Help with these questions thank you

An object of mass 0.50 kg is released from the top of a building of height 8 m. The object experiences a horizontal constant for

liberstina [14]

To solve this problem with the given elements we will apply the linear motion kinematic equations. We will start by calculating the time taken, with the vertical displacement data. Subsequently, with the components of the acceleration, we will obtain the magnitude of the total acceleration, to finally obtain the horizontal displacement with the data already found.

PART A) From vertical movement we know that the acceleration is equivalent to gravity and the displacement is 8m so the time taken to carry out the route would be

$h = \frac{1}{2} gt^2$

Here,

$h = 8 m$

$g = 9.8 m/s^2$

Replacing,

$8 = 0.5 * 9.8 * t^2$

$t = 1.277 sec$

PART B) Now, Magnitude of acceleration

$a = \sqrt{a_x^2 + a_y^2}$

$a_x = \frac{1.9}{0.5} = 3.8 m/s^2$

$a_y = g = 9.8 m/s^2$

Thus, magnitude of net acceleration

$a = \sqrt{3.82^2 + 9.8^2}= 10.51 m/s^2$

PART C) Finally the displacement along horizontal direction is:

$s =v_0 t + \frac{1}{2} a t^2$

$s = 0 + \frac{1}{2} (3.8)(1.277)^2$

$s = 3.098 m$

Therefore the distance traveled along the horizontal direction before it hits the ground is 3.098m

7 0

3 years ago

Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =

Crank

Answer:

Fnet = 12 N

Explanation:

Force on a point charge due to another point charge = kq1q2 / d^2

Force on +32uC = due to + 20uC + due to -60uC

where uC = 1 x 10^-6 C and k = 9 x 10^9 N m^2 / C^2

Net Force =

$= \frac{1}{4\pi \epsilon _0} [\frac{32 \times 10^-^6 \times60\times10^-^6}{(60/100)^2}-\frac{32 \times 10^-^6 \times20\times10^-^6}{ (40/100)^2} ]$

$F_{net}=9 \times10^9\times 10^-^1^2[\frac{32\times60\times10^4}{60\times60} -\frac{32\times20\times10^4}{40\times40} ]$

$=90[32(\frac{80-60}{60\times 80} )]\\\\=90\times32\times0.004167\\\\=12N$

Fnet = 12 N

8 0

4 years ago

(a) Check all of the following that are correct statements, where E stands for γmc2. Read each statement very carefully to make

Reil [10]

Answer:

V=9.60m/s

Attached is the solution

5 0

3 years ago

If two variables have a non-related relationship and one of the variables is changed, how will the other variable change?

PolarNik [594]

If one of the variables is changed, that tells nothing about what happens to the other one, or IF anything happens, or when, or how long it lasts. Because they are UN-RELATED. You just said so yourself.

None of the choices says this.

6 0

3 years ago

Planets that are cold have only a small amount of gravity

Nonamiya [84]

There is no theoretical OR observational evidence for that statement.

4 0

4 years ago