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3 years ago

Which cekk part is found in a plant cells but not animal cells?

Physics

2 answers:

liraira [26]3 years ago

7 0

The correct answer should be c

Citrus2011 [14]3 years ago

3 0

Answer:

cell wall ...

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hope it will help you.....

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One complete expression of a waveform beginning at a certain point, progressing through the zero line to the wave’s highest (cre

elixir [45]

Answer:

wavelength.

Explanation:

One complete expression of a waveform beginning at a certain point, progressing through the zero line to the wave’s highest (crest) and lowest (trough) points, and returning to the same value as the starting point is called a is called wavelength. Its can be also defined as the distance between two successive crests or trough points in wave form.

4 0

2 years ago

A gas in a cylinder expands from a volume of 0.110 m³ to 0.320 m³. heat flows into the gas just rapidly enough to keep the press

Elden [556K]

80000 Joule is the change in the internal energy of the gas.

<h3>In Thermodynamics, work done by the gas during expansion at constant pressure:</h3>

ΔW = -pdV

ΔW = -pd (V₂ -V₁)

ΔW = - 1.65×10⁵ pa (0.320m³ - 0.110m³)

= - 0.35×10⁵ pa.m³

= - 35000 (N/m³)(m³)

= -35000 Nm

ΔW = -35000 Joule

Therefore, work done by the system = -35000 Joule

<h3>Change in the internal energy of the gas,</h3>

ΔV = ΔQ + ΔW

Given:

ΔQ = 1.15×10⁵ Joule

ΔW = -35000 Joule

ΔU = 1.15×10⁵ Joule - 35000 Joule

= 80000 Joule.

Therefore, the change in the internal energy of the gas= 80000 Joule.

Learn more about thermodynamics here:

brainly.com/question/14265296

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3 0

1 year ago

Please help! I'm not sure what equation or the process to do this question.

lions [1.4K]

Answer:

The momentum is 1.94 kg m/s.

Explanation:

To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.

The potential energy $U$ of the compressed spring is given by

$U = \dfrac{1}{2} kx^2$ ,

where $x$ is the length of compression and $k$ is the spring constant.

And the kinetic energy of the ball is

$K.E = \dfrac{1}{2}mv^2.$

When the spring is released all of the potential energy of the spring goes into the kinetic energy of the ball; therefore,

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

solving for $v$ we get:

$v = x \sqrt{\dfrac{k}{m} }.$

And since momentum of the ball is $p=mv$ ,

$p =mx \sqrt{\dfrac{k}{m} }.$

Putting in numbers we get:

$p =(0.5kg)(0.25m) \sqrt{\dfrac{(120N/m)}{0.5kg} }.$

$\boxed{p=1.94kg\: m/s}$

5 0

2 years ago

A car accelerates at a rate of 3.0 m/s2. If its original speed is 8.0 m/s, how many seconds will it take the car to reach a fina

V125BC [204]

Answer:

5.7 s

Explanation:

Given:

a = 3.0 m/s²

v₀ = 8.0 m/s

v = 25.0 m/s

Find: t

v = at + v₀

25.0 m/s = (3.0 m/s²) t + 8.0 m/s

t ≈ 5.7 s

5 0

3 years ago

A car traveling in a straight line has a velocity of 5.0 m/s. After an acceleration of 0.75 m/s/s, the cars velocity is 8.0. In

Bogdan [553]

Vs - velocity on beginning
ve - velocity on ending. You've got:
$v_s = 5 \frac{m}{s} \\ v_e=8 \frac{m}{s} \\ \hbox{Then:} \\ \Delta v=v_e - v_s = 8 \frac{m}{s} - 5\frac{m}{s}=3 \frac{m}{s} \\ a=0,75 \frac{m}{s^2} \\ \hbox{And from formula:} \\ a=\frac{\Delta v}{\Delta t} \qquad \Rightarrow \qquad \Delta t= \frac{\Delta v}{a} \\ \hbox{Substitute:} \\ \Delta t=\frac{3\frac{m}{s}}{0,75 \frac{m}{s^2}}= \frac{3}{\frac{3}{4}} s= 3 \cdot \frac{4}{3} s= 4 s$
So he needed 4 second.

3 0

3 years ago