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Nat2105 [25]
3 years ago
10

How large is theta? Please help!!!!

Physics
1 answer:
Lorico [155]3 years ago
8 0

Answer:

30°

Explanation:

The angle between the normal and the vertical is the same as the angle between the incline and the horizontal.  We can show this with geometry.

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A pulling force is called what?<br> A. normal<br> B. tension<br> C. balanced<br> D. compression
joja [24]
<span>There is no special name for that. Physics is usually just concerned with "forces", and doesn't specify whether the force pushes or pulls. If you want to be more specific, you can just call it a "pulling force".
I hoped this was satisfying!:)</span>
3 0
4 years ago
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Digital thermometers often make use of thermistors, a type of resistor with resistance that varies with temperature more than st
Nezavi [6.7K]

Answer:

The temperature coefficient of resistivity for a linear thermistor is 1.38\times10^{-3}^{\circ}C^{-1}

Explanation:

Given that,

Initial temperature = 0.00°C

Resistance = 75.0 Ω

Final temperature = 525°C

Resistance = 275 Ω

We need to calculate the temperature coefficient of resistivity for a linear thermistor

Using formula for a linear thermistor

R=R_{0}(1+\alpha\Delta T)

R=R_{0}+R_{0}\alpha\Delta T

\alpha=\dfrac{R-R_{0}}{R_{0}\Delta T}

Put the value into the formula

\alpha=\dfrac{275-75}{275\times(525-0)}

\alpha=1.38\times10^{-3}^{\circ}C^{-1}

Hence, The temperature coefficient of resistivity for a linear thermistor is 1.38\times10^{-3}^{\circ}C^{-1}

4 0
3 years ago
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If the magnitude of the moment of F about line CD is 57 N·m, determine the magnitude of F.If the magnitude of the moment of F ab
bazaltina [42]

Answer:

F_ab = 260.17 N

Explanation:

Given:

- Moment of force F about CD, (M)_cd = 57 Nm

Find:

- First we will write down the position vectors of points A, B , C , D:

- We will take left and bottom most corner of cube to be the origin.

- The unit vectors i , j , k are along vertical planes and outside the plane, respectively.

- The position vectors wrt to the origin are:

                             Point A = 0.2 k

                             Point B = 0.4 i + 0.2 j

                             Point C = 0.2 j + 0.4 k

                             Point D = 0.4 i + 0.4 k

- Now we will determine the Force vector F_ab along vector AB.

                             vec (AB) = B - A = 0.4 i + 0.2 j - 0.2 k

                             unit (AB) = 0.4 i + 0.2 j - 0.2 k  / sqrt ( 0.4^2 + 2*0.2^2)

                                            = [5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

Hence,

                              vec(F_ab) = Fab*[5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

- Now, form a unit vector along the line CD:

                             vec(CD) = D - C = 0.4 i - 0.2 j

                             unit (CD) = 0.4 i - 0.2 j / sqrt ( 0.4^2 + 0.2^2)

                                           = [sqrt(5)]*(0.4 i - 0.2 j)

- Now select a point on line CD, lets say C. Find the moment arm from line of action of force along AB and line CD. Hence, vector AC is:

                               vec(AC) =r_ac = C - A = 0.2 j + 0.2 k

- Now the moment about a line CD due to force is:

                              (M)_cd = unit(CD) . ( r_ac x vec(F_ab) )

The cross product of r_ac and vec(F_ab) is as follows:

                               (M)_c =  ( r_ac x vec(F_ab) ) :

                                \left[\begin{array}{ccc}i&j&k\\0&0.2&0.2\\0.8165&0.40824&-0.40824\end{array}\right]

                              (M)_c =  F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k]

The dot product of (M)_c and unit (CD)  is as follows:

                              (M)_cd = unit(CD) . (M)_c :

 (M)_cd = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k] .  [sqrt(5)]*(0.4 i - 0.2 j)

                              (M)_cd = F_ab*(sqrt(30) / 25)

- The given magnitude of the moment is (M)_cd. Calculate F_ab:

                               57 = F_ab*(sqrt(30) / 25)  

                              F_ab = 260.17 N

7 0
3 years ago
Which statement describes a property of a proton?
lesantik [10]

Answer:

They have no charge and are present in the nucleus of an atom. They have a negative charge and travel around the nucleus of an atom. They have a positive charge and travel around the nucleus of an atom

Explanation:

7 0
3 years ago
HELP 50 POINTS
Inga [223]

Answer:

A

Explanation:

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3 years ago
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