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1 year ago

special metal hangers or stirrips called joist hangers are used when joist must be _____ the bottom of the grider or beam

Physics

1 answer:

djyliett [7]1 year ago

7 0

Answer:

Special metal hangers or stirrups called joist hangers are used when joist must be flush with the bottom of the girder or beam.

Explanation:

A joist hanger also known as a beam hanger is a mechanical device which is used to fasten joists and rafters.
The rafters are the carried members to beams and headers are the carrying members.

Thus, special metal hangers or stirrups called joist hangers are used when joist must be flush with the bottom of the girder or beam.

Learn more about construction here:

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Speed has ____ magnitude and _____ direction ??

ArbitrLikvidat [17]

Speed has only magnitude and no direction.

6 0

3 years ago

A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (

Anika [276]

Answer:

$a=9.8 rad/s^{2}$

Explanation:

Torque, $\tau$ is given by

$\tau=Fr$ where F is force and r is perpendicular distance

$R=0.5Lcos\theta$ where $\theta$ is the angle of inclination

Torque, $\tau$ can also be found by

$\tau=Ia$ where I is moment of inertia and a is angular acceleration

Therefore, Fr=Ia and F=mg where m is mass and g is acceleration due to gravity

Making a the subject, $a=\frac {Fr}{I}=\frac {mgr}{I}$ and already I is given as

$I=\frac {mL^{2}}{3} and r is 0.5Lcos\theta$ hence

$a=\frac {0.5mgLcos\theta}{1/3 mL^{2}}$

$a=\frac {3gcos\theta}{2L}$

Taking g as 9.81, $\theta$ is given as 37 and L is 1.2

$a=\frac {3*9.81cos37}{2*1.2}=9.7932679419$

$a=9.8 rad/s^{2}$

4 0

3 years ago

Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,

REY [17]

Answer:

a) 238U, 40K and 87Rb, b) 235U and to a lesser extent 40K , c) he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

Let's write the half-life times of the given materials

87Rb T ½ = 4.75 1010 years

147Sm T ½ = 1.06 1011 years

235U = 7,038 108 years

238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed

f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

8 0

3 years ago

Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal po

zvonat [6]

Answer:

$0.293I_0$

Explanation:

When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.

Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:

$I_1=\frac{I_0}{2}$

Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:

$I_2=I_1 cos^2 \theta$

where

$\theta$ is the angle between the axes of the two polarizer

Here we have

$\theta=40^{\circ}$

So the intensity after the 2nd polarizer is

$I_2=I_1 (cos 40^{\circ})^2=0.587I_1$

And substituting the expression for I1, we find:

$I_2=0.587 (\frac{I_0}{2})=0.293I_0$

5 0

2 years ago

A certain circuit is composed of two series resistors. The total resistance is 10 Ohms. One of the resistors is 4 Ohms. The othe

Mariulka [41]

<h3>Given :- </h3>

A certain circuit is composed of two series resistors
The total resistance is 10 ohms
One of the resistor is 4 ohms

We have to find the value of other resistor?

<h3>Let's Begin :-</h3>

We know that,

In series combination,

When a number of resistances are connected in series, the equivalent I.e resultant resistance is equal to the sum of the individual resistances and is greater than any individual resistance

That is,

Rn in series = R1 + R2 + R3.....So on

Therefore,

According to the question,

We have,

R1 + R2 = 10 Ω

4 + R2 = 10Ω

R2 = 10 - 4

R2 = 6Ω

Hence, The value of R2 resistor in series is 6Ω

4 0

2 years ago