Plant and animal cells are both eukaryotic cells that share many organelles and cell structures in common. However, each have unique organelles that perform specific functions in each cell type. Examine each of the lists of cell structures. Which of these are found in plant cells? Select ALL that apply.
A) nucleus, ribosomes, and chloroplasts
B) mitochondria, cell wall, chloroplasts
C) nucleus, mitochondria, and centrioles
D) cell membrane, ribosomes, small vacuoles
E) cell wall, central vacuole, and cell membrane

Considering the Coulomb's Law, the combination of particle A: 12.0×10⁻⁴ C and particle B: 2.5×10⁻⁴ C would yield the same Coulomb force of 12 N.

Charged bodies experience a force of attraction or repulsion on approach.

From Coulomb's Law it is possible to predict what the electrostatic force of attraction or repulsion between two particles will be according to their electric charge and the distance between them.

From Coulomb's Law, the electric force with which two point charges at rest attract or repel each other is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

[tex]F=k\frac{Qq}{d^{2} }[/tex]

where:

The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.

In this case, you know that two particles are 15 meters apart and the resulting Coulomb force is 12 N.

On the other side, you know:

Replacing in the Coulomb's Law, you get:

Finally, the combination of particle A: 12.0×10⁻⁴ C and particle B: 2.5×10⁻⁴ C would yield the same Coulomb force of 12 N.

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The change in angular momentum of the pulley-block system is given as [tex]\omega = Rm_ogt_o[/tex]

Data;

This is the force that makes object rotate about an axis. The formula is given as the product between the radius about the axis and the force acting upon it.

The torque about point o is given as

[tex]\tau = R * m_og[/tex]

The change in angular momentum about point 'o' is the product between the torque and time.

[tex]\omega = \tau * t_o\\\omega = Rm_ogt_o[/tex]

The change in angular momentum of the pulley-block system is given as

[tex]\omega = Rm_ogt_o[/tex]

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Answer:

0.5 mega grams

Explanation:

Answer:

50 N

Explanation:

Let the natural length of the spring = L

so

100 = k(40 - L)       (1)

200 = k(60 - L)       (2)

(2)/(1):   2 = (60 - L)/(40 - L)

60 - L = 2(40 - L)

60 - L = 80 - 2L

2L - L = 80 - 60

L = 20

Sub it into (1):

100 = k(40 - 20) = 20k

k = 100/20 = 5 N/in

Now

X = k(30 - L) = 5(30 - 20) = 50 N

Answer:

1 m/s^2

Explanation:

The formula for accleration is a=Δv/Δt

where, Δv = final velocity - initial velocity = 4 - 0 = 4

initial velocity = 0 since the car starts form rest and final velovity is 4 as the car goes from rest till 4 m/s

Δt = 4 since the car takes 4 seconds to reach a velocity of 4m/s

Hence, a = 4 m/s / 4 s = 1 m/s^2

Answer:

1m/s

Explanation:

Given

initial velocity(u)=0

final velocity(v)=4m/s

timetaken (t)=4

Aceleration(a)=v-u÷t

now

=4-0÷4

=1m/s

For a tire on a car is rolling smoothly and It's center of mass is moving at 18 m/s, the speed of the top of the tire is mathematically given as

Vtop=36m/s

Generally, the equation for the relationship between the top and the core is mathematically given as

Vtop=2*(vcore)

Threrefore

Vtop=2*18

Vtop=36m/s

In conclusion, the speed of the top

Vtop=36m/s

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i want the answer of this exercise

Answer:

7 minutes

Explanation:

Explanation:

Power = work / time

Power = force × distance / time

500 W = 210 N × 1000 m / t

t = 420 s

t = 7 min

Answer:

c

Explanation:

the question answered itself.

This is defined as the rate of change of position of an object with respect to time. The unit is metres per second(m/s).

Velocity = distance/ time

The velocity for column 1 = 500m/7.4s = 67.57m/s.

The velocity for column 2  = 500m/6.4s = 78.13m/s.

The velocity for column 3  = 500m/5.6s = 89.29m/s.

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Answer:

67.6

78.1

89.3

Explanation:

The balloon will attach to the wall because the balloon's negative charges will drive the electrons in the wall to shift to the other side of their atoms, leaving the wall's surface positively charged.

Electric potential difference is the difference in electric potential between two positions and it is measured in volts.

This is the potential energy of an electric charge has due to its location in a field.

The electric potential of a charged particle is the electric potential energy of the charged particle divided by its charge magnitude.

Electric potential difference is the difference in electric potential between two positions.

Volt is the SI unit of electric potential difference.

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Answer:

-259322.03N

Explanation:

[tex]F=m*(\frac{v}{t})\\ F=4500kg*(\frac{0-102m/s}{1.77s} )\\F=-259322.033898\\\\[/tex]

Answer:

electromagnetic radiation moves at the speed of light

Answer:

It could point her in a direction that best suits her

Explanation:

Hope that helps!

The statements that are true are;

Work is said to be done when te force applied travels a distance in the direction of the force. Now we can see that when the force is applied by shoving the crate, work is done, an additional work is done by friction to bring the crate to a stop.

Hence, the following are true;

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Answer:

Approximately [tex]1.1 \times 10^{4}\; {\rm m\cdot s^{-1}}[/tex] if air friction is negligible.

Explanation:

Let [tex]G[/tex] denote the gravitational cosntant. Let [tex]M[/tex] denote the mass of the earth. Lookup the value of both values: [tex]G \approx 6.67 \times 10^{-11}\; {\rm N\cdot m^{2}\cdot kg^{-2}}[/tex] while [tex]M \approx 5.697 \times 10^{24}\; {\rm kg}[/tex].

Let [tex]m[/tex] denote the mass of the meteor.

Let [tex]v_{0}[/tex] denote the initial velocity of the meteor. Let [tex]r_{0}[/tex] denote the initial distance between the meteor and the center of the earth.

Let [tex]r_{1}[/tex] denote the distance between the meteor and the center of the earth just before the meteor lands.

Let [tex]v_{1}[/tex] denote the velocity of the meteor just before landing.

The radius of planet earth is approximately [tex]6.371 \times 10^{6}\; {\rm m}[/tex]. Therefore:

Note the significant difference between the two distances. Thus, the gravitational field strength (and hence acceleration of the meteor) would likely have changed significant during the descent. Thus, SUVAT equations would not be appropriate.

During the descent, gravitational potential energy ([tex]\text{GPE}[/tex]) of the meteor was turned into the kinetic energy ([tex]\text{KE}[/tex]) of the meteor. Make use of conservation of energy to find the velocity of the meteor just before landing.

Initial [tex]\text{KE}[/tex] of the meteor:

[tex]\displaystyle (\text{Initial KE}) = \frac{1}{2}\, m\, {v_{0}}^{2}[/tex].

Initial [tex]\text{GPE}[/tex] of the meteor:

[tex]\displaystyle (\text{Initial GPE}) &= -\frac{G\, M\, m}{r_{0}}[/tex].

(Note the negative sign in front of the fraction.)

Just before landing, the [tex]\text{KE}[/tex] and the [tex]\text{GPE}[/tex] of this meteor would be:

[tex]\displaystyle (\text{Final KE}) = \frac{1}{2}\, m\, {v_{1}}^{2}[/tex].

[tex]\displaystyle (\text{Final GPE}) &= -\frac{G\, M\, m}{r_{1}}[/tex].
If the air friction on this meteor is negligible, then by the conservation of mechanical energy:

[tex]\begin{aligned}& (\text{Initial KE}) + (\text{Initial GPE}) \\ =\; & (\text{Final KE}) + (\text{Final GPE})\end{aligned}[/tex].

[tex]\begin{aligned}& \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} \\ =\; & \frac{1}{2}\, m\, {v_{1}}^{2} - \frac{G\, M\, m}{r_{1}}\end{aligned}[/tex].

Rearrange and solve for [tex]v_{1}[/tex], the velocity of the meteor just before landing:

[tex]\begin{aligned}{v_{1}} &= \sqrt{\frac{\displaystyle \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} + \frac{G\, M\, m}{r_{1}}}{(1/2)\, m}} \\ &= \sqrt{{v_{0}}^{2} - \frac{G\, M}{r_{0}} + \frac{G\, M}{r_{1}}} \\ &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)}\end{aligned}[/tex].

Substitute in the values and evaluate:

[tex]\begin{aligned}v_{1} &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)} \\ &\approx \sqrt{\begin{aligned}(& 6.5 \times 10^{3}\; {\rm m \cdot s^{-1}}) \\ & - [6.67 \times 10^{-11}\; {\rm N \cdot {m}^{2}\cdot {kg}^{2} \times 5.697\; {\rm kg}}\\ &\quad\quad \times (1 / (6.371 \times 10^{6}\; {\rm m}) - 1 / (1.9371 \times 10^{7}\; {\rm m}))]\end{aligned}} \\ &\approx 1.1 \times 10^{4}\; {\rm m\cdot {s}^{-1}}\end{aligned}[/tex].

(Note that assuming a constant acceleration of [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] would give [tex]v_{1} \approx 1.7\times 10^{4}\; {\rm m\cdot s^{-1}}[/tex], an inaccurate approximation.

Answer:

The question is not complete. Since force is a vector quantity we need to consider the direction of the two forces.

hence if they are acting opposite on the body . then the net force will be  30-10=20N20N

Acceleration=Force/mass

=20/4

=5m/s^2

Explanation:

Answer:

the two types of materials interacting (described by the coefficient μk) and how hard these two surfaces are pushed together (the normal force)

hope this helps and is it needs more let me know

Question :-

Answer :-

Explanation :-

As per the provided information in the given question, The Resistance is given as 330 Ohm . The Voltage is given as 5 Volt . And, we have been asked to calculate the Current .

For calculating the Current , we will use the Formula :-

[tex] \bigstar \: \: \: \boxed{ \sf{ \: Current \: = \: \dfrac{Voltage}{Resistance} \: }} [/tex]

Therefore , by Substituting the given values in the above Formula :-

[tex] \dag \: \: \: \sf {Current \: = \: \dfrac {Voltage}{Resistance} } [/tex]

[tex] \longmapsto \: \: \: \sf {Current \: = \: \dfrac {5}{330} } [/tex]

[tex] \longmapsto \: \: \: \sf {Current \: = \: \dfrac {1}{66} } [/tex]

[tex] \longmapsto \: \: \: \textbf {\textsf {Current \: = \: 66 }} [/tex]

Hence :-

[tex] \underline {\rule {180pt} {4pt}} [/tex]

Additional Information :-

[tex] \Longrightarrow \: \: \: \sf {Voltage \: = \: Current \: \times \: Resistance} [/tex]

[tex] \Longrightarrow \: \: \: \sf {Current \: = \: \dfrac {Voltage}{Resistance} } [/tex]

[tex] \Longrightarrow \: \: \: \sf {Resistance \: = \: \dfrac {Voltage}{Current} } [/tex]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

[tex]\text{Voltage,}~ V = IR =33.72\times 32 = 1079.04~ \text{volt}[/tex]

When the speed of the block is plotted on the vertical axis and displacement of the block plotted on the horizontal axis, a linear graph will be obtained showing the inverse relationship between the speed of the block and the displacement of the block.

Hooke's law states that force or load applied to elastic material is directly proportional to the extension produced in the material. This law can be written as;

F = kx

where;

The speed of the block increases with decreasing displacement of the bock. This is because the kinetic energy of the block is maximum at zero displacement and minimum at maximum displacement of the block.

Thus, when the speed of the block is plotted on the vertical axis and displacement of the block plotted on the horizontal axis, a linear graph will be obtained showing the inverse relationship between the speed of the block and the displacement of the block.

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We know that:

[tex]0.5m(vB^{2} -vA^{2} )=q*[VB-VA]\\== > vB^{2} =vA^{2} +(\frac{2q}{m})*[VB-VA]\\ =(4.30*10^{4} )^{2} +(\frac{2*1.6*10^{-19} }{2*1.67*10^{-27} }) [-10-30]\\[/tex]

[tex]=18.49*10^{8} -76.64*10^{8}[/tex]

[tex]=-58.15*10^{8} \\== > vB=-7.62*10^{4} m/s[/tex]

The proton's speed at point B is  -7.62 x ×10⁴ m/s.

The proton is a subatomic particle lies inside the nucleus of an atom of element. The number of proton gives an idea of the atomic number of the element.

The proton's speed as it passes point A is 4.00×10⁴ m/s . It follows the trajectory shown.

1/2m (vb² - va²) = q (Vb - Va)

vb² =  va² +2q/m (Vb - Va)

where v is the velocity and V is the potential at point A and B, q ia the charge on proton and m is the mass of proton.

Substitute the given values, we get

vb² = (4.00×10⁴ ) + (2x1.6 x 10⁻¹⁹/1.67 x 10⁻²⁷) (-10 - 30)

vb = -7.62 x ×10⁴ m/s

Thus, proton's speed at point B is  -7.62 x ×10⁴ m/s.

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Answer:

Explanation:

Speed of light v = 3 x 10⁸ m/s

wavelength λ = 8.1 x 10⁻⁸ m

frquency f = v/λ = 3.7 x 10¹⁵ Hz

If a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then the frequency of the light would be 3.7 × 10¹⁵ Hz, as the wavelength and the frequency of the photon are inversely proportional to each other.

It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.

C = λν

As given in the problem if a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then we have to find out the frequency of the light,

The frequency of the light = 3 × 10⁸ / 8.1 x 10⁻⁸

                                            =3.7 × 10¹⁵ Hz

Thus, the frequency of the light would be 3.7 × 10¹⁵ Hz

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Answer:

See below.

Explanation:

According to the question, we know that,

work done is given by,  [tex]W=qV[/tex]

and change in kinetic energy is, Δ [tex]KE=W=1=1/2[mv^{2} ][/tex]

therefore equating both the equations we get,

[tex]qV=1/2[mv^{2} ][/tex] ⇒ [tex]V=\frac{mv^{2} }{2q}[/tex]

m= mass of electron =  [tex]9.1*10^{-31} kg[/tex]

q= charge on an electron = [tex]1.6*10^{-19} C[/tex]

v= speed of electron= 700000m/s

substituting the values in the above equation, we get

[tex]V=\frac{9.1*10^{-31} *(700000)^{2} }{2*1.6*10^{-19} } =1.39V[/tex]

(1).  the potential difference that stopped the electron is 1.39 volts.

now the kinetic energy equation is :  2 ways[tex]KE=1/2[mv^{2} ]=\frac{9.1*10^{-31} *700000^{2} }{2} =2.22*10^{-19} J\\[/tex]

or [tex]KE=\frac{2.22*10^{-19} }{1.6*10^{-19} } =1.39eV[/tex]

(2).  the initial kinetic energy of the electron is 1.39eV.

Answer:

Explanation:

A television set is plugged into a 120 V outlet. The television circuit carries a current equal to 0.75 A. What is the overall resistance of the television set?​

V = IR

120 volt = 0.75A * R

R = 160 Ω

Answer:

load

a generator, a light bulb (load) and a closed switch

Explanation:

as explained in the other question, the fan is using generated electric energy to create mechanical movement. as such it is a load on the grid or circuit or net.

and electric power can only flow, if there is a closed (uninterupted) circuit from the power source to a load and back.

any open switch is an interruption of the circuit.

a buzzer is a kind of switch. it closes the circuit (and puts a load on) only when somebody presses it.

by the way, a closed circuit without a load will "destroy" (short circuit) the power source or at least the wires (burn through).

Answer:

Mechanical Energy (initial) = Mechanical energy (final)

Ep(initial) + Ek(initial) = Ep(final) + Ek(final)

mgh(initial) + 1/2mv²(initial) = mgh(final) + 1/2mv²(final)

m being the mass of the penguin (kg)

g being gravitational acceleration (9.80 m/s²)

v(initial) being zero

h(final) being zero,i.e. final height is zero

v(final) being final velocity

solve for h(initial) being initial height at the top of the glacier

answer will be in metres

NB: don't forget to square the velocity

Explanation:

Since they mentioned frictionless, we know that this is a closed system therefore we are free to use this equation of conservation of mechanical energy

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