A small 1240-kg SUV has a wheelbase of 3.2 m. If 67% of its weight rests on the front wheels, how far behind the front wheels is the wagon's center of mass

Answers

Answer 1

Answer:

Explanation:

Let d be the distance to the center of mass from the front wheels

Sum moments about the front wheel contact point to zero

1240(9.8)[d] - 1240(9.8)(1 - 0.67)[3.2] = 0

1240(9.8)[d] = 1240(9.8)(1 - 0.67)[3.2]

                d = (1 - 0.67)[3.2]

                d = 1.056 m


Related Questions

The center of mass of a 1600 kg car is midway between the wheels and 0.7 m above the ground. The wheels are 2.6 m apart. (a) What is the minimum acceleration A of the car so that the front wheels just begin to lift off the ground

Answers

Answer:

Explanation:

I guess we are ASSUMING that this is a rear wheel drive car as a front wheel drive car will never get the front wheel normal force to zero

If we consider it as a statics problem and choose our moment center carefully...say 0.7 m above the rear wheel to ground contact point.

Call the traction force at the rear wheels F

The normal force on the front wheels will be zero, so no moment generated by the front wheels.

Summing moments about our chosen point to zero

1600(9.8)[2.6 / 2] - F[0.7] = 0

F = 291,200

this force will create an acceleration of

a = F/m

a = 291200/1600

a = 182 m/s²

which is about 18.6 times gravity acceleration

A rock is thrown off a cliff with a speed of 5 m/s downward. How far will it fall after 7 seconds have elapsed?

Free-fall Acceleration is -10 m/s^2

I also need the Formula

Answers

Answer:

Explanation:

s = s₀ + v₀t + ½at²

if the throw point is origin and UP the positive direction and ignoring air resistance.

s = 0 + (-5)(7) + ½(-10)(7²)

s = 0 - 35 - 245

s = - 280 m

A racing car on the straight accelerates from 100 km/h to 316 km/h in three seconds.
What is its acceleration?

40m/s2

30m/s2

20m/s2

72m/s2

Answers

Answer:

[tex]20m/s^2[/tex]

Explanation:

Solution is attached. I apologize if it is a little messy.

A car is driving 12m/sec, has to stop suddenly because a pedestrian dashes out in front of the car. If the coefficient of kinetic friction between the tires and parking lot is ∪=60

what is the time, after the breaks are applied, before the car comes to a stop? Sketch the velocity time graph for the car's motion from the instant the breaks are applied until the car comes to a stop.

Answers

Answer:

Approximately [tex]2\; \rm s[/tex], assuming that the floor of this parking lot is level, [tex]\mu_{\rm k} = 0.60[/tex], and [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].

Explanation:

Let [tex]m[/tex] denote the mass of this vehicle. Weight of this vehicle: [tex]m\, g[/tex].

If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight: [tex]N = m \, g[/tex].

Given that [tex]\mu_{\rm k}[/tex], the kinetic friction between this vehicle and the ground would be consistently [tex]\mu_{\rm k} \, N = \mu_{\rm k} \, m \, g[/tex] until the vehicle comes to a stop.

Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be [tex](-\mu_{\rm k} \, m \, g)[/tex] (negative because this force is opposite to the direction of the motion.)

By Newton's second law of motion, the acceleration of this vehicle would be:

[tex]\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= \frac{-\mu_{\rm k} \, m \, g}{m} \\ &= -\mu_{\rm k}\, g \\ &= -0.60 \times 9.81\; \rm m\cdot s^{-2} \\ &= -5.886\; \rm m\cdot s^{-2}\end{aligned}[/tex].

In other words, braking would reduce the velocity of this vehicle by a constant [tex]5.886\; \rm m\cdot s^{-1}[/tex] every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from [tex]v_{0} = 12\; \rm m\cdot s^{-1}[/tex] to [tex]v_{1} = 0\; \rm m\cdot s^{-1}[/tex]:

[tex]\begin{aligned}t &= \frac{v_{1} - v_{0}}{a} \\ &= \frac{0\; \rm m\cdot s^{-1} - 12\; \rm m\cdot s^{-1}}{-5.886\; \rm m \cdot s^{-2}} \\ &\approx 2.0\; \rm s \end{aligned}[/tex].

Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.

a box has a mass of 4 kg and surface area 4 metre square calculate the pressure exerted by the box on the floor​

Answers

- BRAINLIEST answerer ❤️

Answer:

9.8

Explanation:

we know ,

f=m.g(g=9.8m/s^2)

now,

p=force/area

p=m.g/a

p=4×9.8/4

p=4 pascal.

if u put g as 10 then ypu will get 10 pascal

If an electron moves in a direction perpendicular to the same magnetic field with this same linear speed,
what is the radius of its circular orbit?

Answers

Answer:

An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.

Explanation:

Answer:

An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.

Explanation:

given two vector
p= 2i + 2j + 4k
q = i - 4j + 4k
find p+ q​

Answers

Answer:

3i - 2j + 8k

Explanation:

p + q = (2i + i) + (2j - 4j ) + (4k + 4k )

= 3i -2j + 8k

3) A 60. kg person is in an elevator. The elevator starts from rest and then accelerates upwards at 2.0 m/s^2 for 4.0 seconds. Calculate the work done by the normal force on the person. *

Answers

Answer:

WD = 960 J

Explanation:

WD = work done (J)

F = force (N)

s = displacement (m)

m = mass (kg) = 60

a = acceleration (m/s²) = 2

t = time (s) = 4

u = initial velocity (m/s) = 0

The formulas or equations that are relevant ate:

WD = F × s

F = m × a

s = u + at

We want to find WD, so we need to now the force and the displacement (or distance);

We calculate force, in Newtons, with the formula F = ma:

F = 60 × 2

F = 120 N

We also need displacement, which get with the formula s = u + at:

s = 0 + 2(4)

s = 8 m

Now we have F and s, we can calculate WD:

WD = 120 × 8

WD = 960 J

Methodology:

Starting with what you want to find, in this case WD, list the formula/s you could use;

Then, identify the information you need for the formula and whether or not you are given that information;

Next, list the formulas for the information you don't have and once again, identify whether the information you are given is sufficient to use those formulas;

Once you can calculate all necessary information, then proceed to calculate the values and finally, the answer;

I suggest also keeping a list of all the variables as I've done at the top of my working so it is clear for you to see and use.

6) An object is released from rest at the top of a ramp inclined at 30. degrees up from the horizontal. Due to friction, the ramp is only 20. % efficient. What is the object's speed after it slides down ALONG the ramp for 2.0 m? *

Answers

Answer:

Explanation:

I've been doing these types of problems for many years and I don't think I've ever seen an "efficiency" rating on a ramp.

I'm going to ASSUME that 20% efficient means that 80% of the Potential energy that gets converted becomes system internal heat energy.

Potential energy at the start of a 2.0 m slide

PE = mgh = mg2sin30 = mg2(½) = mg J

0.8mg J gets converted to heat and 0.2mg converts to kinetic energy

0.2mg = ½mv²

v² = 0.4g

v = √(0.4(9.8)) = 1.979898... ≈ 2.0 m/s

2. Which of the following contributions did Louie De Broglie do for electronic structure of matter? A. determined the speed of electron of hydrogen atom B. proposed a theory that electrons showed characteristics similar to light C. provided mathematical operation for the characteristics of light D. recorded the movement of proton in the nucleus of an atom

❤️​

Answers

Answer:

In 1924 Louis de Broglie introduced the idea that particles, such as electrons, could be described not only as particles but also as waves. This was substantiated by the way streams of electrons were reflected against crystals and spread through thin metal foils.

Explanation:

I know I probably didn't answer your question, I just used all of my knowledge that I learned about Louie De Broglie. Hope it helps!

A tennis player strikes the tennis ball with an initial velocity of 44.7 m/s horizontally. The ball is initially 1.28 m above the ground and 12.9 m from the 0.914 m tall net. Does the tennis ball make it over the net?

Answers

Hi there!

We can begin by finding the total time taken for the ball to reach the net using the equation:

dₓ = vₓt

12.9 = 44.7t

12.9/44.7 = t = 0.289 s

Now, we can use the following equation to solve for displacement in the Y direction:

d = y₀ + vit + 1/2at²

There is no initial vertical velocity, so:

d = y₀ + 1/2at²

Plug in known values:

d = 1.28 + 1/2(-9.8)(0.289²)

d = 0.87m

Thus, since 0.87 m < 0.914 m, the tennis ball does NOT make it over the net.

please answer this as fast as you can i need it

Answers

Answer:

it says pdf only i dont knowwhat u want me to do

numerical problems:
a.) convert 300K into the celsius scale.
b.) convert 220 centigrade scale into kelvin scale.
c.) convert 20 ventigrade scale into Fahrenheit scale.
d.) convert 260 Fahrenheit into centigrade. pls help me to solve this problems

Answers

The answer is:
A) 300K = 26.85°C

300K - 273.15K = 26.85°C


B) 220 °C = 493.15K

220 °C + 273.15 = 493.15K


C) 20 °C = 68 °F

(20°C x 9/5) + 32 = 68°F


D) 260°F = 126.667°C

(260°F − 32) × 5/9 = 126.667°C

Is electrical energy the same or different from energy it takes to play a soccer game

Answers

Answer:

Diffrent

Explanation:

Electrical energy is electrical charges moving. When you play soccer kinetic energy is used. Kinetic energy is the movement of atoms, objects, and electrons.

Help pls!

A 3 kg mass is raised a distance of 14 m above the earth by a vertical force of 93 N.
The final kinetic energy of the mass, to 3 significant figures, if it was originally at rest is:

Answers

[tex] \large★·.·´¯`·.·★ {Answer}★·.·´¯`·.·★[/tex]

As we know that Kinetic Energy is the Energy that is possessed by a moving object. and if the object is at rest then it doesn't have velocity therefore there is no kinetic Energy.

In the numerical terms we can express it as : -

[tex] \sf0.00 \: \: joules[/tex]

[tex]꧁  \:  \large \frak{Eternal \:  Being } \: ꧂[/tex]

A 115 kg hockey player, Adam, is skating east when he tackles a stationary 133 kg player, Bob. Afterward, they move at 1.35 m/s east. What was Adam's velocity before the collision? (Unit = m/s) ​

Answers

Answer:

Explanation:

Conservation of momentum

115v + 133(0) = (115 + 133)1.35

v = 2.911304...

v= 2.91 m/s east

Answer:

The velocity east is 2.91

Explanation:

Fill in the box

2.91

Two trucks leave at different times (from the same place) headed for the same city. Both trucks arrive at the same time. Based on this information, which of the following sentences is true? Select one:
a. The trucks travelled the same distance in the same amount of time.
b. The trucks were traveling at the same average speed.
c. The trucks travelled different distances.
d. The truck that left later was travelling faster.

Answers

Answer:

d. The truck that left later was travelling faster

Explanation:

Both trucks travelled from the same place to the same place, meaning they both travelled the same distance;

They both arrive at the same time, but the second truck left later so it took less time to travel the distance than the first truck;

The only variable that can account for this difference is speed;

The one that left later, therefore, must have been going faster.

Which is the main gas that makes up the Earth's atmosphere?​

Answers

Answer:

78 percent nitrogen

Explanation:

I hope it's helpful for you

An object is moving with an initial velocity of 3.3 m/s. It is then subject to a constant acceleration of 3.7 m/s2 for 10 s. How far will it have traveled during the time of its acceleration?

I also need the complete Formula (Nothing left out)

Answers

Answer:

Explanation:

s = s₀ + v₀t + ½at²

ASSUMING the acceleration is in the direction of initial motion.

s = 0 + 3.3(10) = ½(3.7)(10²)

s = 218 m

Two blocks are set in a pully system as shown in fig below. Block A sits on the frictionless table while block B hags freely. The pully is light and frictionless towards the light string that runs over it. If the Block A has mass of 3.4 kg and Block has 3.5 kg, what would be the magnitude of the acceleration (in ms-2) of the blocks? [g = 9.8 ms=2]​

Answers

Answer:

Explanation:

F = ma

a = F/m

a = mBg / (mB + mA)

a = 3.5(9.8)/(3.5 + 3.4)

a = 4.971014...

a = 5.0 m/s²

If you want to use individual Free Body Diagrams

mass A will have downward weight and upward normal forces equal at mAg

and a horizontal force of string tension T

F = ma

T = mAa

mass B will have a downward force of mBg and an upward force of T

mBg - T = mBa

substitute for T

mBg - mAa = mBa

mBg = a(mB + mA)

a = mBg / (mB + mA)   which is identical to the above answer.

Which properties make a metal a good material to use for electrial wires

Answers

Answer:

Most importantly metals can pass an electric current without being affected and changed by the electricity. Electrical conductivity combined with ductility makes metals the most suitable materials for electrical transmission wires.

How much did the pressure drop in the storm's center from November 9, 1200z, until November 11, 0000z

Answers

The pressure dropped by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.

Using the attached map below, in the morning of November 9, the pressure situated at the storm's center = 1000 MB isobar located at the center.

Meanwhile, on November 11, 0000z the pressure situated at the storm's center = 976 MB isobar  

The difference in this pressure is regarded as the pressure drop in the storm's center and it is determined as follows;

= 1000 MB - 976 MB

= 24 MB

Therefore, we can conclude that the pressure drop by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.

Learn more about pressure drop in an isobaric process here:

https://brainly.com/question/13089696?referrer=searchResults

Conservation of Energy Roller Coaster A roller coaster cart of mass 100kg travels on a track with one loop. Fill in blanks A-H. А. KE=OJ PE=120000J h= А. V= B B KE=___CE PE=60000J h= _D V= E KE=__F PE=40000J h=__G_ V= KE= PE= h=Om v= K D E F G H K​

Answers

(a) The height of the roller coaster at 120,000 potential energy is 122.45 m.

(b) The velocity of the roller coaster at 0 J kinetic energy is 0.

(c) The height of the roller coaster at 60,000 potential energy is 61.23 m.

(d) The velocity of the roller coaster at 60,000 J kinetic energy is 34.64 m/s.

(e) The height of the roller coaster at 40,000 potential energy is 40.82 m.

(f) The velocity of the roller coaster at 80,000 J kinetic energy is 40 m/s.

The given parameters:

mass of the roller coaster, m = 100 kg

When the kinetic energy = 0 and potential energy = 120,000 J

The height of the roller coaster is calculated as follows;

P.E = mgh

[tex]h = \frac{P.E}{mg}\\\\h = \frac{120,000}{100 \times 9.8} \\\\h = 122.45 \ m[/tex]

Since the kinetic energy = 0, the velocity of the roller coaster = 0

When the potential energy, P.E = 60,000 J, the kinetic energy, K.E is calculated as;

P.E + K.E = M.A

P.E + K.E = 120,000

60,000 + K.E = 120,000

K.E = 120,000 - 60,000

K.E = 60,000 J

The height of the roller coaster at 60,000 potential energy is calculated as follows;

[tex]h = \frac{P.E}{mg} \\\\h = \frac{60,000}{100 \times 9.8} \\\\h =61.23 \ m[/tex]

The velocity of the roller coaster at 60,000 J kinetic energy is calculated as follows;

[tex]K.E = \frac{1}{2} mv^2\\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{ \frac{2K.E}{m}} \\\\v = \sqrt{ \frac{2\times 60,000}{100}}\\\\v = 34.64 \ m/s[/tex]

When the potential energy, P.E = 40,000 J, the kinetic energy, K.E is calculated as;

P.E + K.E = M.A

40,000 + K.E = 120,000

K.E = 120,000 - 40,000

K.E = 80,000

The height of the roller coaster at 40,000 potential energy is calculated as follows;

[tex]h = \frac{P.E}{mg} \\\\h = \frac{40,000}{100 \times 9.8} \\\\h = 40.82 \ m[/tex]

The velocity of the roller coaster at 80,000 J kinetic energy is calculated as follows;

[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 80,000}{100} } \\\\v = 40 \ m/s[/tex]

Learn more here:https://brainly.com/question/19969393

a boat's engine can give it a velocity of 25m/s. if the boat heads east across a river which of the following due south with a velocity of 8.5m/s; what is the resultant velocity of the boat? (remember you must find both a magnitude and direction!)

Answers

Answer:

Explanation:

v = √(25² + 8.5²) = 26.40549... = 26 m/s

θ = arctan(8.5/25) = 18.77803... = 19° S of E

What type of equilibrium maintains body position during sudden motion?

dynamic
rotational
static
balanced

Answers

I think static is the correct answer

Why is it important for the community to take action

Answers

Answer:

Why is community action important? ... Involving communities in the design and delivery of services can help to achieve a number of objectives, including: Building community and social capacity – helping the community to share knowledge, skills and ideas. Community resilience – helping the community to support itself

An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.​

Answers

Answer:

The distance traveled before takeoff is 1720 m

Explanation:

Given:

a = + 3.2 m/s²

t = 32.8 s

Vᵢ = 0 m/s

To Find:

d = ?

Now,

d = Vᵢ × t + 0.5 × a × t²

d = (0 m/s) × (32.8 s) + 0.5 × (3.20 m/s²) × (32.8 s)²

d = 1720 m

Thus, The distance traveled before takeoff is 1720 m

-TheUnknownScientist 72

calculate the mass of a block of ice having volume 5m³. (density of ice≈920 kg/m³)​

Answers

Answer:

4600kg

Explanation:

Density=mass÷volume

920=m/5

m=920×5=4600kg

plz answer the question.

Answers

Answer:

a

Explanation:

sana po makatulong <3♡♡

Need help with dot product

Answers

[tex]\textbf{A}\cdot\textbf{B} = 11.5[/tex]

Explanation:

The dot product between two vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B}[/tex] is defined as

[tex]\textbf{A}\cdot\textbf{B} = AB\cos{\theta}[/tex]

where A and B are the magnitudes of the vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B},[/tex] respectively and [tex]\theta[/tex] is the angle between the two. Since A = 3, B = 5 and [tex]\theta = 40°,[/tex] the dot product [tex]\textbf{A}\cdot\textbf{B}[/tex] is

[tex]\textbf{A}\cdot\textbf{B} = (3)(5)(0.766) = 11.5[/tex]

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