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sergij07 [2.7K]
4 years ago
5

Which experimental result led to the inference that atoms contained electrons?

Physics
2 answers:
masha68 [24]4 years ago
4 0

I learned this 2 years ago, and I'm pretty sure it's D :)

Otrada [13]4 years ago
3 0

Well most of the particles did pass through and a few were deflected. but i think the answer is A

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A rectangular tank is filled with water to a depth of 4m. Calculate the pressure at the bottom of the tank. PLEASEEE HELP!!
Paul [167]

The pressure in a liquid at a given depth is called the hydrostatic pressure. This can be calculated using the hydrostatic equation:

P = ρgh

P = (1000 kg/m³)(9.8 m/s²)(4 m)

<h2>P = 39200 Pa</h2>

4 0
3 years ago
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1. what factor does work depends on?<br>2. what factor does power depends on? ​
Harrizon [31]
1.motivation 2. ????
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3 years ago
What is the volume in millilitre of a 250 cm3 beaker
makkiz [27]

Answer:

250 mL

Explanation:

6 0
3 years ago
There is a basketball with a diameter of 9in. and a softball with a diameter of 4in. Find how many times greater the volume of t
erica [24]

Answer: #.37 times greater

Explanation: Volume of a sphere is: V = \frac{4}{3}·π·r³

Since it is given the diameter of the basketball and the softball, their ratios are r₁ = 3in and r₂ = 2in, in which index 1 represents the basketball and index 2, the softball.

Calculating the volume of basketball:

V₁ = \frac{4}{3} . 3.14 . 3³

V₁ = 113.04 in³

Calculating the volume of softball:

V₂ = \frac{4}{3} . 3.14 . 2³

V₂ = 33.5 in³

Comparing one volume with the other:

\frac{V1}{V2} = \frac{113.04}{33.5} = 3.37

We have that V₁ = 3.37V₂

This means the volume of a basketball is 3.37 times greater than the volume of a softball.

4 0
3 years ago
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a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
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