Ask question

Ask question

All categories

3 years ago

11

Knowing that the board is 4 meters long, one child weighs 300 n and the other 250 n, where should one of the children sit so tha

t the board remains horizontal? (lever)

1 answer:

Neko [114]3 years ago

8 0

Answer:v nxfgdjngdnmgndjfnncnfndngndsnbxzmnfn

Explanation:

You might be interested in

What is the result of two displacement vectors having opposite directions?

jonny [76]

<u>Answer:</u>

<em>Resultant of two vectors having opposite direction is the difference of the two displacements having the same direction as the larger vector. </em>

<u>Explanation:</u><u> </u>

Resultant of two vectors is obtained by performing the vector addition operation. When the directions of both vectors are same the resultant’s direction will also be the same as the inputs. When two vectors have opposite directions, one direction will be taken positive making one vector positive and the other negative.

By performing addition of a positive and negative number we are actually taking the difference between both. Thus performing vector addition of two vectors with opposite directions is equivalent to finding the difference between the vectors. Consider a system consisting of a solid block, on which two forces F1 and F2 act in the opposite direction.

One force will be considered positive and the other is considered negative. The resultant is given by the difference of two force vectors. Displacement of the block will be in the direction of the greater force.

5 0

3 years ago

Read 2 more answers

Which type wire should be used to increase resistance

8_murik_8 [283]

Nichrome wire. That's the stuff that toasters are made from. The resistance is pretty high, considering the diameter. 1 meter is at about the same guage as that listed below for copper is about 96 ohms.

Most of the time you are trying to use wire with the least resistance.

A meter of copper has a listed resistance of 0.024 ohms / meter. The wire is a 19 guage wire which makes it pretty thin.

===============

I'm not sure what you are asking. If want the resistance of something in terms of what would increase the resistance of the same material for both calculations then

Rule 1: It you decrease the wire diameter, you increase the resistance

Rule 2: If you increase the length of the wire, you increase the resistance.

Both rules assume you are using something like copper.

7 0

3 years ago

Read 2 more answers

A 2320 pound roller coaster starts from rest and is launched such that it creates a 110 ft high hill with a speed of 65 mph. The

Sergeu [11.5K]

Answer: E = 941738.537J

Explanation:

to begin,

given that the mass = 2320 pound = 1052.334 kg

Δh = 110 ft = 33.528 m

given that distance (d) = 1283 ft = 391.058 m

also the speed (v) is 65 mph = 29.058 m/s

force (F) = 87 pounds = 386.995 N

from our knowledge in work energy theory;

E = Fd + 1/2mv² + mgh

E = (386.995 × 391.058) + (1/2×1052.334×29.058²) + (1052.334×9.81×33.528)

E = 151337.491 + 444278.2 + 346122.84

E = 941738.537J

i hope this helps, cheers.

3 0

3 years ago

Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an

8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 $m/s^2$ where as radial the component of acceleration is 8.77 $m/s^2$

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

$r=e^u$

So the transverse component of acceleration are given as

$a_{\theta}=(ru''+2r'u')\\$

Here

$r=e^u\\r=e^{\pi/4}\\r=2.1932 m$

$r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m$

So

$a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\$

The transverse component of acceleration is 26.32 $m/s^2$

The radial component is given as

$a_r=r''-r\theta'^2$

Here

$r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m$

So

$a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2$

The radial component of acceleration is 8.77 $m/s^2$

6 0

3 years ago

A boat sails south with the help of a wind blowing in the direction S36°E with magnitude 300 lb. Find the work done by the wind

goldfiish [28.3K]

Answer:

The work done by the wind as the boat moves 130 ft is (rounded) W= 31,550 ft-lb.

Explanation:

F= 300 lb < -54º

Fsouth= 300 lb * cos(36º)

Fsouth= 242.7 lb

d= 130 ft

W= F*d

W= 31551 ft-lb

6 0

3 years ago