A worker is holding a filled gas cylinder still. Which two sentences are true about the energy of the filled gas cylinder?

A 1.00-kilogram ball is dropped from the top of a building. just before striking the ground, the ball's speed is 12.0 meters per

Anarel [89]

During the fall, the potential energy stored in the ball is converted into kinetic energy.
Thus,
PE = KE before hitting the ground
= 1/2 • mv^2
= 1/2 • 1 • 12^2
= 72J

6 0

2 years ago

You are participating in a NASA traineeship, working with a group planning a new landing on Mars. Your supervisor has come up wi

aivan3 [116]

Answer:

$h=17005.8 km$

Explanation:

Newton's law of universal gravitation states that the force experimented by a satellite of mass m orbiting Mars, which has mass $M=6.39\times10^{23} kg$ at a distance r will be:

$F=\frac{GMm}{r^2}$

where $G=6.67\times10^{-11}Nm^2/kg^2$ is the gravitational constant.

This force is the centripetal force the satellite experiments, so we can write:

$F=ma_{cp}=mr\omega^2=mr(\frac{2\pi}{T})^2=\frac{4\pi^2mr}{T^2}$

Putting all together:

$\frac{GMm}{r^2}=\frac{4\pi^2mr}{T^2}$

which means:

$r=\sqrt[3]{\frac{GM}{4\pi^2}T^2}$

Which for our values is:

$r=\sqrt[3]{\frac{(6.67\times10^{-11}Nm^2/kg^2)(6.39\times10^{23} kg)}{4\pi^2}(1.026\times24\times60\times60s)^2}=20395282m=20395.3km$

Since this distance is measured from the center of Mars, to have the height above the Martian surface we need to substract the radius of Mars R=3389.5 km , which leaves us with:

$h=r-R=20395.3km-3389.5 km=17005.8 km$

6 0

2 years ago

A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu

melomori [17]

Answer:

please read the answer below

Explanation:

The angular momentum is given by

$|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)$

By taking into account the angles between the vectors r and v in each case we obtain:

a)

v=(2,0)

r=(0,1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

b)

r=(0,-1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

c)

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

d)

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

e)

r=(1,1)

angle = 45°

$L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}$

f)

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

g)

r=(-1,-1)

angle = 135°

$L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}$

h)

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

4 0

2 years ago

A stiff wire 50.0 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the l

zloy xaker [14]

Answer:

Magnitude of the force is 2.135N and the direction is 41.8° below negative y-axis

Explanation:

The stiff wire 50.0cm long bent at a right angle in the middle

One section lies along the z axis and the other is along the line y=2x in the xy plane

$\frac{y}{x} = 2$

tan θ = 2

Therefore,

slope m = tan θ = y / x

$\theta=\tan^-^1(2)=63.4^0$

Then length of each section is 25.0cm

so, length vector of the wire is

$\hat I= (-25.0)\hat k +(25.0) \cos 63.4^0 \hati +(25.0) \sin63.4^0 \hatJ\\\\\hat I = (11.2) \hat i + (22.4) \hat j - (23.0) \hat k$

And magnetic field is B = (0.318T)i

Therefore,

$\bar F = \hat I (\bar l \times \bar B)$

$\bar F = (20.0)[(0.112m)i +(0.224m)j-(0.250m)k \times 90.318T)i]$

$= (20.0)(i(0)+j(-0.250)(0.318T)+k[0-(0.224m)(0.318T)]\\\\=(20.0)(-0.250)(0.318)j-(20.0)(0.224)(0.318T)\\\\=-(1.59N)j-(1.425N)k$

Magnitude of the force is

$F = \sqrt{(-1.59N)^2+(-1.425N)^2\\} \\F = 2.135N$

Direction is

$\alpha = \tan^-^1(\frac{-1.425N}{-1.59N} )\\\\= 41.8^0$

Magnitude of the force is 2.135N and the direction is 41.8° below negative y-axis

5 0

2 years ago

Read 2 more answers

A pump jack scaffold must be fitted with two ______ gripping mechanisms to prevent slippage.

FromTheMoon [43]

A pump jack scaffold must be fitted with two positive gripping mechanisms to prevent slippage. Pump jacks are a uniquely designed scaffold consisting of a platform supported by movable brackets on vertical poles. The brackets are designed to be raised and lowered in a manner similar to an automobile jack. It is important to make sure that pump jack brackets have two positive gripping mechanisms to prevent any failure or slippage.

5 0

2 years ago