Question 2 (1 point)

n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend

Kitty [74]

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

$p_f=p_i\\mv_1+Mv_2=(m+M)v$

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

$(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}$

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

$E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2$

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

$(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m$

hence, the heigth is 68cm

4 0

3 years ago

A quarterback takes the ball from the line of scrimmage, runs backward for 12.1 yds, then runs sideways parallel to the line of

Over [174]

Answer:

The answer is 35.45 yds

Explanation:

You have to picture this to be able to understand it better (see attachment).

Start at the origin which is when the quarterback (QB) takes the ball. He runs backwards 12.1 yards, runs sideways for 19.8 yards (it doesn´t matter if he runs right or left), then he throws the ball forward 41.5 yards. If you look at the attachment, you can see I drew the path that the football followed. And then connected the dots from the origin and finish. The distance between those two points is the magnitude of the resultant displacement.

In order to calculate it, all you need to do is use the Pythagoream theorem, which says that the square of the hypotenuse equals the sum of the squares of the legs a and b of the triangle rectangle.

$R^{2} = a^{2} + b^{2}$ then solve for R
$R = \sqrt{a^{2}+b^{2} }$

In this case, you know the length of leg a to be 19.8 yards which how much it moves sideways. And then, to get the length of leg b, all you need to do is substract how much it moved backwards from the 41.5yards forward displacement. This results in b leg being 29.4 yards long.

Now you have a triangle with:

a = 19.8 yards
b = 29.4 yards

Substituting this numbers in the equation:

$R = \sqrt{19.8^{2}+29.4^{2} }$
R = 35.45 yards

3 0

3 years ago

The teacher measured the maximum height and the minimum height of the plastic duck above the screen as the wave passed. The teac

fenix001 [56]

The teacher measured the maximum height and the minimum height of ...

5 days ago — ... screen as the wave passed . The teacher repeated his measurements. Table 4 shows the teacher's measurements. Calculate the mean amplitude of the water wave.

4 0

3 years ago

A scientific law is a statement that describes some aspect of the world. It is based on repeated experimental observations. What

Doss [256]

I’m pretty sure it is C third law of motion
I looked it up on google and went through tons of facts about Isaac Nuton

5 0

3 years ago

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

Salsk061 [2.6K]

Answer:

The fluids speed at a) $0.105\,m^{2}$ and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$ and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$ (1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to $A_{2}=0.105\,m^{2}$ , so now we can solve (2) for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is $Q=\frac{V}{t}$ , so we can write:

$A_{1}v_{1}=\frac{V}{t}$ , solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$

3 0

4 years ago