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3 years ago

How did Rutherford's experiments demonstrate that Thomson's model of the atom was incorrect?

2 answers:

8 0

Rutherford's Gold foil experiment involved the firing of atoms at a piece of gold foil only a few atoms thick. Rutherford expected that the particles would go straight through, but some were deflected. This experiment proved that atoms have a dense positively charged nucleus.

Answer: d. They showed that atoms have a dense nucleus.

mestny [16]3 years ago

8 0

Answer:

d) They showed that atoms have a dense nucleus.

Explanation:

Thomson's model predicted that the atom is composed of negatively charged electrons, which is surrounded by positively charged cloud. This was called Plum pudding model of the atom.

But Rutherford proved it to be incorrect . His gold foil experiment showed that the atom's central part is heavy and positively charged. The deflections of the alpha particles showed that the heavy central region is surrounded by negatively charged particles called electrons.

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A steel ball has a diameter of 0.125 m. How much do you need to raise its temperature to make it expand by 0.000200 m (unit=C)?

Ivanshal [37]

Answer:

The temperature raise is = 0.14 [C]

Explanation:

In order to find the temperature difference we must use the following equation:

$V=V_{0} *\beta *DT\\where:\\V = final volume [m^{3} ]\\V_{0} = original volume [m^{3} ]\\\beta =34*10^-6[\frac{1}{C} ]\\DT= temperature difference [C]$

But first we need to find the original volume V:

$V0 = \frac{4}{3} *\pi *r^{3} \\D = diameter = 0.125 [m]\\r = radius =0.0625[m]\\replacing\\V=\frac{4}{3} *\pi *(0.0625^{3} )\\V0=1.022[m^{3} ]\\$

When the volume is increased the diameter will be larger:

D1 = 0.125[m] + 0.0002[m] = 0.1252[m]

Therefore the new radius will be:

r1=0.0626[m]

The new volume is:

$V=\frac{4}{3} *\pi *(0.0626)^{3} \\V=1.027*10^-3[m^{3} ]$

Replacing all the calculated values in the equation:

$(1.027*10^-3 - 1.022*10^-3)= 34*10^-6*DT\\DT= 0.14[C]$

3 0

3 years ago

Question 1 (1 point) There were 3 friends that decided to race their bikes down a hill without peddling. They all approached the

sattari [20]

Answer:

Explanation:

For the questions we will be using the formula;

acceleration = change in velocity/Time

acceleration = v-u/t

v is the final velocity

u is the initial velocity

t is the time

1) v = 15m/s, u = 2m/s t = 15s

a = 15-2/15

a = 13/15

a = 0.87m/s²

Their acceleration was 0.87m/s²

2) u = 10m/s, v = 45m/s, t = 1.75s

a = 45-10/1.75

a = 35/1.75

a = 35/(175/100)

a = 35×100/175

a = 3500/175

a = 20m/s²

3) u = 35m/s, v = 0m/s (comes to a stop), t = 1.25m/s

a = 0-35/1.25

a = -35/1.25

a = -28m/s²

The acceleration of the car is -28m/s²

4) v =55m/s, u = 0m/s t = 12s

a = 55-0/12

a = 55/12

a = 4.58m/s²

His acceleration when he got to the turn is 4.58m/s²

5) u = 17m/s (starting velocity)

v = 0m/s (stopping velocity)

t = 4.5s

a = 0-17/4/5

a = -17/4.5

a = -3.78m/s²

The the acceleration of the ball was -3.78m/s²

7 0

3 years ago

Describe each type of biotechnology:

Dovator [93]

Answer:

Genetic Engineering: the direct manipulation of an organisms genes using biotechnology (ex. genetically engineering corn to be insect resistant)

Cloning: The process of producing genetically identical individuals of an organism either natually or artificially,cloning allows for the creation of multiple copies of genes, expression of genes, and study of specific genes (ex. Cloning cells for untreatable diseases)

Artificial Selection: Also known as Selective breeding it is the process by which humans use animal breeding and plant breeding to selectively develop particular phenotypic traits/ characteristics (ex. Dog breeding)

4 0

3 years ago

an object has a constant acceleration of 3.2 m/s^2.at a certain instant its velocity is 10 m/s . what is its velocity after 5 se

aleksandrvk [35]

So here, you're looking for distance. The formula is D=vt+1/2at^2.

Lets plug in the informations.

10 m/s is our v (initial velocity)

5 second is out t(time)

3.2 m/s is our a(acceleration)

10m/s(5)+1/2(3.2m/s+5^2)

50m/s+1/2(28.2)

50m/s+14.1

Answer =64.1 m

Glad to help you out buddy. Let me know if you need help.

5 0

3 years ago

A wire of length 6cm makes an angle of 20° with a 3 mT

Crazy boy [7]

Answer:

Approximately $7.3 \times 10^{-3}\; \rm A$ (approximately $7.3\; \rm mA$ ) assuming that the magnetic field and the wire are both horizontal.

Explanation:

Let $\theta$ denote the angle between the wire and the magnetic field.

Let $B$ denote the magnitude of the magnetic field.

Let $l$ denote the length of the wire.

Let $I$ denote the current in this wire.

The magnetic force on the wire would be:

$F = B \cdot l \cdot I \cdot \sin(\theta)$ .

Because of the $\sin(\theta)$ term, the magnetic force on the wire is maximized when the wire is perpendicular to the magnetic field (such that the angle between them is $90^\circ$ .)

In this question:

$\theta = 20^\circ$ (or, equivalently, $(\pi / 9)$ radians, if the calculator is in radian mode.)
$B = 3\; \rm mT = 3 \times 10^{-3}\; \rm T$ .
$l = 6\; \rm cm = 6 \times 10^{-2}\;\rm m$ .
$F = 1.5\times 10^{-4}\; \rm N$ .

Rearrange the equation $F = l \cdot I \cdot \sin(\theta)$ to find an expression for $I$ , the current in this wire.

$\begin{aligned} I &= \frac{F}{l \cdot \sin(\theta)} \\ &= \frac{3\times 10^{-3}\; \rm T}{6 \times 10^{-2}\; \rm m \times \sin \left(20^{\circ}\right)} \\ &\approx 7.3 \times 10^{-3}\; \rm A = 7.3 \; \rm mA\end{aligned}$ .

5 0

3 years ago