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4 years ago

11

masses of 3kg on a smooth horizontal table.It is connected by a light string passing at the edge of the table to another mass of

2kg hanging vertically.When to the system is released from rest with what acceleration do the mass move

1 answer:

allochka39001 [22]4 years ago

7 0

Answer:

aaawwwwwwwsssaaaasasss

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The heat flux for a given wall is in the x-direction and given as q^n = 11 W/m^2, the walls thermal conductivity is 1.7 W/mK and

MrMuchimi

Answer:

$\frac{dT}{dx} = 6.47 ^oC/m$

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards

Explanation:

As we know that heat flux is given by the formula

$q^n = K\frac{dT}{dx}$

here we know that

K = thermal conductivity

$\frac{dT}{dx}$ = temperature gradient

now we know that

$q^n = 11 W/m^2$

also we know that

K = 1.7 W/mK

now we have

$11 = 1.7 \frac{dT}{dx}$

so temperature gradient is given as

$\frac{dT}{dx} = \frac{11}{1.7} = 6.47 K/m$

also in other unit it will be same

$\frac{dT}{dx} = 6.47 ^oC/m$

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards

5 0

3 years ago

In (A) Water has same mass and great volume

aniked [119]

Answer:

I didn't get pic from you but I got the type of answer that u want.

Explanation:

Water molecules are always moving. But on the average they are packed the same throughout. Therefore, the ratio between the mass and volume is the same, making the density the same. This is true no matter the size of the sample or where you select your sample from.

7 0

3 years ago

An 8 g bullet leaves the muzzle of a rifle with

Elena-2011 [213]

Answer: 1872 N

Explanation:

This problem can be solved by using one of the Kinematics equations and Newton's second law of motion:

$V^{2}=V_{o}^{2} + 2ad$ (1)

$F=ma$ (2)

Where:

$V=611.9 m/s$ is the bullet's final speed (when it leaves the muzzle)

$V_{o}=0$ is the bullet's initial speed (at rest)

$a$ is the bullet's acceleration

$d=0.8 m$ is the distance traveled by the bullet before leaving the muzzle

$F$ is the force

$m=8 g \frac{1 kg}{1000 g}=0.008 kg$ is the mass of the bullet

Knowing this, let's begin by isolating $a$ from (1):

$a=\frac{V^{2}}{2d}$ (3)

$a=\frac{(611.9 m/s)^{2}}{2(0.8 m)}$ (4)

$a=234013.5063 m/s^{2} \approx 2.34(10)^{5} m/s^{2}$ (5)

Substituting (5) in (2):

$F=(0.008 kg)(2.34(10)^{5} m/s^{2})$ (6)

Finally:

$F=1872 N$

4 0

3 years ago

How do igneous rocks differ from each other.

olya-2409 [2.1K]

Although hand specimen identification is some what of an art; igneous rocks are classified using two critiria:
•Texture
•Chemical and mineral composition
Most of the time the rock name, such as basalt or granite, reflects the chemical composition. The minerals and textures are used as adjectives that modify the rock name. You might see a name like an olivine basalt or a quartz porphyry rhyolite.

8 0

4 years ago

Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum

Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

$\frac{1}{2}mv^2 = mgh$

where

m is the mass of the man

v is the speed after jumping

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s$

In the acceleration phase, we know that the initial velocity is

$u=0$

And the force exerted on the floor is 2.3 times the gravitational force, so

$F=2.3 mg$

This means the net force on you is

$F_{net} = F-mg=2.3mg-mg=1.3 mg$

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

$a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g$

Now we can use the following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

$v^2-u^2=2as$

where s is the quantity we want to find. Solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m$

2)

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

$\Delta h = -0.77 m$

This means that the lowest point reached by the center of mass above the floor during the crouch is

$h'=h+\Delta h = 1.0 - 0.77 = 0.23 m$

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

3 0

3 years ago