All categories

2 years ago

Frank has a paperclip. It has a mass of 12g and a volume of 3cm3. What is its density?

Physics

1 answer:

Mkey [24]2 years ago

4 0

Answer:

D = 4 g/cm³

Explanation:

Density = Mass / Volume

Step 1: Define

D = x

M = 12 g

V = 3 cm³

Step 2: Substitute

D = 12g/3 cm³

Step 3: Simplify

D = 4g / cm³

You might be interested in

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago

If transpiration didn't take place, would water be able to enter the roots of the plant? Explain your reasoning

s2008m [1.1K]

Answer:

If transpiration didn't take place water would still be able to enter the roots of a plant

Explanation:

transpiration is the process of water leaving from living organisms to the atmosphere, therefore, if transpiration didn't occur the water would not transpire to the atmosphere and would remain in the root but water absorption would not change because it is a biological need for the living organism as such

3 0

3 years ago

No force is necessary to

igor_vitrenko [27]

No force is necessary to keep a moving object moving (in a straight line at a constant speed).

4 0

3 years ago

N LC circuit has an oscillation frequency of 105 Hz. If C = 0.1 F , then L must be about:

Umnica [9.8K]

Answer:

L = 22.97 H

Explanation:

Given that,

Capacitance, $C=0.1\ \mu F=0.1\times 10^{-6}\ F$

Oscillation frequency, f = 0.5 Hz

The frequency of an AC circuit is given by :

$f=\dfrac{1}{2\pi \sqrt{LC} }$

Where

L is impedance

$f^2=\dfrac{1}{4\pi ^2LC}\\\\L=\dfrac{1}{4\pi ^2 f^2 C}\\\\\text{Putting all the values}\\\\L=\dfrac{1}{4\pi^2 \times (105)^2\times 0.1\times 10^{-6}}\\\\L=22.97\ H$

So, the impedance of LC circuit 22.97 H.

7 0

2 years ago

If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy

ruslelena [56]

The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

distance of the galaxy, x = 2,000 Mpc

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

$T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\ \times \ years}{3600 \ s \ \times\ 24\ h\ \times \ 365.25 \ days} \\\\T = 1.397 \times 10^{10} \ years\\\\T = 13.97 \ billion \ years$

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

8 0

2 years ago

Frank has a paperclip. It has a mass of 12g and a volume of 3cm3. What is its density?​

Frank has a paperclip. It has a mass of 12g and a volume of 3cm3. What is its density?