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dimulka [17.4K]

3 years ago

10

Interactive LearningWare 4.1 reviews the approach taken in problems such as this one. A 1800-kg car is traveling with a speed of

14.0 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 36.9 m?

1 answer:

Lubov Fominskaja [6]3 years ago

6 0

Answer:

F= 4788 N

Explanation:

Because the car moves with uniformly accelerated movement we apply the following formula:

vf²=v₀²+2*a*d Formula (1)

Where:

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Data

d=36.9 m

v₀=14.0 m/s m/s

vf= 0

Calculating of the acceleration of the car

We replace dta in the formula (1)

vf²=v₀²+2*a*d

(0)²=(14)²+2*a*(36.9)

-(14)²= (73.8) *a

a= - (196) / (73.8)

a= - 2.66 m/s²

Newton's second law of the car in direction horizontal (x):

∑Fx = m*ax Formula (2)

∑F : algebraic sum of the forces in direction x-axis (N)

m : mass (kg)

a : acceleration (m/s²)

Data

m=1800 Fkg

a= - 2.66 m/s²

Magnitude of the horizontal net force (F) that is required to bring the car to a halt in a distance of 36.9 m :

We replace data in the formula (2)

-F= (1800 kg) * ( -2.66 m/s² )

F= 4788 N

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In the given figure, weight of stone inside water

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The actual weight of the stone is 11 N. It is based on the Archimedes principles.

<h3>What is Archimedes principle?</h3>

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2 years ago

According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this

e-lub [12.9K]

Answers:

a) $\rho_{cylinder}= 0.55 g/cm^{3}$

b) $\rho_{liq}= 1.48 g/cm^{3}$

c) When we divided both volumes (sumerged and displaced) the factor $\pi r^{2}$ is removed during calculations.

Explanation:

a) According to <u>Archimedes’ Principle:</u>

<em>A body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.</em>

In this case we have a wooden cylinder floating (partially immersed) in water. <u>This object does not completely fall to the bottom because the net force acting on it is zero, this means it is in equilibrium.</u> This is due to Newton’s first law of motion, that estates if a body is in equilibrium the sum of all the forces acting on it is equal to zero.

Hence:

$W_(cylinder)=B$ (1)

Where:

$W_(cylinder)=m.g$ is the weight of the wooden cylinder, where $m$ is its mass and $g$ gravity.

$B$ is the Buoyant force, which is the force the fluid (water in this situation) exert in the submerged cylinder, and is directed upwards.

We can rewrite (1) as follows:

$m_{cylinder}g=m_{water}g$ (2)

On the other hand, we know density $\rho$ establishes a relationship between the mass of a body andthe volume it occupies. Mathematically is expressed as:

$\rho=\frac{m}{V}$ (3)

isolating the mass:

$m=\rho V$ (4)

Now we can express (2) in terms of the density and the volume of cylinder and water:

$\rho_{cylinder} V_{cylinder} g=\rho_{water} V_{water} g$ (5)

In this case $V_{water}$ is the volume of water displaced by the wooden cylinder (remembering Archimedes's Principle).

At this point we have to establish the total volume of the cylinder and the volume of water displaced by the sumerged part:

$V_{cylinder}=\pi r^{2} h$ (6)

Where $r$ is the radius and $h=30 cm$ the total height of the cylinder.

$V_{water}=\pi r^{2} (h-h_{top})$ (7)

Where $h_{top}=13.5 cm$ is the height of the top of the cylinder above the surface of water and $(h-h_{top})$ is the height of the sumerged part of the cylinder.

Substituting (6) and (7) in (5):

$\rho_{cylinder} \pi r^{2} h g=\rho_{water} \pi r^{2} (h-h_{top}) g$ (8)

Clearing $\rho_{cylinder}$ :

$\rho_{cylinder}=\frac{\rho_{water}(h-h_{top})}{h}$ (9)

Simplifying;

$\rho_{cylinder}=\rho_{water}(1-\frac{h_{top}}{h}$ (10)

Knowing $\rho_{water}=1g/cm^{3}$ :

$\rho_{cylinder}=1g/cm^{3}(1-\frac{13.5 cm}{30cm})$ (11)

$\rho_{cylinder}= 0.55 g/cm^{3}$ (12) This is the density of the wooden cylinder

b) Now we have a different situation, we have the same wooden cylinder, which density was already calculated ( $\rho_{cylinder}= 0.55 g/cm^{3}$ ), but the density of the liquid $\rho_{liq}$ is unknown.

Applying again the Archimedes principle:

$\rho_{cylinder} V_{cylinder} g = \rho_{liq} V_{liq} g$ (13)

Isolating $\rho_{liq}$ :

$\rho_{liq}= \frac{\rho_{cylinder} V_{cylinder}}{V_{liq}}$ (14)

Where:

$V_{cylinder}=\pi r^{2} h$

$V_{liq}=\pi r^{2} (h-h_{top})$

Then:

$\rho_{liq}= \frac{\rho_{cylinder} \pi r^{2} h}{\pi r^{2} (h-h_{top})}$ (15)

$\rho_{liq}= \frac{\rho_{cylinder} h}{h-h_{top}}$ (16)

$\rho_{liq}= \frac{0.55 g/cm^{3} (30 cm)} {30 cm - 18.9 cm}$ (17)

$\rho_{liq}= 1.48 g/cm^{3}$ (18) This is the density of the liquid

c) As we can see, it was not necessary to know the radius of the cylinder (we did not need to knoe its length and width), we only needed to know the part that was sumerged and the part that was above the surface of the liquid.

This is because in this case, when we divided both volumes (sumerged and displaced) the factor $\pi r^{2}$ is removed during calculations.

6 0

4 years ago

A person pushes a 15.7-kg shopping cart at a constant velocity for a distance of 25.9 m on a flat horizontal surface. She pushes

valentinak56 [21]

Answer:

a) F = 35.7 N, b) W = 846.7 J, c) W = - 846.9 J, d) W=0

Explanation:

a) For this exercise let's use Newton's second law, let's set a reference frame with the x-axis horizontally

let's break down the pushing force.

cos (-23,7) = Fₓ / F

sin (-237) = F_y / F

Fₓ = F cos 23.7 = F 0.916

F_y = F sin (-23.7) = - F 0.402

Y axis

N- W - F_y = 0

N = W + F 0.402

X axis

Fₓ - fr = 0

F 0.916 = fr

F = fr / 0.916

F = 32.7 / 0.916

F = 35.7 N

It is asked to calculate several jobs

b) the work of the pushing force

W = fx x

W = 35.7 cos 23.7 25.9

W = 846.7 J

c) friction force work

W = F x cos tea

friction force opposes movement

W = - fr x

W = - 32.7 25.9

W = - 846.9 J

d) The work of the force would gravitate, as the displacement and the force of gravity are at 90º, the work is zero

W = 0

6 0

3 years ago