All categories

3 years ago

What type of pipe wrenches are designed for turning and holding where marrying is not objectionable?

1 answer:

3 0

The type of pipe wrenches that are designed for turning and holding where marrying is not objectionable is called the chain wrench. This tool has two serrated that are adjusted to grip the pipe. This is very useful for those objects where marrying is not objectionable since this can be adjusted for that purpose.

You might be interested in

Sjskskszpd no djdkzsnkakzx d

Mazyrski [523]

Answer:

jadjsakdjadkjadhfmafkj

Explanation:

4 0

3 years ago

Pics pls you know what I mean

V125BC [204]

boiiii I wont show you nun

4 0

3 years ago

Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time–when they\'re not sleep

Anit [1.1K]

Answer:

mice total momentum (-0.000250, 0.00639) Kg m

Explanation:

To calculate the moment of the mice we must multiply their mass by their velocities, remember that the moment is a vector quantity, so we use the components of velocity

mouse 1

m1 = 0.0225 Kg

V1 = (0.869, -0.283) m / s

Px = m Vx

Px1 = 0.0225 0.869

Px1 = 0.01955 Kg m

Py = m Vy

Py1 = 0.0225 (-0.283)

Py1 = -0.006368 Kg m

P1 = (0.0196, -0.00637) Kg m

Mouse 2

m2 = 0.0223 Kg

Px2 = 0.0223 (-0.883) = -0.0196 Kg m

Py2 = 0.0223 (-0.253) = -0.00564 Kg m

P2 = (-0.0196, -0.00564) Kg m

Mouse 3

m3 = 0.0197

Px3 = 0.0197 0.345 = 0.00680 Kg m

Py3 = 0.0197 0.803 = 0.0158 Kg m

P3 = (0.00680, 0.0158) Kg m

Mouse4

m4 = 0.0127 Kg

Px4 = 0.0127 (-0.555) = -0.00705 Kg m

Py4 = 0.0127 0.205 = 0.00260 Kg m

P4 = (-0.00705, 0.00260) Kg m

To find the total momentum we must add each component of the individual moments

Px = Px1 + Px2 + Px3 + Px4

Py = py1 + Py2 + Py3 + Py4

Px = 0.0196 -0.0196 +0.00680 -0.00705

Px = -0,000250 Kg m

Py = -0.00637 -0.00564 +0.0158 +0.00260

Py = 0.00639 Kg m

P = (-0.000250, 0.00639) Kg m

7 0

3 years ago

A newly discovered planet is found to have a density if 2/3pe and a radius of 2RE, where PE and RED are the density and radius o

Liono4ka [1.6K]

Missing question in the text of the exercise. Found on internet:
"What is the acceleration due to gravity on the surface of the planet?"

Solution:
The gravitational acceleration at Earth's surface is given by:
$g= \frac{GM}{r^2}$
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius

The Earth's mass can be rewritten also as the product between the Earth's density, $d$ , and its volume (the volume of a sphere of radius r):
$M=dV=d ( \frac{4}{3} \pi r^3)= \frac{4}{3} \pi d r^3$

Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
$g' = \frac{GM'}{r'^2}$ (1)
so we need to find M' and r'.

The problem says the radius of the new planet is twice the Earth's radius:
$r'=2r$ (2)
and that its density is 2/3 of Earth's density:
$d'= \frac{2}{3} d$
so the mass M' of the new planet is, with respect to the Earth's mass:
$M' = d'V' = \frac{4}{3} \pi d' (r')^3 = \frac{4}{3} \pi ( \frac{2}{3}d) (2r)^3 = ( \frac{4}{3} \pi d r^3 )( \frac{16}{3}) = \frac{16}{3} M$ (3)

And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
$g'= \frac{G( \frac{16}{3}M) }{(2r)^2}= \frac{GM}{r^2} \frac{4}{3} = \frac{4}{3}g$
And since $g=9.81 m/s^2$ , we find
$g'= \frac{4}{3}(9.81 m/s^2)=13.1 m/s^2$

8 0

3 years ago

Bryce has selected a major in political science but has a little interest in understanding public relations

Dominik [7]

Hi, you've asked an incomplete question. However, I assumed you want additional comments about this scenario.

Explanation:

Note, the term Major as used in academic contexts refers to a person's preferred interest or specialization at college or university.

Hence, the statement about Bryce appears to be contradictory, since we are told he "has little interest in understanding public relations".

8 0

2 years ago