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3 years ago

Hola puntos gratis !!!!!!!!!!!!!!!!!

Physics

2 answers:

mina [271]3 years ago

7 0

Answer:

gracias

Explanation:

Sloan [31]3 years ago

4 0

Hello.

Brainly points can be used for better situations, brainly points are experience that you receive by answering well.

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The diagram below shows a 5.00-kilogram block

bixtya [17]

The name and strength of the force holding the block up is 50 N upward - Normal force.

The given parameters:

Mass of the block, m = 5 kg

The weight of the block acting downwards due to gravity is calculated as follows;

W = mg

where;

g is acceleration due to gravity = 10 m/s²

W = 5 x 10

W = 50 N (downwards)

Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.

Fₙ = 50 N (upwards)

Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.

Learn more about Normal force here: brainly.com/question/14486416

4 0

2 years ago

Read 2 more answers

A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle

expeople1 [14]

Explanation:

Given that,

Wavelength of the light, $\lambda=691\ nm=691\times 10^{-9}\ m$

(a) Slit width, $a=3.8\times 10^{-4}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}$

$\theta=0.104^{\circ}$

(b) Slit width, $a=3.8\times 10^{-6}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}$

$\theta=10.47^{\circ}$

Hence, this is the required solution.

7 0

3 years ago

Two sealed 1 l containers full of gas are at room temperature. Container a has a pressure of 4 atm and container b has a pressur

loris [4]

Answer:

The number density of the gas in container A is twice the number density of the gas in container B.

Explanation:

Here we have

P·V =n·R·T

n = P·V/(RT)

Therefore since V₁ = V₂ and T₁ = T₂

n₁ = P₁V₁/(RT₁)

n₂ = P₂V₂/(RT₂)

P₁ = 4 atm

P₂ = 2 atm

n₁ = 4V₁/(RT₁)

n₂ =2·V₁/(RT₁)

∴ n₁ = 2 × n₂

Therefore, the number of moles in container A is two times that in container B and the number density of the gas in container A is two times the number density in container B.

This can be shown based on the fact that the pressure of the container is due to the collision of the gas molecules on the walls of the container, with a kinetic energy that is dependent on temperature and mass, and since the temperature is constant, then the mass of container B is twice that of A and therefore, the number density of container A is twice that of B.

5 0

3 years ago

You want to make a ride so you do not want to exceed 1.1g’s, if the radius of the turns are 10m, then what is the maximum speed

Citrus2011 [14]

The maximum speed is 10.4 m/s

Explanation:

For a body in uniform circular motion, the centripetal acceleration is given by:

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

where

v is the linear speed

r is the radius of the circular path

In this problem, we have the following data:

- The maximum centripetal acceleration must be

$a=1.1 g$

where $g=9.8 m/s^2$ is the acceleration of gravity. Substituting,

$a=(1.1)(9.8)=10.8 m/s^2$

- The radius of the turn is

r = 10 m

Therefore, we can re-arrange the equation to solve for v, to find the maximum speed the ride can go at:

$v=\sqrt{ar}=\sqrt{(10.8)(10)}=10.4 m/s$

Learn more about centripetal acceleration:

brainly.com/question/2562955

#LearnwithBrainly

8 0

3 years ago

g A bowling ball with a mass of 3.86 kg and a radius of 0.161 m starts from rest at a height of 2.5 m and rolls down a 48.4 o sl

Fynjy0 [20]

Answer:

$v=1.5m/s$

Explanation:

The gravitational potential energy gets transformed into translational and rotational kinetic energy, so we can write $mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}$ . Since $v=r\omega$ (the ball rolls without slipping) and for a solid sphere $I=\frac{2mr^2}{5}$ , we have:

$mgh=\frac{mv^2}{2}+\frac{2mr^2\omega^2}{2*5}=\frac{mv^2}{2}+\frac{mv^2}{5}=\frac{7mv^2}{10}$

So our translational speed will be:

$v=\sqrt{\frac{10gh}{7}}=\sqrt{\frac{10(9.8m/s^2)(0.161m)}{7}}=1.5m/s$

6 0

2 years ago