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3 years ago

Help this was due yesterday...

Physics

2 answers:

lyudmila [28]3 years ago

7 0

Answer:

Explanation:

plz help me with my most recent post if u can

insens350 [35]3 years ago

3 0

Answer:

Concept: Periodic Trends

So we have Li and we have C
Li has is lower in the periodic table and as you progress from s group to p group and eventually d group, the traction and ability to attract electrons increases.
Hence option D

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The graph represents the force applied on an 3.00kg crate while it moved 5.0m. A. How much total work is done on the crate? B. I

jeka57 [31]

a. We can calculate the amount of work by calculating the area under the graph.

first area (rectangular): 2.5 x 6 = 15

second area(trapezoid): 1/2 x (6+10) x 2.5 =20

total work done: 35 J

b. the force was first applied = 6 N

F = m.a

a = 6 : 3 = 2 m/s²

vf²=vi²+2as

vf²=6²+2.2.5

vf²=56

vf=7.5 m/s

8 0

3 years ago

A Roller Derby Exhibition recently came to town. They packed the gym for two

arlik [135]

Answer:

14.36m/s

Explanation:

From the law of conservation of linear momentum

m1u1 + m2u2 = v(m1 + m2)

68×17 + 76×12= v(68+76)

1156+912 = 144v

2068 = 144v

v = 2068/144

=14.36 m/s

7 0

4 years ago

The gravitational attraction between two objects will what is the object move further apart

finlep [7]

The farther apart the two objects, the weaker the gravitational attraction between them.

3 0

3 years ago

I need to Swap the convergent lens to a menisc Congergent one And make a graph like the attached photo

weeeeeb [17]

A convergent meniscus lens is a lens that is composed of two spherical surfaces, like the on shown next:

The imaginary line that runs through the middle of the lens is the "symmetry axis".

In this type of lenses incident parallel beams of light converge in one point, as follows:

And thus we get the diagram.

4 0

1 year ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

5 0

3 years ago