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3 years ago

What do you need to know to be able to determine how far a projectile travels horizontally?

1 answer:

5 0

As for me, there are two suitable answers for the question represented above and here is a short explanation why I consider these two to be correct :
D. The horizontal velocity of the projectile and B. The length of time before it lands
 $v_x = v*cos(x) [

d_x = v_x(t) + ½at²$ -- this led me to answers! Hope everything is clear! Regards!

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A car is moving in the positive direction along a straight highway and accelerates at a constant rate while going from point A t

rusak2 [61]

Answer:

The time where the avergae speed equals the instaneous speed is T/2

Explanation:

The velocity of the car is:

v(t) = v0 + at

Where v0 is the initial speed and a is the constant acceleration.

Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

$v_{ave} = \frac{1}{T}\int\limits^T_0 {v(t)} \, dt$

v_ave = v0+a(T/2)

We can esaily note that when t=T/2

v(T/2)=v_ave

Now we want to know where the car should be, the osition of the car is:

$x(t) = x_A + v_0 t + \frac{1}{2}at^2$

Where x_A is the position of point A. Therefore, the car will be at:

x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2

8 0

3 years ago

In Newton’s second law, if the net force acting on object doubles. The object’s Will also double

Anna71 [15]

Answer: Explanation:

If the net force on an object is doubled, its acceleration will double If the mass of an object is doubled, the acceleration will be halved .

3 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

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

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

PLEASEEE HELPPPP

solniwko [45]

Answer:

Explanation:

average speed more than 25.0m/s.

5 0

3 years ago

Read 2 more answers

Which waves require a material medium for transmission? a.) light waves b.) radio waves c.) sound waves d.) microwaves

shepuryov [24]

Answer:

mechanical waves require a medium

5 0

3 years ago