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sergiy2304 [10]

3 years ago

11

what car experiences more impulse - one stopped by a wood block fixed in place, or one stopped by a foam noodle that is allowed

to move?

1 answer:

forsale [732]3 years ago

4 0

What do you mean with this question?

You might be interested in

What net force acting on a 14 KG wagon produces an acceleration of 1.5 MS^2?

valentina_108 [34]

21N

Explanation:

Given parameters:

Mass of wagon = 14kg

Acceleration = 1.5m/s²

Unknown:

Net force on wagon = ?

Solution:

Force is a pull or push on a body that causes a body to change its state. It is expressed as:

Force = mass x acceleration

Force on wagon = 14 x 1.5 = 21N

Learn more:

Force brainly.com/question/10470406

#learnwithBrainly

7 0

3 years ago

Which statement about elements and atoms is true?

Vika [28.1K]

Atoms are the smallest unit of an element

8 0

3 years ago

A ball is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the wa

vovangra [49]

Answer:

a) 48.5 ft/s

b) 36.5 ft

c) -80.3 ft/s

Explanation:

a)

The equation of motion of the ball is :

y(t) = -16.1 ft/s^2 * t^2 + Vo*t

Where Vo is the initial velocity

If y(5s) = - 160 ft:

-160 ft = -16.1 ft/s^2 * (5 s)^2 + Vo*(5s)

Solving for Vo

Vo = (16.1*25- 160) ft / 5s = 48.5 ft/s

b)

To answer this question we must first know when the velocity became zero, at this time is when the ball was at its highest point.

v(t) = -32.2 ft/s^2 * t + Vo

t = Vo/32.2ft/s^2 = 1.5 s

And now, the highest point which the ball reached is given by:

y(1.5s) = -16.1 ft/s^2 * (1.5)^2 + Vo*(1.5s)

y(1.5s) = 36.52 ft

c)

We now need the time at which y(t') = -64 ft

-64 = -16.1*t'^2 + 48.5*t'

By means of the quadratic formula, we find that

t' = 4.00498 s ≈ 4 s

And the velocity at t = 4s is:

v(4s) = -32.2 ft/s^2 * 4s +48.5 ft/s = -80.3 ft/s

3 0

3 years ago

2. A Se lanza un electrón con rapidez inicial v0 = 1.60×106 m/s hacia el interior de un campo uniforme entre las placas paralela

Vanyuwa [196]

Answer:

A) E = 145.6 N / C , B) y= 2,8 10-7 m with a downward direction

C) he shape of the trajectory of the two particles is to simulate a parabola,

D) F_{e} /F_{g} = 10³⁴

Explanation:

A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric

F = m a

- e E = m a

a = - e E / m

with the field directed downward, the acceleration is in the vertical upward direction.

We look for how much the electron moves with kinematics, in the x direction there is no acceleration,

x axis (parallel to plates)

x = v₀ t

t = x / v₀

y axis (perpendicular to plates)

y = y₀ + $v_{oy}$ t + ½ a t²

Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known

y = ½ a t²

we substitute

y = ½ (e E /m) (x / v₀)²

y = ½ e x2 /m v₀² E

we look for the electric field

E = 2 m y v₀² / e x²

where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m

let's calculate

E = 2 9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)

E = 145.6 N / C

B) The electron is exchanged for a proton

Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates

y = ½ e x² / m v₀² E

y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)² 145.6

y = 28.4375 10⁻⁸ m

since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small

In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)

C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards

D) The force of gravity

$F_{g}$ = G m M / R²

electron

Between the electron and the positive charges of the conducting plate

F_{g}= 6.67 10⁻¹¹ 1.67 10⁻²⁷ 9.1 10⁻³¹ / (0.5 10⁻³)²

F_{g} = 4.1 10⁻⁵¹ N

electric force

F_{e} = -e E

F_{e} = - 1.6 10⁻¹⁹ 145.6

F_{e} = 2.3 10⁻¹⁷ N

let's look for the reason between these two forces

F_{e} / F_{g} = 2.3 10⁻¹⁷ / 4.1 10⁻⁵¹

F_{e} /F_{g} = 10³⁴

We see that the electric force is many orders of magnitude higher than the gravitational force.

5 0

3 years ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago