How many 5 cm squares

Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c

Mariulka [41]

Answer:

A. $v_{3}=12.17m/s$

B. $v_{car}=6.3m/s\\v_{truck}=-6.3m/s$

C. ΔK $=-4.13x10^3J$

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

$v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg$

A. Since the two vehicles become entangled the final mass is:

$m_{3}=102kg+103kg=205kg$

From linear momentum we got that:

$p_{1}=p_{2}$

$m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}$

$v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}$

$v_{3}=12.17m/s$

B. The change in velocity of both vehicles are:

For the car

$v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s$

For the truck

$v_{truck}=12.17m/s-18.5m/s=-6.3m/s$

C. The change in kinetic energy is:

ΔK= $K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})$

ΔK= $\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J$

ΔK $=-4.13x10^{3}J$

6 0

3 years ago

28. Which of the following correctly shows the order of highest amount of friction to the lowest amount of

Bad White [126]

Answer:

$\mathrm{d.\:Static,\: Sliding,\:Rolling}$

Explanation:

Static friction occurs when an object initially starts at rest. When the surfaces of the materials touch, the microscopic unevenness interlock greatest with each other, causing the most friction out of the three.

During sliding friction, an object is already moving or in motion. The microscopic surfaces still interlock, but because the object is in motion, it has a momentum. Therefore, the magnitude of sliding friction is less than that of static friction.

Rolling friction occurs when an object rolls across some surface. Rather than surfaces interlocking, rolling friction is caused by the constant distortion of surfaces. As it rolls, the surfaces of the object are constantly wrapping and changing. This distortion causes the rolling friction. However, it is much less in magnitude when compared to static or sliding friction.

4 0

3 years ago

A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m

Levart [38]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

$a=2.92164g$

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

$\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s$

Angular velocity is 1.3823 rad/s

Linear velocity is given by

$v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s$

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

$a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2$

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

$\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g$

$a=2.92164g$

Centripetal force is given by

$F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N$

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

3 0

4 years ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

3 years ago

What is #6<br><br> IM GIVING 40 POINTS

frosja888 [35]

Instantaneous velocity, on the other hand, describes the motion of a body at one particular moment in time. Acceleration is a vector which shows the direction and magnitude of changes in velocity. Its standard units are meters per second per second, or meters per second squared. (this is for number 3)

4 0

3 years ago