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4 years ago

Ive tried numerous methods, but I cant seem to get this right. Please help, and explain step by step.

1 answer:

3 0

Well, just for fun, let's see hat the form of the limit is
v = dx/dt = b + 3 c t^2 
= 1.5 + 1.92 t^2 
========================= 
now 
from 1 to 3 
x(3)= 1.5*3 +.64*27=4.5+17.28 
= 21.78 
x(1) = 1.5+.64 = 2.14 
so 
x(3)-x(1) = 19.64 
delta t = 3 - 1 = 2 seconds 
so 
v during interval = 19.64/2 
= 9.82 m/s 
now check that with calculus v in the middle of that range 
v(2) =1.5+ 1.92 (4) = 9.18 m/s 
that makes sense, the parabola is headed up so the instant speed i the middle should be a bit less than average.

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Answer:

6.0 m/s

Explanation:

According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.

Therefore, we can write:

$KE_i+PE_i =KE_f+PE_f$

$\frac{1}{2}mu^2+0=\frac{1}{2}mv^2+mgh$

where:

m is the mass of the athlete

u is the initial speed of the athlete (at the bottom)

0 is the initial potential energy of the athlete (at the bottom)

v = 0.80 m/s is the final speed of the athlete (at the top)

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 1.80 m is the final height of the athlete (at the top)

Solving the equation for u, we find the initial speed at which the athlete must jump:

$u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s$

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4 years ago

A car is traveling at the bottom of a 9.00-meter-radius circular hill with a constant speed v. The moment the car is at the bott

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Answer:

Explanation:

reading of scale = reaction force of surface R

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given ,

m v² / R = .80 mg

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3 years ago

Two balls move away from each other, both traveling at 7 m/s. One has a mass of 2 kg and the other has a mass of 3 kg

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4 years ago

The helicopter was deformed and destroyed in the __?___ collision.

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Answer:

The helicopter was deformed and destroyed in the inelastic collision.

Explanation:

When two object collide there exist two way of colliding: elastic collision and inelastic collision.
Two terms are considered during the collision: kinetic energy and momentum.
If both of these terms are conserved in any collision then there is no significant loss of property, this is called as elastic collision.

If only momentum is conserved but kinetic energy is converted into other forms then it is inelastic collision. In inelastic collision, the energy is lost in the form of vibration, sound etc. causing the damage to colliding object.

Hence the deformation of helicopter was due to inelastic collision.

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4 years ago

Two particles, each with charge 55.3 nC, are located on the y axis at y 24.9 cm and y -24.9cm (a) Find the vector electric field

ser-zykov [4K]

Answer:

Ex = kq 2x / ∛ (x² + y²)² and Ex = 2008 N / C

Explanation:

a) The electric field is a vector quantity, so we must find the field for each particle and add them vectorially, as the whole process is on the X axis,

The equation for the electric field produced by a point charge is

E = k q / r²

With r the distance between the point charge and the positive test charge

We look for each electric field

Particle 1. Located at y = 24.9 m, let's use Pythagoras' theorem to find the distance

r² = x² + y²

E1 = k q / (x² + y²)

Particle 2. located at x = -24.9 m

r² = x² + y²

E2 = k q / (x² + y²)

We can see that the two fields are equal since the particles have the same charge and coordinate it and that is squared.

In the attached one we can see that the Y components of the electric fields created by each particle are always the same and it is canceled, so we only have to add the X components of the electric fields. Let's use Pythagoras' theorem to find

Let's measure the angle from axis X

cos θ = CA / H = x / (x2 + y2) ½

E1x = E1 cos θ

E2x = = E1 cos θ

The resulting field

Ey = 0

Ex = E1x + E2x 2 E1x

Ex = 2 k q / (x² + y²) cos θ) = 2 k q / (x² + y²) x / √(x² + x²)

Ex = kq 2x / ∛ (x² + y²)²

b) For this part we substitute the numerical values

Ex = 8.99 10⁹ 55.3 10⁻⁹ x / (x² + 0.249 2) ³/₂

Ex = 497.15 x / (x² + 0.062) ³/₂

Point where can the value of the electric field x = 38.1 cm = 0.381 m

Ex = 497.15 0.381 / (0.381² + 0.062) ³/₂

Ex = 497.15 0.381 / (0.1452 + 0.062) 3/2 = 189.41 / 0.2072 3/2

Ex= 189.41 /0.0943

Ex = 2008 N / C

c) E = 1.00 kN / C = 1000 N / C

To solve this part we must find x in the equation

Ex = 497.15 x / (x² + 0.062) ³/₂

Let's use some arithmetic

Ex / 497.15 = x / (x² + 0.062) ³/₂

[Ex / 497.15] ²/₃ = [x / (x² + 0.062) 3/2] ²/₃

∛[Ex / 497.15]² = (∛x²) / (x² + 0.062) (1)

The roots of this equation are the solution to the problem,

For Ex = 1.00 kN / C = 1000 N / C

[Ex / 497.15] 2/3 = 1000 / 497.15) 2/3 = 1,312

1.312 = (∛x² ) / (x² + 0.062)

1.312 (x² + 0.062) = ∛x²

1.312 X² - ∛x² + 1.312 0.062 = 0

1.312 X² - ∛x² + 0.0813 = 0

We need used computer

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3 years ago