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3 years ago

Which of Franz Mesmer’s claims about his mesmerism abilities eventually led to his downfall?

Physics

2 answers:

MrMuchimi3 years ago

7 0

He can mesmerize as well as magnatize trees, this is what he claims would lead to his downfall.

Angelina_Jolie [31]3 years ago

5 0

Answer:

the correct answer is b

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Given the following lens combination:

natulia [17]

Given:

Lens.........diameter ...fl#

eyepiece...2cm............5

objective...40cm........15

focal length of eyepiece = 2*5 = 10cm

focal length of objective = 40*15 = 600cm

magnification = FL obj / FL eyp = 600/10 = 60x

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3 years ago

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Identify two types of simple machines in this compund machine.

sleet_krkn [62]

the answer would be "screw and lever!", thats the answer because the middle is made into a rod of metal rapped in metal to screw into the cork, and the lever part is the two handles on the side to lift it out of the bottle.

4 0

3 years ago

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2. What makes Newton's 3rd Law an unbalanced force?

LiRa [457]

Answer:

Explanation:

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2 years ago

A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are

slava [35]

Answer:

The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

Explanation:

Given that,

Mass $m = 1.81\times10^{-3}\ kg$

Velocity $v = (3.00\times10^{4}\ m/s)j$

Charge $q = 1.22\times10^{-8}\ C$

Magnetic field $B= (1.63\hat{i}+0.980\hat{j})\ T$

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

$F = ma=q(v\times B)$

$a =\dfrac{q(v\times B)}{m}$

We need to calculate the value of $v\times B$

$v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})$

$v\times B=4.89\times10^{4}$

Now, put the all values into the acceleration 's formula

$a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}$

$a= -0.3296\ \hat{k}\ m/s^2$

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

7 0

3 years ago

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A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a

sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0

3 years ago