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Neporo4naja [7]
3 years ago
10

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 ...

please mark me as brainlist .....

hope it will help you.....

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Where are the two lenses located in a compound microscope use in most classrooms today?
Fittoniya [83]
They stay with the microscope as it moves around to different schools, and they are always located in the same classroom where the rest of the microscope is being used.
8 0
3 years ago
What is the likely identity of a metal if a sample has a mass of 63.5 g when measured in air and an apparent mass of 60.2 g when
Gre4nikov [31]

Answer:

Gold

Explanation:

Given:

Mass of sample = 63.5 g

Mass of water = 60.2 g

Find:

Object

Computation:

Mass of water displaced = 63.5 g - 60.2 g

Mass of water displaced = 3.3 g

So, volume in water = 3.3 cm³

Density = Mass / Volume

Density = 63.5 g / 3.3

Density = 19.24

So,

Object ,must be gold.

7 0
3 years ago
Question:
exis [7]

Answer:

She can swing 1.0 m high.

Explanation:

Hi there!

The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

And the potential energy:

PE = m · g · h

Where:

m = mass of Jane.

v = velocity.

g = acceleration due to gravity (9.8 m/s²).

h = height.

Then:

ME = KE + PE

Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:

ME = KE + PE      (PE = 0)

ME = KE

ME = 1/2 · m · (4.5 m/s)²

ME = m · 10.125 m²/s²

When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:

ME = KE + PE      (KE = 0)

ME = PE

ME = m · 9.8 m/s² · h

Then, equallizing both expressions of ME and solving for h:

m · 10.125 m²/s² =  m · 9.8 m/s² · h

10.125 m²/s² / 9.8 m/s²  = h

h = 1.0 m

She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).

6 0
3 years ago
You are on a cruise ship traveling north at a speed of 13 m/s with respect to land. 1) if you walk north toward the front of the
Brrunno [24]

In order to find the our own velocity with respect to land,we need to apply the theory of relative velocity.


Now consider the velocity of the ship traveling towards the north with respect to land as A.Consider our own velocity headed northwards as B.

The relative velocity is the velocity that the body A would appear to an observer on the body B and vice versa.


In this case the relative velocity would be arrived by summing up our velocity with the velocity of the ship as the object (I) is travelling in the ship.


Relative velocity = Velocity of Body A+ Velocity of Body B.


Velocity of the ship traveling towards the north with respect to land(A)= 13.0m/s. (Given)

Our own velocity headed northwards(B)= 2.8 m/s.



Relative velocity = Velocity of Body A+ Velocity of Body B.


Relative velocity= 13.0 + 2.8 = 15.8m/s.


Thus our own velocity with respect to the land is 15.8 m/s.



5 0
3 years ago
A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS
Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux  GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;  

transferred energy + generated energy = 0

(radiation + convection) +  generated energy = 0

[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0

[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s

(a) the electrical power generated for still summer day

T_s = T_{\infty} = 35 ^oC = 308 \ k

[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \  \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k

P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W

(b)the electrical power generated for a breezy winter day

T_s = T_{\infty} = -15 ^oC = 258 \ k

[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \  \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k

P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W

3 0
3 years ago
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