Ask question

Ask question

All categories

2 years ago

8

Matter is defined as anything that a. can be seen and touched. b. has mass and takes up space. can be weighed. d. contains kinet

ic or potential energy. Please select the best answer from the choices provided a. can be seen and touched by. has mass and takes up space c. can be weighed. or d.contains kinetic or potential energy. I will give branliy.

1 answer:

Nookie1986 [14]2 years ago

6 0

Answer:

b. has mass and takes up space.

Explanation:

Matter, in physics and chemistry, is defined as any substance that has mass and takes up space by having volume. The rest of the following options do not match this definition.

hope this helps!

You might be interested in

A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo

m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 . t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

p₀ = m1 v₀₁

Moment after shock

$p_{f}$ = (m1 + m2) $v_{f}$

p₀ = $p_{f}$

m1 v₀₁ = (m1 + m2) $v_{f}$

$v_{f}$ = v₀₁ m1 / (m1 + m2)

$v_{f}$ = 4.4 600 / (600 + 500)

$v_{f}$ = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

Em₀ = K = ½ (m1 + m2) $v_{f}$ ²

After compressing the spring

$E_{mf}$ = Ke = ½ k x²

As there is no rubbing the energy is conserved

Em₀ = $E_{mf}$

½ (m1 + m2) $v_{f}$ ² = = ½ k x²

x = $v_{f}$ √ (k / (m1 + m2))

x = 2.4 √ (11/3000)

x = 0.396 m

7 0

3 years ago

C'est quoi le lien entre l'atmosphere et les plantes?

Rus_ich [418]

Answer:

Les plantes produisent de l'oxygène et ont contribué à faire de la Terre une planète habitable. Grâce au processus de photosynthèse pendant la journée, les plantes absorbent le dioxyde de carbone de l'air, le convertissent en sucre et libèrent de l'oxygène dans l'atmosphère.

Les plantes consomment du dioxyde de carbone - un gaz à effet de serre important - au cours du processus de photosynthèse. La réduction du dioxyde de carbone dans l'atmosphère a un effet de refroidissement indirect. Les plantes refroidissent également l'atmosphère car elles libèrent de la vapeur d'eau lorsqu'elles deviennent chaudes, un processus similaire à la transpiration.

la température, l'humidité et l'intensité lumineuse autour de la plante; la concentration de dioxyde de carbone dans l'air autour des feuilles. La relation est inverse; autrement dit, à mesure que la concentration de CO2 augmente, le nombre de stomates produits diminue, et vice versa.

Explanation:

3 0

3 years ago

Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of

viktelen [127]

Answer:

$W=2.76\times 10^{-6}\ J$

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

$U=\dfrac{Q^2}{2C}$

Put all the values,

$U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J$

So, the work done is $2.76\times 10^{-6}\ J$ .

8 0

3 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

7 0

2 years ago

11. Will the cube in#10 float in water? Will it float in benzene?

densk [106]

Where is the cube I don't see any picture?

5 0

3 years ago