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

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.

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

Answer:

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

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:

Answer:

Explanation:

I hope it's helpful for you

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.

I think static is the correct answer

Answer:

The distance traveled before takeoff is 1720 m

Explanation:

a = + 3.2 m/s²

t = 32.8 s

Vᵢ = 0 m/s

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

Answer:

4600kg

Explanation:

Density=mass÷volume

920=m/5

m=920×5=4600kg

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

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.

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.

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

Answer:

3i - 2j + 8k

Explanation:

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

= 3i -2j + 8k

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.

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

Answer:

a

Explanation:

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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

Answer:

Explanation:

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

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

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) 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:

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

- 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

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

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!

[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]

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.

Answer:

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

Explanation:

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

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

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.