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4 years ago

Did thomson conclude from his experiment ?

Physics

1 answer:

Law Incorporation [45]4 years ago

6 0

It would be A because he wanted to know more about the physical properties of these particles like their mass

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Force f⃗ =−10j^n is exerted on a particle at r⃗ =(7i^+5j^)m. part a what is the torque on the particle about the origin?

cluponka [151]

Answer:

Torque, $\tau=0i+0j-70k$

Explanation:

It is given that,

Force acting on the particle, $F=-10j\ N$

Position of the particle, $r=(7i+5j)\ m$

We need to find the torque on the particle about the origin. It is equal to the cross product of position and the force. Its formula is given by :

$\tau=r\times F$

$\tau=(7i+5j)\times (-10j)$

The cross product of vectors is given by :

$\tau=\begin{pmatrix}0&0&-70\end{pmatrix}$

$\tau=0i+0j-70k$

So, the torque on the particle about the origin $0i+0j-70k$ . Hence, this is the required solution.

6 0

3 years ago

Aang and Appa are flying over Ba Sing Se. They covered 500 m in 30 seconds while heading east. What was their velocity?

barxatty [35]

Answer:

Explanation:

I need help to marysol;)

8 0

4 years ago

Consider the electric field lines drawn below for a configuration of two charges. Several locations are labeled on the diagram.

geniusboy [140]

Answer:

The magnitude or strength of an electric field in the space surrounding a source charge is related directly to the quantity of charge on the source charge and inversely to the distance from the source charge. The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source charge.

Explanation:

I really hope this helps you!

6 0

2 years ago

A 1.5 kg copper block is given an initial speed of3.00m/s on a

ankoles [38]

Answer:

a. 0.01 C

b. dissipated to outside environment

Explanation:

Let the specific heat of copper be 0.3846 kJ/kg-K or 384.6 J/kg-C

(a)The original kinetic energy of the block is:

$E_k = \frac{mv^2}{2} = \frac{1.5*3^2}{2} = \frac{1.5*9}{2} = 6.75 J$

As 85% of this kinetic energy is converted to block internal heat energy, with specific heat we can calculate the rise in temperature:

$E_h = 0.85E_k = 0.85*6.75 = 5.7375 J$

$mc_c\Delta T = 5.7375$

$1.5*384.6\Delta T = 5.7375$

$\Delta T = \frac{5.7375}{1.5*384.6} \approx 0.01^oC$

(b) the remaining 15% energy would probably be dissipated to outside environment as heat energy

6 0

3 years ago

Simone is walking her dog on a leash. The dog is pulling with force of 34 N to the right and simone is pulling backward with a f

e-lub [12.9K]

I do not understand the full question, however if you are wondering which way Simone and the dog will go, they will go right because the force of 34 N from the dog is higher than the force of 16 N from Simone.

4 0

3 years ago