completely surrounds a submerged object. For a solid it is usually a single flat surface - but it can be a collection of surfaces, which do not need to be flat or regular, and which can surround the object

Upthrust occurs at a fluid-solid interface whereas normal reaction occurs at a solid-solid surface. However, it is possible to generate the same fluid-like phenomenon of upthrust by immersing a solid object in sand or small beads and agitating them to simulate the pressure of atoms. With

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Dos cargas iguales de 10^-7C estan separadas por una distancia de 2m. calcule la fuerza con que se repelen

m_a_m_a [10]

Answer:

$F=2.25\times 10^{-5}\ N$

Explanation:

According to question,

Charge 1 and charge 2 are $10^{-7}\ C$

The distance between charges is 2 m

We need to find the force with which two positive charges repel. It is called electrostatic force of repulsion. It can be given by :

$F=\dfrac{kq^2}{r^2}\\\\F=\dfrac{9\times 10^9\times (10^{-7})^2}{2^2}\\\\F=2.25\times 10^{-5}\ N$

So, the electric force of repulsion is $2.25\times 10^{-5}\ N$ .

8 0

2 years ago

RC time constant circuit if R 50 KOC-21 a TOSS c. 1.05 s . what is the expected RC value b. 10.55 d. 0.105 s

Afina-wow [57]

Answer:

Time constant of RC circuit is 0.105 seconds.

Explanation:

It is given that,

Resistance, $R=50\ K\Omega=5\times 10^4\ \Omega$

Capacitance, $C=2.1\ \mu F=2.1\times 10^{-6}\ F$

We need to find the expected time constant for this RC circuit. It can be calculated as :

$\tau=R\times C$

$\tau=5\times 10^4\times 2.1\times 10^{-6}$

$\tau=0.105\ s$

So, the time constant for this RC circuit is 0.105 seconds. Hence, this is the required solution.

7 0

2 years ago

A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple

aliina [53]

Answer:

T= 1 s

Explanation:

Given that

When x= cm ,T= 1

we know that time period of spring mas system given as

$T=2\pi \sqrt{\dfrac{m}{k}}$

T= Time period

m= mass

k=spring constant

So from above equation we can say that time period of system does not depends on the value of x.

So when x= 10 cm ,still time period will be 1 s.

T= 1 s

5 0

3 years ago

In Newton’s second law, if the net force acting on object doubles. The object’s Will also double

Anna71 [15]

Answer: Explanation:

If the net force on an object is doubled, its acceleration will double If the mass of an object is doubled, the acceleration will be halved .

3 0

3 years ago

n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend

Kitty [74]

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

$p_f=p_i\\mv_1+Mv_2=(m+M)v$

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

$(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}$

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

$E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2$

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

$(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m$

hence, the heigth is 68cm

4 0

2 years ago