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zhuklara [117]
3 years ago
14

a scientist is creating a new synthetic material. the material’s density is 6.1 g/cm3. which sentences describe how the scientis

t can lower the material’s density?
Physics
2 answers:
icang [17]3 years ago
6 0
<span>A.) The scientist can decrease the mass while keeping the volume constant.
B.) The scientist can increase the mass while keeping the volume constant.
C.) The scientist can decrease the volume while keeping the mass constant.
D.) The scientist can increase the volume while keeping the mass constant.
</span>-Plato answer choices for this question 
Oksana_A [137]3 years ago
5 0

Answer:

A,D. Decrease of Mass, and increase of Volume.

:D Hope this helps. Stay amazing.

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The pressure of the earth's atmosphere at sea level is 14.7 lb/in2. What is the pressure when expressed in g/m2? (2.54 cm = 1 in
miss Akunina [59]
<h2>Answer:</h2>

14.7 lb / in² = 10333018.166 g / m²

<h2>Explanation:</h2>

Given from the question;

pressure = 14.7lb/in²

<u>To convert from lb / in² to g / m²</u>

Note;

2.54 cm = 1 in       ------------(i)

But;

100 cm = 1m

=> 2.54cm = 2.54 cm x 1 m / 100 cm = 0.0254m

=>2.54cm = 0.0254m

Substitute 2.54cm = 0.0254m into equation(i)

=> 0.0254m = 1 in

<em>Note also;</em>

2.205 lb = 1 kg            ------------------(ii)

But;

1kg = 1000g

Substitute 1kg = 1000g into equation (ii)

=> 2.205 lb = 1000g   ---------------------(iii)

<em>From equation(iii);</em>

if, 2.205 lb = 1000g

then, 1 lb = 1 lb x 1000 g / 2.205 lb = 453.5g

=> 1 lb = 453.5 g

<em>Now let's convert 14.7 lb/in² to g / m²;</em>

=> 14.7 lb / in² could be written as 14.7 x 1 lb / (1 in x 1 in)

i.e

=> 14.7 lb / in² = 14.7 x 1 lb / (1 in x 1 in)     ------------------(iv)

<em>Substitute the values of 1 lb = 453.5g and 1 in = 0.0254m into equation(iv)</em>

=> 14.7 lb / in² = 14.7 x 453.5g / (0.0254m x 0.0254m)

=> 14.7 lb / in² = 6666.45 g / (0.00064516m²)

=> 14.7 lb / in² = 10333018.166 g / m²

<em>Therefore, 14.7 lb / in² = 10333018.166 g / m²</em>

5 0
3 years ago
Use the data from Table D of your Student Guide to answer the questions. Be sure to round your answers to the
Vikentia [17]

Answer:

2kg cart = 16.0

4kg cart =8.1

6kg cart =5.3

Explanation:

8 0
3 years ago
Read 2 more answers
A small rock is thrown vertically upward with a speed of 27.0 m/s from the edge of the roof of a 21.0-m-tall building. The rock
lbvjy [14]

Answer:

A. 33.77 m/s

B. 6.20 s

Explanation:

Frame of reference:

Gravity g=-9.8 m/s^2; Initial position (roof) y=0; Final Position street y= -21 m

Initial velocity upwards v= 27 m/s

Part A. Using kinematics expression for velocities and distance:

V_{final}^{2}=V_{initial}^{2}+2g(y_{final}-y_{initial})\\V_{f}^{2}=27^{2}-2*9.8(-21-0)=33.77 m/s

Part B. Using Kinematics expression for distance, time and initial velocity

y_{final}=y_{initial}+V_{initial}t+\frac{1}{2} g*t^{2}\\==> -0.5*9.8t^{2}+27t+21=0\\==> t_1 =-0.69 s\\t_2=6.20 s

Since it is a second order equation for time, we solved it with a calculator. We pick the positive solution.

8 0
3 years ago
The ISS was the first space station to be built; it is the largest space station and is actually still under construction as of
const2013 [10]

Thank you for that exciting news bulletin.

Here are a few more to keep you up to date:

-- The armistice has been signed, as of November 11.

-- Lindy made it !

-- The Japanese have surrendered to General Macarthur.

-- Secretary Clinton has won in the popular vote.

8 0
3 years ago
Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was
tigry1 [53]

Answer:

\Delta V=316167V

Explanation:

The difference of electric potential between two points is given by the formula \Delta V=Ed, where <em>d</em> is the distance between them and<em> E</em> the electric field in that region, assuming it's constant.

The electric field formula is E=\frac{F}{q}, where <em>F </em>is the force experimented by a charge <em>q </em>placed in it.

Putting this together we have \Delta V=\frac{Fd}{q}, so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight <em>W</em>, giving a net force N=W-F. On the first drop only <em>W</em> acts, while on the second drop is <em>N</em> that acts.

Using the equation for accelerated motion (departing from rest) d=\frac{at^2}{2}, so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

a_1=\frac{2d}{t_1^2}

a_2=\frac{2d}{t_2^2}

These relate with the forces by Newton's 2nd Law:

W=ma_1

N=ma_2

Putting all together:

N=W-F=ma_1-F=ma_2

Which means:

F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})

And finally we substitute:

\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})

Which for our values means:

\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V

7 0
3 years ago
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