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Which part of john dalton’s theory was disproved by j. J. Thomson’s cathode-ray tube experiments?.

Leni [432]

Answer:

While controversial at first, Thomson's discoveries were gradually accepted by scientists. Eventually, his cathode ray particles were given a more familiar name: electrons. The discovery of <em>the electron</em> disproved the part of Dalton's atomic theory that assumed atoms were indivisible.

5 0

3 years ago

Physics question, please answer this ASAP, thanks!

miv72 [106K]

(a) The gravitational force received by each 1 kg mass is 8.66 N.

(b) The magnitude of gravitational acceleration is 8.66 m/s².

(c) The orbital speed of the ISS is 7,663.6 m/s.

(d) The time take for the ISS to orbit round the Earth is 5,558.75 s = 1.54 hours.

<h3>Gravitational force received by each 1 kg mass</h3>

The gravitational force received by each 1 kg mass is calculated as follows;

F = Gm₁m₂/r²

where;

m₁ is mass of Earth
m₂ is mass of ISS
r is the distance between the ISS and center of Earth

F = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ x 1) / (6780,000)²

F = 8.66 N

<h3>Magnitude of gravitational acceleration</h3>

mg = GMm/r²

g = GM/r²

where;

M is mass of Earth

g = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ ) / (6780,000)²

g = 8.66 m/s²

<h3>Orbital Speed of the ISS</h3>

v = √GM/r

v = √(6.67 x 10⁻¹¹ x 5.97 x 10²⁴ / 6780,000)

v = 7,663.6 m/s

<h3>Time of motion of the ISS round the Earth</h3>

T = 2πr/v

T = (2π x 6780,000) / (7663.6)

T = 5,558.75 seconds

1 hour = 3600 seconds

= 5,558.75/3600

= 1.54 hours

Thus, the gravitational force received by each 1 kg mass is 8.66 N.

The magnitude of gravitational acceleration is 8.66 m/s².

The orbital speed of the ISS is 7,663.6 m/s.

The time take for the ISS to orbit round the Earth is 5,558.75 s = 1.54 hours.

Learn more about orbital speed here: brainly.com/question/22247460

#SPJ1

4 0

2 years ago

Stokes' law describes sedimentation of particles in liquids and can be used to measure viscosity. Particles in liquids achieve t

anygoal [31]

Answer: $\eta =325.73\times 10^{-3}=0.325 kg/m-s$

Explanation:

Given

density $(\rho )=7.8\times 10^3 kg/m^3$

Diameter(d)=2.4 mm

time taken=10 s

Distance moved(h)=0.75 m

At terminal velocity Drag force is equal to Weight

$F_D=mg$

Volume of ball $=\frac{4\pi r^3}{3}=7.23 mm^3$

Mass of ball $=\rho v=7.23\times 7.8\times 10^3\times 10^{-9}=56.39\times 10^{-6} kg$

$F_D=56.39\times 10^{-6}\times 9.8=552.66\times 10^{-6} N$

Also for spherical bodies drag force is equal to Stock Force

$F_s=6\times \pi \times \eta \times r\times v_r$

Where $v_r$ = Terminal velocity

$v=\frac{h}{t}=\frac{0.75}{10}=0.075 m/s$

$552.66\times 10^{-6}=6\times pi\times \eta \times \1.2\times 10^{-3}\times 0.075$

$\eta =325.73\times 10^{-3}=0.325 kg/m-s$

4 0

4 years ago

Please help me find a) and b):

Triss [41]

Answer:

A=3.33m/s

B=3.11m/s

Explanation:

Speed = total distance/time taken to cover distance

hence,

S=d/t

s=32.0/9.60

s=3.33m/s

acceleration due to gravity

v=u+at

that is (v=final velocity,u=initial velocity,a=acceleration and t= time)

from the formula

2.10=32.0+a*9.60

2.10=32.0+9.60a

CLT

9.60a=32.0-2.10

9.60a=29.9

a=29.9/9.60

a=3.11m/s

hope it correct!?

7 0

3 years ago

Use vectors to decide whether the triangle with vertices P(1, −2, −1), Q(2, 1, −3), and R(6, −1, −4) is right-angled. Yes, it is

Andreas93 [3]

Answer:

Yes, it is right-angled

Explanation:

Two vectors are orthogonal if the scalar product between them is zero. Then, we will match each pair of vertices with a vector, wich is formed with the following formula:

Given two points A and B, the vector going from A to B is

$AB=B-A=(B_{x}-A_{x},B_{y}-A_{y},B_{z} -A_{z})$