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Name the three major parts of a cell and their functions

Answer:

Explanation:

a gravitational field is the region of space surrounding a body in which another body experiences a force of gravitational attraction.

The formula is: weight/mass = gravitational field strength. On Earth the gravitational field strength is 10 N/kg.

an example of how you can use it if you need to find the gravitational force from earth.

Answer:

In space

Explanation:

I took the quiz

Answer:

in space

Explanation:

Answer:

A

Explanation:

I believe the correct answer is atmosphere (D).

Answer:

D) atmosphere

Explanation:

but the correct answer is troposhare

Answer:

The correct option is;

a. 107 m, 10.1°, east of south

Explanation:

The given question can be answered using vectors as follows;

The initial displacement south by Bradley = 65.0 meters

In vector form, we have

The initial displacement south by Bradley = -65·j

The displacement 25.0° east of south = 44.0 meters

In vector form, we have

The displacement 25.0° east of south = 44 × sin(25°)·i - 44 × cos(25°)·j

In vector, form, we have;

The total displacement = The sum of the vectors of the two displacement

∴ The total displacement = -65·j + 44 × sin(25°)·i - 44 × cos(25°)·j

The total displacement = 44 × sin(25°)·i - (44 × cos(25°) + 65)·j

We use the i and j components to find the magnitude and the direction of the total displacement as follows;

The magnitude of the total displacement = √((44 × sin(25°))² + (44 × cos(25°) + 65)²) = 106.5132881 ≈ 107

The magnitude of the total displacement ≈ 107 m

The direction of the total displacement = arctan((44 × sin(25°))/((44 × cos(25°) + 65)) = 10.0542797419° ≈ 10.1° east of south

Answer:

Kinetic energy is energy an object has because of its motion and is equal to one-half multiplied by the mass of an object multiplied by its velocity squared (KE = 1/2 mv2). Kinetic energy is greatest at the lowest point of a roller coaster and least at the highest point.

Explanation:

here is an example Where is the apple when it has the most kinetic energy? When the apple is falling and is right before it hits the ground. As an object falls under the influence of gravity, kinetic energy and potential energy are equal everywhere/at the halfway point only. 10. As an object falls under the influence of gravity, potential energy is greater than kinetic energy after halfway point/ before the halfway point

In the roller coaster, the kinetic energy is greatest at point X.

The kinetic energy of a body is given by 1/2mv^2.

Where;

m = mass of the object

v =velocity of the object.

For a roller coaster, the velocity is maximum at the lowest points of the roller coaster. If the velocity is maximum at the lowest point, it also follows that the kinetic energy is also maximum at the lowest point. Hence point X has the highest kinetic energy.

Learn more: https://brainly.com/question/999862

Explanation:

Ribosomes are not quite the same as different organelles since they have no layer around them that isolates them from different organelles, they comprise of two subunits, and when they are creating sure proteins they can become film bound to the endoplasmic reticulum, however they can likewise be free drifting while at the same time performing

Answer:

Dead sea is full of salt.

Explanation:

The water of the Dead Sea is full of salt, which makes it much denser and heavier than freshwater. If you swim in it, you float very easily.

When salt is dissolved in water, for example sea water, dissolved salt adds to the mass of the water and makes the water denser than it would be without salt. Because objects float better on a dense surface, they float better on salt water than fresh water.

Answer:3

Explanation:

Answer:

[tex]\boxed {\boxed {\sf 500 \ Joules}}[/tex]

Explanation:

Work can be found by multiplying the force by the distance.

[tex]W=F*d[/tex]

The force is 100 Newtons and the distance is 5 meters. Therefore,

[tex]F= 100 \ N \\d= 5 \ m[/tex]

We can substitute the values into the formula.

[tex]W= 100 \ N * 5 \ m[/tex]

Multiply.

[tex]W= 500 \ N*m[/tex]

[tex]W= 500 \ J[/tex]

You did 500 Joules of work to push a shopping cart 5 meters with a force of 100 Newtons.

Answer:

Work done

As we know that Work done is the product of Force and displacement

[tex] \tt \: w = force \: \times displacement[/tex]

[tex] \tt \: w = 100 \times 5[/tex]

[tex] \tt \: w = 500 \: j[/tex]

Therefore, Work done is 500 Joule

Answer:

1176N

Explanation:

Given parameters:

Mass of bench  = 100kg

Forward force  = 980N

Coefficient of static friction = 0.2

Acceleration  = 1.84m/s²

Unknown:

Magnitude of the applied force  =?

Solution:

Frictional force is a force that opposes motion, for a body to move, the applied force must be greater than the frictional force;

The Force applied;

  Force applied  = Frictional force + Weight of the body

   Frictional force  = umg

     u is the coefficient of static friction

      m is the mass

        g is the acceleration due to gravity

    Force applied  = 0.2 x 100 x 9.8 + 980 = 1176N

Answer:

Archimedes discovered the buoyancy laws

Explanation:

when asked by King Hiero of Syracuse to determine whether his new crown was pure gold (SG = 19.3). Archimedes measured the weight of the crown in air to be 11.8 N and its weight in water to be 10.9 N.

Answer:

The correct answer is C

Explanation:

can tigger sudden mass movement

Answer:

Magnetic Domain

Explanation:

Answer:

63.8454kgm/s

Explanation:

According to Newton's second law;

Force = mass ×acceleration

Fore = m(v-u)/t

Force = mv-mu/t

MV is the final momentum

mu is the initial momentum = 38kgm/s

Force = 88.3N

Time = 0.338s

Substitute and get mv

88.3 = mv-38/0.338

88.3×0.338 = mv-38

29.8454 = mv-38

MV = 29.8454+34

mv = 63.8454kgm/s

Hence the final momentum is 63.8454kgm/s

Answer:

67.8

Explanation:

im an acellus student, and this is the answer that i got when doing the problem.

Answer:

it's 220 J

the second one

Answer:

We're like children who move away from their parents - then back,

HUMANS LOVE to travel. Mentally or physically, they escape to other places. They crave adventure. They seek the unknown. They want to see new places, learn new things, enjoy new experiences and then return home to the familiar, the predictable, the secure, before taking off again on their travels.

This departure-return behaviour has its origins, perhaps, in the early exploratory behaviour of the child. Research has shown that children who are "securely attached" to their parents, sure of their affection and protection, and who know that their parents will respond to their needs, are more independent, adventurous and exploratory in their play and behaviour.

Observations of infants show that from the safe base of having a parent present and available, the infant dares to move away a short distance and then return to the parent, to move a little further and return, until in incremental and amazingly measurable distances it explores the environment that surrounds it and has the courage to move beyond it.

This may be the first symbolic travel: the first departure, the first expedition and the possible beginning of the travel bug for those for whom their initial independent voyaging was successful.

With these first independent steps away from the parental presence lies the atavistic or recurring wish to reach out beyond the current confines of space and relationship

and see what is further away. And yet further again to see what is not visible, what is around the next corner, in the next room and eventually on

the other side of our geographical or psychological world. This is a human need. It is why we have uncovered our world, travelled its length, hiked to the top of its mountains and the depth of its oceans. It is why we became discontent with exploring our own plant but had to move beyond it into the unknown. It is why, as poet TS Eliot reminds us, we will "not cease from exploration".

It is our nature to travel. It is our joy to travel. It is our paradoxical psychological disposition to wish for the contradictory conditions of stability and change: to stay and to go, to be and to explore, to rest and to travel unceasingly.

Travel is more than going somewhere. It is a psychological event. It requires motivation and imagination. It begins in childhood with faltering first steps and with imagination fired by the tales of "long, long ago in a faraway land".

We emerge from these fantasies of childhood determined to cross the oceans, to find those faraway lands, the characters they containand the magic they hold. Determined to find the adventures they promise and the prospect of returning home triumphant with trophies having vanquished all fear, overcome all obstacles and live happily eve after.

Answer:

to see the world and experience new things. it opens your mind and you can meet more people. you have a new view on the world after seeing it from other places, not only new views on the world but new cultures, and languages.

Explanation:

Explanation:

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pass : frmCV5

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I am boy :)

Answer:

v₃ = 2 [m/s]

Explanation:

To solve such problems we must use the principle of conservation of momentum. That is, the linear momentum is conserved before and after the collision.

P = m*v

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

The momentum is conserved before and after the collision, in this way we can obtain the following equation.

[tex](m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}[/tex]

where:

m₁ = mass of the cart moving = 4 [kg]

v₁ =  velocity of the cart moving before the collision = 3 [m/s]

m₂ = mass of the cart initially at rest = 2 [kg]

v₂ = velocity of the cart at rest = 0

v₃ = velocity of the two carts combined (carts stick together) after the collision [m/s]

[tex](4*3)+(2*0)=(4+2)*v_{3}\\v_{3}=12/6\\v_{3}=2[m/s][/tex]

Answer:

The correct option is;

b. 7.94 m

Explanation:

The given parameters of the jump of the long jumper are;

The angle above the horizontal with which the long jumper leaves the ground, θ = 20.0°

The speed with which the long jumper leaves the ground, u = 11.0 m/s

The furthest horizontal distance the long jumper jumps, given that the motion is equivalent to that of a particle, is given by the formula for the range, R, of a projectile motion as follows;

[tex]R = \dfrac{u^2 \times sin \left (2 \cdot \theta \right )}{g}[/tex]

Where;

g = The acceleration due to gravity ≈ 9.8 m/s²

u = The initial velocity of the long-jumper = 11.0 m/s

θ = The angle of the direction above the horizontal the long-jumper jumps = 20.0°

Plugging in the values, gives;

[tex]R = \dfrac{(11.0 \ m/s)^2 \times sin \left (2 \times 20.0 ^{\circ} \right )}{9.8} = \dfrac{121 \ m^2/s^2 \times sin \left (40.0 ^{\circ} \right )}{9.8 \ m/s^2} \approx 7.94 \ m[/tex]

How far the long-jumper goes = The range, R, of the projectile motion ≈ 7.94 m.

Answer:

theyll accelerate towards the floor :)

Answer:

4.9949

Explanation:

M = 3 kg

P.E = 147 joules

G = 9.81 ms^-2

H = ?

P.E = MGH

147 = 3 x 9.81 x H

147 = 29.43H

147/29.43 = 29.43H/29.43

4.9949 = H

:. H = 4.9949

Answer:

The donkey might be dead

Answer:

a. s = 0.96 m

b. s = 2.15 m

Explanation:

a.

The relationship between the linear displacement and the angular displacement is given as follows:

[tex]s = r\theta[/tex]

where,

s = linear distance covered = ?

r = radius of wheel = (6 in)(0.0254 m/1 in) = 0.1524 m

θ = angular displacement = (1 rev)(2π rad/1 rev) = 2π rad

Therefore,

[tex]s = (0.1524\ m) 2\pi[/tex]

s = 0.96 m

b.

Assuming, we have to find linear displacement here, as well:

[tex]s = r\theta[/tex]

where,

s = linear distance covered = ?

r = radius of wheel = (13.5 in)(0.0254 m/1 in) = 0.3429 m

θ = angular displacement = (1 rev)(2π rad/1 rev) = 2π rad

Therefore,

[tex]s = (0.3429 \ m) 2\pi[/tex]

s = 2.15 m

Answer:

20 meters

Explanation:

Given the following parameters

Time t = 2 seconds

Acceleration a = 10m/s²

Required

Distance covered S

Since the body started from rest, u = 0m/s

Using the equation of motion S = ut + 1/2 at²

Substitute

S = 0 + 1/2(10)(2)²

S = 5 * 2²

S = 5 * 4

S - 20m

Hence he travels 20 meters

Answer: 321.027 m/s

Explanation:

First let´s analyze the vertical problem:

Always when an object is above the ground and nothing is holding it (like the bullet after being fired) the gravitational force will start acting on it.

Then the vertical acceleration of the bullet will be the gravitational acceleration, we can write this as:

a(t) = -g

Where g = 9.8m/s^2

and the negative sign is because this acceleration pulls the bullet downwards.

To get the vertical velocity we need to integrate over time, this will lead to:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity, as the bullet is fired horizontally, there is no initial vertical velocity, then we have v0 = 0m/s

And the velocity equation is:

v(t) = (-9.8m/s^2)*t

Now for the vertical position, we need to integrate again, to get:

p(t)  =(1/2)*(-9.8m/s^2)*t^2 + p0

Where p0 is the initial vertical position, in this case, is 1.9 meters above the ground, then p0 = 1.9m

And the vertical position equation will be:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + 1.9 m

Now we want to find the time such that the vertical position is equal to zero, this will mean that the bullet it the ground.

p(t) = 0m = (1/2)*(-9.8m/s^2)*t^2 + 1.9 m

    (1/2)*(9.8m/s^2)*t^2 = 1.9m

    (4.9m/s^2)*t^2 = 1.9m

                  t = √(1.9m/(4.9m/s^2)) = 0.623 seconds.

This means that the bullet will travel for 0.623 seconds before hitting the ground.

Now we also can ignore the air friction for the horizontal motion, then we can assume that the horizontal speed does not change.

Then we can use the relationship:

Distance = speed*time

We know that:

time = 0.623 seconds

distance = 200m

now we can replace that in the equation to find the horizontal speed.

200m = speed*0.623s

200m/0.623s = speed = 321.027 m/s

Answer:

It is that kind of permeable membrane that allows certain molecules to pass through. Not every molecules can pass through it

Explanation:

Answer:

[tex]a=2.4\ m/s^2[/tex]

Explanation:

Given that,

Initial speed of a car, u = 13 m/s

Final speed of a car, v = 25 m/s

Time, t = 5 s

We need to find the acceleration of the car during this 5.0 second time interval. Let a is the acceleration. It can be calculated as :

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{25-13}{5}\\\\=2.4\ m/s^2[/tex]

So, the acceleration of the car is [tex]2.4\ m/s^2[/tex].