There are 8 players in a chess tournament. after the tournament each player has to shake hands with e

Answers

Answer 1

Answer: There are 8 players in a chess tournament. After the tournament each player has to shake hands with every other player. Which number sentence below is the best way to show the number of handshakes?
Answer :
7+6+5+4+3+2+1= 28

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Find the lengths of two segments whose sum is 45 and whose ratio is 2:3

Answers

Let the two segments be x and y
x+y=45...1
X/y = 2/3...2
From1, y=45-x
Sub this into 2
x/(45-x)=2/3
3x=90-2X
5x=90
x=18
y=45-18=27

Which percent is equal to the given fraction?3/5

A.
3.5%

B.
30%

C.
35%

D.
60%

Answers

D) 60%. an easy way to do this would be to multiply the denominator and the numerator by the same thing to make the denominator equal 100, and the numerator is the percent. For example, this question, how many time can 5 go into 100? 20 times. multiply 20 by 3, because what you do to the denominator you have to do to the numerator; so 20 x 3 = 60, 60%.

3/8, multiply by 12.5, since 8 goes into a hundred 12.5 times, to get 8 x 12.5 = 100 and 3 x 12.5 = 37.5, 37.5%

The anwer is d,because a persent is always out of hundred so,multiply the denominator by 100

write the equation if the line in slope- intercept form that passes through (7,-4) with a slope of -1.

Answers

9514 1404 393

Answer:

y = -x +3

Step-by-step explanation:

The point-slope form can be a useful place to start.

y -k = m(x -h) . . . . . line with slope m through point (h, k)

You require the line ...

y -(-4) = -1(x -7)

y = -x +7 -4 . . . . . . . . eliminate parentheses, add -4

y = -x +3 . . . . . . . . . slope-intercept form

Two blocks are connected by a very light string passing over a massless and frictionless pulley. The 20.0 N block moves 75.0cm to the right and the 12.0 N block moves 75.0cm downward.Find the total work done on 20.0N block if there is no friction between the table and the 20.0N block.

Answers

The string is assumed to be massless so the tension is the sting above the 12.0 N block has the same magnitude to the horizontal tension pulling to the right of the 20.0 N block. Thus,
1.22 a = 12.0 - T (eqn 1)
and for the 20.0 N block:
2.04 a = T - 20.0 x 0.325 (using µ(k) for the coefficient of friction)
2.04 a = T - 6.5 (eqn 2)

[eqn 1] + [eqn 2] → 3.26 a = 5.5
a = 1.69 m/s²

Subs a = 1.69 into [eqn 2] → 2.04 x 1.69 = T - 6.5
T = 9.95 N

Now want the resultant force acting on the 20.0 N block:
Resultant force acting on the 20.0 N block = 9.95 - 20.0 x 0.325 = 3.45 N
Units have to be consistent ... so have to convert 75.0 cm to m:

75.0 cm = 75.0 cm x [1 m / 100 cm] = 0.750 m
work done on the 20.0 N block = 3.45 x 0.750 = 2.59 J

Final answer:

The total work done on the 20.0 N block as it moves 75.0 cm to the right, with no friction between the table and the block, is 15 Joules.

Explanation:

In physics, work done by a force is given by the formula Work = Force x Distance x Cos θ, where θ is the angle between the force and the direction of motion. Here, the force is the same as the weight of the 20.0 N block (because weight = mass * gravity, and the mass is the weight divided by gravity, so mass * gravity = weight), and because the block moves horizontally (rightward), θ is 0 degrees. The Cos of 0 degrees is 1.

So, the work done on the 20.0 N block as it moves 75.0 cm (or 0.75 meters, because 1 m = 100 cm) to the right is 20.0 N * 0.75 m * 1 = 15 Joules.