I need help ASAP please.....

Explanation:

the acceleration will double because force is directly proportional to the acceleration

Let's see

[tex]\\ \sf\longmapsto \overrightarrow{drag}[/tex]

[tex]\\ \sf\longmapsto \overleftarrow{Thrust}[/tex]

[tex]\\ \sf\longmapsto Lift\:Force\uparrow[/tex]

[tex]\\ \sf\longmapsto Acceleration\:due\:to\: gravity\downarrow[/tex]

Net force is 0

Answer:

D) 19.8 lbs

Explanation:

1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.

1kg is also equal to 2.205 lbs. 9*2.205=19.8416

9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system

a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.

This leaves us with 19.8 lbs

Answer:

Explanation:

a) A is accurate because the half of the Moon that is facing the sun is it by the sun, and the other half is dark.

The most accurate diagram of the Sun and Moon is: a. A is accurate because the half of the Moon that is facing the sun is it by the sun, and the other half is dark.

A Moon can be defined as the natural satellite of planet Earth that shines due to the Sun's reflected light (sunlight), as it revolves around the Earth from west to east in 29½ days.

Based on astronomical information, there are four (4) primary Moon phases and these are:

In conclusion, diagram A is the most accurate because half of the Moon that faced the Sun is lit by the Sun while the other half is dark.

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Answer:

D - 1.52 m/s2

Explanation:

force: 125 N

Mass: 82 kg

you just divide the force by the mass which would be approximately 1.52

Answer:

10N right

Explanation:

First, we need to find the base of the triangle to find the force acting on the right of the object.

Next, sub in the values (we don't know the adjacent so i've called it x):

To find x, we need to multiply both sides by 60

Now we know all the forces, we can work out the net force:

Weight=Applied force

[tex]\\ \sf\longmapsto F=mg[/tex]

[tex]\\ \sf\longmapsto F=20(10)[/tex]

[tex]\\ \sf\longmapsto F=200N[/tex]

Answer:

Weight=200

Explanation:

Given,

[tex]\sf\longmapsto \: mass = 20 \: kilograms[/tex]

We know that–

[tex]\sf\longmapsto \: acceleration \: \: because \: of \: gravity = 10m/s^2[/tex]

[tex]\sf\longmapsto \: Weight= \: is  \: F = m•g[/tex]

Let's start solving!

Weight = 20×10

Weight=200

Answer:

Explanation:

85 = ½(0.43)t²

t = √(2(85)/0.43)

t = 19.883380...

t = 20 s

v→ 8.55 m/s initial, 0 m/s final

a← 0.43 m/s²

Answer:

No because you don't have to be black or white you don't have to be a pacific race to just get it like lets say for example a asian person come in and they want to get a abortion but the doctor say no then that person can end up killing her self or when the baby is born she could end you know doing something bad. But no you do not have to be a pacific person to just get it

Explanation:

Answer:

what do you need help with I can try to help you

Answer:

Abcdfeghijklmnopqrstuvwxyz

Explanation:

They are in order

According to the picture, the chess piece is the least dense. The correct option is A.

The amount of mass that a substance has in relation to its volume is referred to as its density. It is a measurement of how closely together the molecules or particles are arranged within a given object or material.

Typically, floating objects are less dense than the liquid they are in. The fact that the chess piece is floating on the water suggests that it is less dense than the liquid.

It might be an alternative, though, if the toy soldier is also floating and not submerged. More visual context is necessary for precise determination.

Thus, the correct option is A.

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Answer:

Enthalpy

The quantity of heat contained in a material as measured from a particular base or point.

Newton's second law allows us to find the results for the acceleration of the blocks are:

1) The acceleration is a = 559 m / s²

2) The acceleration is a = 7.88 m / s²

Newton's second law states that the net force is equal to the product of the mass and the acceleration of the body.

       ∑ F = m a

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the body.

The reference system is a coordinate system with respect to which the decomposition of the forces is carried out, in the attached we have a free body diagram of the system.

1) They indicate that the body mass is 25 kg.

y-axis

       N - W = 0

       N = W = m g

x-axis

        F -fr = ma

The friction force is a macroscopic force that results from the sum of all the microscopic interactions between the two surfaces, it has the formula

        fr = μ N

Where fr is the friction force, N the normal and very the friction coefficient.

This friction coefficient has two values:

We substitute.

         F - μ m g = m a

a) The system moves which is the acceleration.

Suppose that the force that star to move the system keeps constant, just before the system begins to move the coefficient of friction is static, let's find the applied force.

         F = μ m g

         F = 0.73 25 9.8

         F = 178.85 N

The block begins to move and the friction coefficient decreases to the dynamic value, we look for the acceleration.

         a = [tex]\frac{F - \mu \ m g}{m}[/tex]  

         a = [tex]\frac{178.85 - 0.16 \ 25 \ 9.8 }{25}[/tex]  

         a = 5.59 m / s²

2) In this case the mass of the block is 79 kg and the applied force is

        F = 909 N

We look for acceleration.        

        a = [tex]\frac{909 - 0.37 \ 79 \ 9.8 }{79}[/tex]  

        a = 7.88 m / S²

In conclusion using Newton's second law we can find the results for the acceleration of the blocks are:

         1) The acceleration is a = 559 m / s²

         2) The acceleration is a = 7.88 m / s²

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A 25.0 kg box is moved 4.50 m by a force of 134 N applied at 35.0° above horizontal. The work done by the pulling force is 4.9 × 10² J.

Starting from rest, a student pulls a 25.0 kg box a distance of 4.50 m across the floor using a 134 N force applied at 35.0 degrees above horizontal.

We can calculate the work  (W) done by the pulling force using the following expression.

[tex]W = F \times d \times cos\theta = 134 N \times 4.50 m \times cos35.0\° = 4.9 \times 10^{2} J[/tex]

where,

The friction between the floor and the box. does not influence the work done by the pulling force.

A 25.0 kg box is moved 4.50 m by a force of 134 N applied at 35.0° above horizontal. The work done by the pulling force is 4.9 × 10² J.

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The definition of moment of inertia and kinetic energy we can find the results for the questions are:  

          a) The moment of inertia is: I = 32.49 kg m²

          b) The kinetic energy is: K = 509.4 J

a) The moment of inertia is a scalar quantity that has the same function as mass, it is given by

             I = m r²

They indicate that the length of the bar is y₁ = 5.70 m, if the two rigid bars form the square that supports the masses, the distance from the center to the end where each mass is:  

            r = y₁ / 2

             r = 5.70 / 2 = 2.85 m

The moment of inertia of the system is.

         I = 4 m r²

Let's calculate.

        I =4 m 2.85²

        I = 32.49m kg m²

If we assume that the mass of each point particle is:

        m = 1 kg

         I = 32.49 kg m²

b) They ask for the rotational kinetic energy, if the angular velocity is.

           w = 5.60 rad / s

The kinetic energy for rotational motion is.

         K = ½ I w²

Let's calculate.

          K = ½ 32.49 5.60²

          K = 509.4 J

In conclusion with the definition of moment of inertia and kinetic energy we can find the results for the questions are:

          a) The moment of inertia is: I = 32.49 kg m²

          b) The kinetic energy is: K = 509.4 J

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hope it helps

Answer:

Option A

Explanation:

N is Chloroplast, P is a large centralized vacuole and R is the cell wall which are found only in plant cells. Hence, the given cell is a plant cell.

S = Rough Endoplasmic Reticulum

P = Vacuole

O = Nucleus

T = Smooth Endoplasmic Reticulum

U = Golgi Apparatus

N = Chloroplast

U = Cell Membrane

M = Mitochondria

R = Cell wall

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

Answer:

what are the choices

Explanation:

don't worry i'll edit it

From your previous question

Mass=30kg

Acceleration=12m/s^2

[tex]\\ \sf\longmapsto F=ma[/tex]

[tex]\\ \sf\longmapsto F=30(12)[/tex]

[tex]\\ \sf\longmapsto F=360N[/tex]

Given,

Acceleration = 12 m/s²

According to your previous question you posted,

Mass = 30 KG

Solution:

We know that the formula of force according to the Newton's second law is:

[tex]\longmapsto \sf \red{f = m \times a}[/tex]

[tex]\sf \longmapsto \ 30 \times 12[/tex]

[tex]\sf \longmapsto \: \boxed {\sf \blue3 \red60}[/tex]

Answer:

[tex]\sf \boxed{ \bf \: force = 360\: N}[/tex]

Answer:

d. Ice

Explanation:

Answer:

yes

Explanation:

Bouncing is a way to increase impulse. Because an object that bounces changes directions the force of impulse must be absorbed then generated by the target object. (Impulse is nearly doubled.)

Answer:

According to Newton's first law of motion, an object maintains its state unless a force acts on it. Therefore, a moving car does not change its direction and keeps its speed unless a force acts on it.

Answer:

Gas

Explanation:

Hot air rises because gases expand as they heat up. When air heats up and expands, its density also decreases.

Answer:

Explanation:

Remark

The question really is how long does the projectile stay in the air if it was just dropped from a height of 46 meters.

Givens

vi = 0

a = 9.81 m/s^2

h = 46 m

t = ?

Solution

d = vi*t + 1/2 a t^2

46 = 0 + 1/2 * 9.81 * t^2       multiply both sides by 2

92 = 9.81 * t^2                     divide by 9.81

t^2 = 9.38                            Take the square root of both sides

sqrt(t^2) = sqrt(9.38)

t = 3.06 seconds.

In 3.06 seconds the object will hit the ground, thus stopping it. How far horizontally does it go in that time.

d = r * t

d=?

r = 102 m/s

t = 3.06 seconds

d = 102 * 3.06

d = 312.4  meters.

Answer:

low pressure

Explanation:

The velocity of the rock after 2s is 12.56 m/s.

The velocity of the rock at 25 m upwards is 14.68 m/s.

The velocity of the rock at 25 m downwards is -15.97 m/s.

The given parameters;

The velocity of the rock after 2s is calculated as follows;

[tex]v = \frac{dh}{dt}= 20 -2(1.86)t \\\\v = 20 - 3.72t\\\\at \ t= 2 \ s \\\\v = 20 - 3.72(2) = 12.56 \ m/s[/tex]

The velocity of the rock when the height is 25 m:

h = 20t - 1.8t²

25 = 20t - 1.8t²

1.8t² - 20t + 25 = 0

solve for the time of motion "t" using quadratic formula

a = 1.8,  b = -20,  c = 25

[tex]t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-(-20) \ \ +/- \ \ \sqrt{(-20)^2 - 4(1.8\times 25)} }{2(1.8)}\\\\t = 9.67s, \ \ or \ 1.43 \ s[/tex]

The value of t that will give 25 m upwards using the motion model is 1.43 s;

h = 20(1.43) - 1.86(1.43)² = 25 m

The velocity of the rock at 25 m upwards (t = 1.43 s) is calculated as follows;

[tex]v = \frac{dh}{dt} = 20 -3.72t\\\\v = 20 - 3.72(1.43) \\\\v = 14.68 \ m/s[/tex]

The velocity of the rock at 25 m downwards (t = 9.67 s) is calculated as;

[tex]v = \frac{dh}{dt} = 20 - 3.72t\\\\v = 20 - 3.72(9.67)\\\\v = -15.97 \ m/s[/tex]

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