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svet-max [94.6K]

1 year ago

10

Help me please forget about you and she was

1 answer:

ra1l [238]1 year ago

5 0

The figure is showing a volume of 2.4 mL becuase it's feel 4 little segments.

Therefore, the answer is 2.4 mL.

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Which description explains the flow of heat?

Lemur [1.5K]

Answer:

the answer is b

Explanation:

7 0

3 years ago

Read 2 more answers

Assume four 1 kohm resistors are connected so that they form a square. what is the equivalent resistance if the resistance is me

borishaifa [10]

If the resistors are arranged in a shape of a square, then they are in a series type of circuit. This circuit arrangement is a non-branching, one-way flow of electrons. The total resistance in a series circuit is the summation of the individual resistances, If you place the ohmmeter (measures resistance) on two non-adjacent sides, then, you are measuring the resistance of two of the resistors.

Resistance = 2(1 kΩ) = 2 kΩ

8 0

3 years ago

The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top

leva [86]

Answer:

44.1613858478 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 99.4

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 99.4+0^2}\\\Rightarrow v=44.1613858478\ m/s$

If air resistance was absent Dan Koko would strike the airbag at 44.1613858478 m/s

6 0

3 years ago

All of the following are ways in which sedimentary rocks form EXCEPT

Anna71 [15]

Except A. Lava cooling

6 0

3 years ago

A light spring stretches 0.13 m when a 0.35 kg mass is hung from it. The mass is pulled down from this equilibrium position an a

Alenkasestr [34]

Answer:

v = 1.30 m/s

Explanation:

given,

mass hung = 0.35 Kg

spring stretched when load is hanged (x)= 0.13 m

now,

weight of the mass attached = Kx

m g = k x

0.35 x 9.8 = k x 0.13

k = 26.38 N/m

now, using conservation of energy

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx'^2$

$v = \sqrt{\dfrac{kx'^2}{m}}$

$v = \sqrt{\dfrac{26.38 \times 0.15^2}{0.35}}$

$v = \sqrt{1.6958}$

v = 1.30 m/s

6 0

3 years ago