A 1000-kg car comes to a stop without skidding. The car's brakes do 50,000 J of work to stop the car. Which of the following was the car's velocity when the brakes were initially applied?

final velocity = 0

acceleration = - 10 m/ s 2

distance. = 20 m

u = ?

v^2 - u ^2 = 2 a s

0^2 - u^ 2 = 2 * -10 * 20

-u^2 = -400

u = √ 400

u = 20m / s

Answer:

D. Downshift to allow you to turn more sharply

D. downshift to allow you to turn more sharply

The purpose of "downshifting", or shifting the transmission down from a higher gear to a lower gear, is so that a driver can accelerate their car as fast as possible when exiting a corner that they had to slow down for.

The reason for ensuring your downshift is completed before turning into the corner is because letting the clutch out and engaging a lower gear while turning can cause the car to spin.

Two special conditions where you should downshift

Before Starting Down a Hill.

Slow down and shift down to a speed that you can control without using the brakes hard. Otherwise the brakes can overheat and lose their braking power.

Before Entering a Curve.

Slow down to a safe speed, and downshift to the right gear before entering the curve. This lets you use some power through the curve to help the vehicle be more stable while turning. It also allows you speed up as soon as you are out of the curve.

Therefore,

In the middle of a turn, you should downshift to allow you to turn more sharply .

Learn more about Downshift here:

https://brainly.com/question/10600958

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Answer:

A) a nerve cells long extensions enable it to transmit electrical impulses from one body part to another

B) the sheet like organization of epithelial cells enables them to protect underlying cells

C) the alignment of contractile proteins within muscle cells enables them to contract, pulling closer together the structures to which they attach

Explanation:

Answer:

r = mv^2/ Fc

Explanation:

Got it right

Answer: C. The time taken for the light to travel would have been too little to have been  measured by them.

Explanation:

Light travels at a speed of 300,000 km per second which is much too fast for the naked eye to even comprehend. This is why light shinning in a distance is picked up instantaneously by out eyes.

This experiment done by Galileo would therefore be unable to measure the speed of light because the light from either lantern would have traveled too fast for Galileo to be able to measure the time taken.

Answer:

Friction

Explanation:

Answer:

It causes electrostatic charging, which makes it stick to your hair,

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:

1655.5 N

Explanation:

We resolve the forces on the rope into horizontal and vertical components. Since the only vertical force exerted on the rope is the weight of the man, this weight equals the vertical components of the tensions on both sides of the rope .

The only horizontal forces acting on the rope are horizontal components of both tensions on both sides. Since the hiker doesn't move sideways, they cancel out and he is balanced. So, the net horizontal force is zero.

Let T be the tension in the rope. So, each vertical component on side A is Tsin35° and that on side B is Tsin65°. Let the weight of the man be W = mg where m = mass of hiker = 250 kg and g = acceleration due to gravity = 9.8 m/s².

So, since the hiker balances, Tsin35° + Tsin65° = mg

0.5736T + 0.9063T = 250 kg × 9.8 m/s²

1.4799T = 2450 N

dividing both sides by 1.4799, we have

T = 2450 N/1.4799

T = 1655.52 N

T ≅ 1655.5 N

So, the tension in the rope equals T = 1655.5 N

Answer:

a = 1.428 [m/s²]

v₀ = 5 [m/s]

Explanation:

To solve this problem we must use the following equation of kinematics.

[tex]x=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}[/tex]

where:

x = final point [m]

x₀ = initial point [m]

v₀ = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

But we need to use this additional equation.

[tex]v_{f}=v_{o}+a*t[/tex]

where:

vf = final velocity = 15 [m/s]

Now we can use this equation, replacing it, in the first one. We must bear in mind that the difference among x - x₀ is equal to 70 [m]

[tex]x-x_{o}=v_{o}*t+\frac{1}{2}*a*t^{2} \\x-x_{o}=(v_{f}-a*t)*t+\frac{1}{2} *a*t^{2}\\70=(15-a*t)*t+\frac{1}{2}*a*t^{2}\\70=15*t-a*t^{2} +\frac{1}{2}*a*t^{2} \\70=15*t-\frac{1}{2}*a*t^{2}\\70=15*(7)-\frac{1}{2} *a*(7)^{2}\\105-70=0.5*a*49\\35=24.5*a\\a=1.428[m/s^{2} ][/tex]

Now replacing this value in the second equation, we can find the initial velocity.

[tex]15=v_{o}+1.428*7\\v_{o}=5[m/s][/tex]

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

Explanation:

The mass of a ball, m = 2 kg

It is traveling with a speed of 10 m/s

The ball's kinetic energy just as it leaves the boy's hand is calculated as follows :

[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 2\times (10)^2\\\\=100\ J[/tex]

The ball's kinetic energy just as it leaves the boy's hand is 100 J. The potential energy of the ball when it reaches the highest point is same as the kinetic energy as it leaves the boy's hand.

Hence, the required kinetic and potential energy is 100 J.

Explanation:

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Answer:

a) α = -65,2 rad/s².

b) t = 2,57 s.

Explanation:

a) La aceleración angular se puede calcular usando la siguiente ecuación:

[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta [/tex]

En donde:

[tex]\omega_{f}[/tex]: es la velocidad angular final =  2000 rpm = 209,4 rad/s

[tex]\omega_{0}[/tex]: es la velocidad angular inicial = 3600 rpm = 377,0 rad/s

α: es la aceleración angular=?

θ: es el desplazamiento o número de vueltas = 120 rev = 754,0 rad

Las conversiones de unidades se hicieron sabiendo que 1 revolución = 2π radianes y que 1 minuto = 60 segundos.  

Resolviendo la ecuación (1) para α, tenemos:

[tex]\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2\theta} = \frac{(209,4 rad/s)^{2} - (377,0 rad/s)^{2}}{2*754,0 rad} = -65,2 rad/s^{2}[/tex]  

Entonces, la aceleración angular es -65,2 rad/s². El signo negativo se debe a que el motor está desacelerando.  

b) El tiempo transcurrido se puede encontrar como sigue:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Resolviendo para t, tenemos:

[tex]t = \frac{\omega_{f} - \omega_{0}}{\alpha} = \frac{209,4 rad/s - 377,0 rad/s}{-65,3 rad/s^{2}} = 2,57 s[/tex]

Por lo tanto, el tiempo transcurrido fue 2,57 s.

Espero que te sea de utilidad!

Hello!!

For calculate the volume of the liquid, lets applicate the formula:

[tex]\boxed{V=m/ p }[/tex]

[tex]\textbf{Being:}[/tex]

[tex]\sqrt{}[/tex] V = Volume = ?

[tex]\sqrt{}[/tex] m = Mass = 6 g

[tex]\sqrt{}[/tex] p = Density = 25 g/ml

[tex]\text{Let's \textbf{replace} and resolve: }[/tex]

[tex]V = 6 \ g / 25 \ g / m l[/tex]

[tex]V = 0,24 \ m l[/tex]

[tex]\textbf{Result:}[/tex]

The volume of the liquid is 0,24 ml.

Answer:

120V

Explanation:

Given parameters:

Current  = 6A

Resistance  = 20Ω

Unknown:

Voltage = ?

Solution:

According to ohms law;

           V = IR

Where V is the voltage

            I is the current

            R is the resistance

Now, insert the parameters and solve;

    V  = 6 x 20  = 120V

Answer:

[tex]Mass = 50kg[/tex]

Explanation:

Required

Which has the larger inertia

Inertia (I) is calculated as follows:

[tex]I = mr^2[/tex]

Where m represents the mass of the object and r represents the distance from the axis.

Take for instance m = 10 and r = 4

The inertia would be:

[tex]I = 10 * 4^2[/tex]

[tex]I = 160[/tex]

Assume r is constant and m is increased to 25

The inertia becomes:

[tex]I = 25 * 4^2[/tex]

[tex]I = 25 * 16[/tex]

[tex]I = 400[/tex]

Notice that there is an increment in inertia, when the mass increased.

Hence, mass of 50kg will have the larger inertia.

Answer:

105 m

Explanation:

Given that the initial velocity of the body, u= 50 m/s in upward direction.

Acceleration due to gravity, g= 10 m / [tex]s ^ 2[/tex] in the downward direction.

Let h be the height of the body when the velocity of the body is 20 m/s.

As the gravitational force acting in the downward direction, so taking it negative as

a= - 10 m/[tex]s ^ 2[/tex]  in the upward direction.

By using the equation of motion [tex]v^2=u^2+2as[/tex] ...(i)

Here, u= 50 m/s

v= 20 m/s

[tex]a=- 10 m/s^2[/tex]

s=h

Putting all the values in the equation (i), we have

[tex]20^2=50^2+2(-10)h \\\\400=2500-20h \\\\h=\frac{2500-400}{20} \\\\h=105 m[/tex]

Hence, the height of the body is 105 m.

Answer:

h₁ = 18 [cm]

Explanation:

This problem can be solved by applying the principles of the static pressure of a liquid column. The key to the solution is to know that the pressure on the horizontal line should be equal for both liquids, this must be met regardless of their density.

[tex]P_{2}=P_{1}\\[/tex]

where:

P₂ = Pressure exerted by the liquid 2 in the given point L₂

P₁ = Pressure exerted by the liquid 2 in the given point L₁

Now we can calculate the pressure in each point using the following expression:

[tex]P_{1}=Ro_{1}*g*h_{1}\\P_{2}=Ro_{2}*g*h_{2}[/tex]

Where:

Ro₁ = density of the liquid 1 = 0.8 [g/cm³]

Ro₂ = density of the liquid 2 = 1.8 [g/cm³]

g = gravity acceleration = 9.81 [m/s²]

h₁ = column of the liquid 1 [m]

h₂ = column of the liquid 2, = 8 [cm] = 0.08 [m]

But first we must convert the units from grams per cubic centimeter to kilograms per cubic meter.

[tex]0.8[\frac{g}{cm^{3}}]*[\frac{1kg}{1000g} ]*[\frac{100^{3}cm^{3} }{1m^{3} } ]=800[kg/m^{3} ]\\1.8[\frac{g}{cm^{3}}]*[\frac{1kg}{1000g} ]*[\frac{100^{3}cm^{3} }{1m^{3} } ]=1800[kg/m^{3} ][/tex]

Now replacing in the first equation:

[tex]800*9.81*h_{1}=1800*9.81*0.08\\h_{1}=\frac{1800}{800} *0.08\\h_{1}=0.18 [m] = 18 [cm][/tex]

Answer:

[tex]h_1 = 18 \: cm[/tex]

Explanation:

See image for explanation

I hope my working is correct.

Answer:

the SI unit of work or energy, equal to the work done by a force of one newton when its point of application moves one meter in the direction of action of the force, equivalent to one 3600th of a watt-hour.

Explanation:

Joule, unit of work or energy in the International System of Units (SI); it is equal to the work done by a force of one newton acting through one metre. Named in honour of the English physicist James Prescott Joule, it equals 107 ergs, or approximately 0.7377 foot-pounds.

Answer:

The laws of physics which apply when you are at rest on the earth also apply when you are in any reference frame which is moving at a constant velocity with respect to the earth. For example, you can toss and catch a ball in a moving bus if the motion is in a straight line at constant speed.

Explanation:

Răspuns:

0,98 N

Explicaţie:

Dat fiind:

µ = Coeficientul de frecare = 0,02

Masa (m) = 5 kg

Folosind relația:

Forța de frecare (F) = µ * N

N = reacție normală = masă * accelerație datorată gravitației

Prin urmare,

F = u * m * g

g = 9,8 m / s²

F = 0,02 * 5 kg * 9,8 m / s²

F = 0,98 kgm / s²

F = 0,98N

Answer:

Flakes of brick: Athens, Ohio

Cracks in rocks: Sandston, Mudcracks

Lichen growing on rocks: Antarctica living below the surface

Roots or plant growth making cracks in concrete: In the driveway, sidewalk, or road.

Answer:

a. Please find attached the diagram of the disc, having arrows that represent the velocity and the acceleration of the particle placed on it

b. The angular speed is approximately 4.488 rad/s

The linear speed is approximately 0.987 m/s

c. The largest distance from the center of the disc where the particle can be placed and still not move is approximately 0.399 m from the center of the disc

d. i The maximum angular speed of the disk so that the particle does not move relative to the disk is approximately 6.044 rad/sec

ii When the angular speed with which the disc rotates is more than the the answer of question d i  above, the particle slips on the disc, and the disc begins to rotate faster than the particle, while the particle is swung in an outward radial direction off the disc due to the centrifugal forces

Explanation:

The given parameters are;

The radius of the horizontal disc, r = 45 cm = 0.45 m

The time the disc makes one full revolution, T = 1.40 s

The mass of the particle placed on the disc = 0.054 kg

The location on the disc the particle is placed = 22 cm from the disc's center

a. Please find attached the diagram of the disc created with Microsoft Visio, with arrows representing the velocity and the acceleration of the particle placed on the disk

b. The angular speed, ω = 2·π/T = 2 × π/1.4 ≈ 4.488 rad/s

The linear speed, v = ω × r = 4.488 rad/s × 0.22 m ≈ 0.987 m/s

The linear speed, v ≈ 0.987 m/s

c. The given coefficient of static friction = 0.82

Therefore;

The frictional force that prevents motion = Weight of the particle × The coefficient of static friction

The frictional force that prevents motion is [tex]F_f[/tex] = 0.054 × 9.8 × 0.82 ≈ 0.434 N

[tex]F_f[/tex] ≈ 0.434 N

Therefore, for the largest distance from the center of the disc where the particle can be placed and still not move, r, is given by the formula for the centripetal force, [tex]F_c[/tex], acting on the particle as follows;

For static equilibrium, no movement of the particle relative to the disc, we have;

[tex]F_f[/tex] = [tex]F_c[/tex]

Where;

[tex]F_c = \dfrac{m \times v^2}{r} = m \times \omega ^2 \times r[/tex]

Which gives;

[tex]F_c = {0.054 \ kg \times (4.488 \ rad/s)^2} \times r = F_f = 0.434 \ N[/tex]

r = 0.434 N/(0.054 kg × (4.488 rad/s)²) ≈ 0.399 m

The largest distance from the center of the disc where the particle can be placed and still not move, r = 0.399 m from the center of the disc

d. i From the static equilibrium equation where r = 0.22 m, we have;

[tex]F_c = {0.054 \ kg \times \omega ^2} \times 0.22 \ m = F_f = 0.434 \ N[/tex]

ω =  √(0.434 N/(0.054 kg × (0.22 m))) ≈ 6.044 rad/sec

The maximum angular speed of the disk so that the particle does not move relative to the disk, ω ≈ 6.044 rad/sec

ii When the angular speed with which the disc rotates is more than the the answer of question d i  above, we have

The particle begins to slip on the disc such that the disc rotates faster than the particle and the particle tends to rotate slower than the speed pf the disc and is swung off the disc by centripetal force acting on the particle due to the rotational motion of the disc.

Answer:

boron and neon.

Explanation:

two element compounds are usually ionic when one element is a metal and the other is a non-metal. boron is a metal, whereas neon is a non-metal. i might be wrong but hope it helps.

Answer:

  48 ft/s

Explanation:

Acceleration due to gravity decreased the upward velocity to zero in 1.5 seconds (half the total time), so the initial velocity must have been ...

  (32 ft/s^2)(1.5 s) = 48 ft/s

Answer:

[tex]\frac{dA}{dt} = 28800 \ m^2/year[/tex]

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Geometry

Algebra I

Calculus

Derivatives

Differentiating with respect to time

Basic Power Rule:

Explanation:

Step 1: Define

Area is A = lw

2w = l

w = 300 m

[tex]\frac{dw}{dt} = 24 \ m/year[/tex]

Step 2: Rewrite Equation

Step 3: Differentiate

Differentiate the new area formula with respect to time.

Step 4: Find Rate

Use defined variables

Answer:

28,800 m²/yr

Explanation:

This rectangle has dimensions such that:

We are given [tex]\displaystyle \frac{dw}{dt} = \frac{24 \ m}{yr}[/tex] and want to find [tex]\displaystyle \frac{dA}{dt} \Biggr | _{w \ = \ 300 \ m} = \ ?[/tex] when w = 300 m.

The area of a rectangle is denoted by Area = length * width.

Let's multiply the width and length (with respect to w) together to have an area equation in terms of w:

Differentiate this equation with respect to time t.  

Let's plug known values into the equation:

Simplify this equation.

The area is changing at a rate of 28,800 m²/yr at this point in time.