What happens when both sides of the tug-of-war rope are equal in force?What is the net force in this scenario?

Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat

Alja [10]

Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

Note: the mass doesn't change, it is only the velocity that change.

Also,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Let assume that if is constant acceleration

Then, m and a is constant,

Then,

K.E is directly proportional to d

So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

K.E_1 = E_k

d_1 = d

d_2 = d/3

K.E_2 = K.E_1 / d_1 × d_2

K.E_2 = E_k × ⅓d / d

Then,

K.E_2 = ⅓E_k

So, the new kinetic energy is one third of the E_k

7 0

2 years ago

You are driving downhill on a rural road with a 3% grade at a speed of 45 mph. While playing on the side of the road, a child ac

Dennis_Churaev [7]

Answer: a) 95.07m b) 81.88 m

Explanation:

a)

For finding the distance when vehicle is going downhill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31-0.03)}$

Stop sight distance= 95.07 m

b)

For finding the distance when vehicle is going uphill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31+0.03)}$

Stop sight distance= 81.88 m

5 0

2 years ago

A 100-kg box is sitting on a 10 degree incline with a coefficient of friction of 0.5. At what angle must the incline be raised t

Marta_Voda [28]

Answer:

26.56°

Explanation:

μ= 0.5

F = ?

N =?

Fs = μN

Fs = force acting due to friction (on an inclined plane)

N = normal force.

μ= coefficient of friction

μ= tanΘ

μ= tan 0.5

Θ = tan⁻¹ 0.5

Θ = 26.56°

The angle in which is required for the box to start moving across the inclined plane is 26.56°

5 0

2 years ago

Three solid, uniform, cylindrical flywheels, each of mass 65.0 kg and radius 1.47 m, rotate independently around a common axis t

Mkey [24]

Answer:

the angular momentum is 1015.52 kg m²/s

Explanation:

given data

mass of each flywheel, m = 65 kg

radius of flywheel, r = 1.47 m

ω1 = 8.94 rad/s

ω2 = - 3.42 rad/s

to find out

magnitude of the net angular momentum

solution

we get here Moment of inertia that is express as

I = 0.5 m r² .................1

put here value and we get

I = 0.5 × 65 × 1.47 × 1.47

I = 70.23 kg m²

and

now we get here Angular momentum that is express as

L = I × ω ...........................2

and Net angular momentum will be

L = 2 × I x ω1 - I × ω2

put here value and we get

L = 2 × 70.23 × 8.94 - 70.23 × 3.42

L = 1015.52 kg m²/s

so

the angular momentum is 1015.52 kg m²/s

3 0

2 years ago

Consider an elevator carrying Kermit the frog weighing 4000.0 N is held 5.00 m above a spring with a force constant of 8000.0 N/

Brums [2.3K]

Answer:

Maximum compression distance (x) = 2.236 m (Approx)

Explanation:

Given:

Weight of frog = 4,000 N

Height = 5 m

Constant force = 8,000 N/m

Frictional force = 1,000 N

Find:

Maximum compression distance (x)

Computation:

Using Law of conservation;

mgh = 1/2(k)(x)²

4,000(5) = 1/2(8,000)(x)²

Maximum compression distance (x) = 2.236 m (Approx)

7 0

1 year ago