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3 years ago

10

a teacher performing a demonstration finds that a piece of cork displaces 23.5 mL of water. the piece of cork has a mass of 5.7g

. what is the density of the cork?

2 answers:

8090 [49]3 years ago

6 0

D = M/V.... it can be thought of as
D = ♡ with a line through it
............
D = .0057 kg / .0235 L
= .2425531915 kg/L
= .24 kg/L

____ [38]3 years ago

6 0

Explanation:

It is given that,

Mass of piece of cork, m = 5.7 g

Water displaced by the piece of cork, volume of the bar, $V=23.5\ mL=23.5\ cm^3$

We need to find the density of the platinum bar. The density of any substance is given by :

$d=\dfrac{m}{V}$

$d=\dfrac{5.7\ g}{23.5\ cm^3}$

$d=0.242\ g/cm^3$

So, the density of the platinum bar $0.242\ g/cm^3$ . Hence, this is the required solution.

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If I have a picosecond laser how would that be expressed in terms of zeptoseconds? In terms of petaseconds?

kolbaska11 [484]

Picosecond = 10 ^ -12 seconds.
Zeptosecond = 10^ -18 seconsds
Petaseonds = 10^15 seconds

To express Picoseconds into any of other two, you have to divide 10^-12 by the power index of the one in question

1Picosecond : 10^-12 / 10^-18 = 10^ (-12- 18) = 10^ (-12+18)= 10^6 zeptoseconds

1Picosecond : 10^-12 / 10^15 = 10^ (-12-15) = 10^-27 Petaseconds.

1Picosecond = 10^6 zeptoseconds

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3 years ago

In which of the following elevator situations would the acceleration be positive. Select TWO answers

kap26 [50]

Both options 5 and 6

Explanation:

Let us consider option 5,

In option 5 body is moving up with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.

Let us consider option 6,

In option 6 body is moving down with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.

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3 years ago

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him

nataly862011 [7]

Answer:

a) $u_c=0\ m.s^{-1}$ & $m_c.v_c=m_b.v_b\times \cos\theta$

b) $v_c=0.0566\ m.s^{-1}$

c) $p_e=2.9218\ kg.m.s^{-1}$

Explanation:

Given:

mass of the book, $m_b=1.35\ kg$

combined mass of the student and the skateboard, $m_c=97\ kg$

initial velocity of the book, $v_b=4.61\ m.s^{-1}$

angle of projection of the book from the horizontal, $\theta=28^{\circ}$

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

$u_c=0\ m.s^{-1}$

where:

$u_c=$ initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

$m_c.v_c=m_b.v_b\times \cos\theta$

where:

$v_c=$ final velcotiy of the student after throwing the book

b)

$m_c.v_c=m_b.v_b\times \cos\theta$

$97\times v_c=1.35\times 4.61\cos28$

$v_c=0.0566\ m.s^{-1}$

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

$p_e=m_b.v_b\sin\theta$

$p_e=1.35\times 4.61\times \sin28^{\circ}$

$p_e=2.9218\ kg.m.s^{-1}$

6 0

3 years ago

A car is traveling at a velocity of 22 m/s when the driver puts on the brakes

Brums [2.3K]

The car’s velocity at the end of this distance is <em>18.17 m/s.</em>

Given the following data:

Initial velocity, U = 22 m/s

Deceleration, d = 1.4 $m/s^2$

Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

$V^2 = U^2 + 2dS$

Substituting the values into the formula, we have;

$V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}$

<em>Final velocity, V = 18.17 m/s</em>

Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>

<em></em>

Read more: brainly.com/question/8898885

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2 years ago

A cyclist accelerates from a velocity of 10 miles/hour east until reaching a velocity of 20 miles/hour east in 5 seconds. What w

Sliva [168]

Answer:

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Explanation:

<u>Motion with Constant Acceleration</u>

A body moves with constant acceleration when the speed changes uniformly in time. The equation used to find the final speed vf is

$v_f=v_o+at$

Where vo is the initial speed, a is the acceleration, and t is the time.

The cyclist has an initial speed of vo=10 miles/hour and ends up at vf=20 miles/hour in t=5 seconds.

Both speeds are given in miles/hour and we must convert it to m/s:

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20 mile/hour = 8.94 m/s

The acceleration is calculated by solving for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

$\displaystyle a=\frac{8.94-4.47}{5}$

$a = 0.894\ m/s^2$

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2 years ago