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3 years ago

Can anyone help me in this question?

Physics

1 answer:

Anna71 [15]3 years ago

3 0

Answer:

I guess the answer is charging by friction

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When did ernest rutherford make his discovery

oee [108]

Answer:

1911

Explanation:

"In 1911, he was the first to discover that atoms have a small charged nucleus surrounded by largely empty space, and are circled by tiny electrons, which became known as the Rutherford model (or planetary model) of the atom."

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3 years ago

In the periodic table, the most reactive metals are found a. in Group 1, the first column on the left. b. in Period 1, the first

Gnesinka [82]

Answer:

Bottom left corner for whatever group that is

Lithium, sodium, and potassium all react with water

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3 years ago

Read 2 more answers

The graph shows projected changes in the populations of the world.

malfutka [58]

Answer: C. Africa

Explanation:

The data given on the graph shows that:

Asia will grow from around 1,300 million to 5,000 million in 2050 which is an increase of:

= 5,000 /1,300 = 3.84 times

Europe will decrease over that period.

Africa will go from around 300 million to 2,000 million which is an increase of:

= 2,000 / 300

= 6.67 times

Mexico

, Central America, Caribbean Islands:

= 900 / 120

= 7.5 times

United States, Canada, and Greenland:

= 400/120

= 3.33

Oceania:

= 50 / 13

= 3.85

From the options give, Africa will see the greatest increase at 6.67 times its population in 1950.

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3 years ago

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Calculate the acceleration of the object from 3 seconds to 7 seconds (1pt). Show your work (1pt) and make sure to include the co

NARA [144]

Answer: 2m/s^2

Explanation: a=v-u/t2-t1

V=8m/s

U=0m/s

T1=3s

T2=7s

a=8-0/7-3

a=8/4

a= 2m/s^2

7 0

3 years ago

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A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta

Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, $\theta = 54^{\circ}$

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, $R = \frac{v^{2}sin2\theta}{g}$

$R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m$

Time of flight, t' = $\frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s$

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

$s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m$

Distance traveled vertically:

$s_{y} = vcos\theta - \frac{1}{2}gt^{2}$

$s_{y} = 12.4sin54^{\circ}\times 1.42 - \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m$

Now,

$s_{x} = \frac{mx + m'x'}{M}$

$10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}$

x' = 3.535 m

Similarly,

$s_{y} = \frac{my + m'y'}{M}$

$4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m$

The location of the other fragment is at (3.535, 1.162)

5 0

3 years ago