All categories

2 years ago

Can someone help me match the words?

Physics

1 answer:

Hatshy [7]2 years ago

6 0

Answer:

a - 16

b - 2

c - 8

d - 13

e - 5

f - 11

g - 10

h - 17

i - 9

j - 3

k - 6

l - 12

m - 4

n - 14

o - 15

p - 1

Sorry if u get some of them wrong I did my best, for some reason number 7 has no match because there are 17 defenitions (1,2,3..etc) and 16 words (a,b,c..etc). Good Luck! Hope this helps :))

You might be interested in

A ball is thrown up into the air with an initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed

kaheart [24]

Answer:

B) t = 1.83 [s]

A) y = 16.51 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f} =v_{o} -g*t$

where:

Vf = final velocity = 0

Vo = initial velocity = 18 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.

A) The maximum height is reached when the final velocity of the ball is zero.

0 = 18 - (9.81*t)

9.81*t = 18

t = 18/9.81

t = 1.83 [s], we found the answer for B.

Now using the following equation.

$y = y_{o} + v_{o}*t - 0.5*g*t^{2}\\$

where:

y = elevation [m]

Yo = initial elevation = 0

y = 18*(1.83) - 0.5*9.81*(1.83)²

y = 16.51 [m]

7 0

2 years ago

I NEED AT LEAST A LIST OF 10 SONGS OF BEETHOVEN'S<br><br>

DanielleElmas [232]

Answer:

<u>Here are some of the songs of Beethoven's</u>:–

Septet.
Moonlight Sonata.
Pathetique Sonata.
Adelaide (Most popular).
Eroica Symphony.
Fifth Symphony.
Fidelio.
Emperor piano concerto.

3 0

2 years ago

Read 2 more answers

How long tg does it take for the balls to reach the ground? use 10 m/s2 for the magnitude of the acceleration due to gravity.

erik [133]

Answer:

1.0s

Explanation:

distance = 1/2 × acceleration × time2 + intial speed × time

5 0

2 years ago

Convert mm3 into m3.

bixtya [17]

Answer:

$1 \times 10 { - }^{9}$

cubic metre or 1e-9

Explanation:

•By division. Number of cubic millimetre divided(/) by 1000000000, equal(=): Number of cubic metre.

•By multiplication. 83 mm3(s) * 1.0E-9 = 8.3E-8 m3(s)

4 0

2 years ago

A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with

gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm$

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s$

The block's speed is 31.422 cm/s

4 0

3 years ago