Answer:
It is an educated guess
Explanation:
let me know if the bottom ones need to be answered too.
-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless. I guess that's what "smooth" means.
-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.
-- Your force is 6.4 N.
Acceleration = (force)/(mass) = 6.4/2.43 m/s²<em>
</em> That's about <em>2.634 m/s²</em> <em>
</em>(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)
-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is
Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = <em>2.45 N</em>.
That's the force that's accelerating the little block, so that must be the tension
in the string.
To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m
The potential difference is related to the electric field by:
∆V=Ed
where,
∆V is the potential difference
E is the electric field
d is the distance
what is potential difference?
The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.
We want to know the distance the detectors have to be placed in order to achieve an electric field of
E=1v/cm=100v/cm
when connected to a battery with potential difference
∆v=1.5v
Solving the equation,we find
![d = \frac{ \:Δv}{e}](https://tex.z-dn.net/?f=d%20%3D%20%20%5Cfrac%7B%20%5C%3A%CE%94v%7D%7Be%7D%20)
![= \frac{1.5v}{100v/m}](https://tex.z-dn.net/?f=%20%3D%20%20%5Cfrac%7B1.5v%7D%7B100v%2Fm%7D%20)
![= 1.5 \times 10 {}^{ - 2} m](https://tex.z-dn.net/?f=%20%3D%201.5%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%202%7D%20m)
learn more about potential difference from here: brainly.com/question/28166044
#SPJ4
The refractive index for glycerine is
![n_g=1.473](https://tex.z-dn.net/?f=n_g%3D1.473)
, while for air it is
![n_a = 1.00](https://tex.z-dn.net/?f=n_a%20%3D%201.00)
.
When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
![\theta_c = \arcsin ( \frac{n_2}{n_1} )](https://tex.z-dn.net/?f=%5Ctheta_c%20%3D%20%5Carcsin%20%28%20%5Cfrac%7Bn_2%7D%7Bn_1%7D%20%29)
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
Answer:
![|\vec r|=339.82\ m](https://tex.z-dn.net/?f=%7C%5Cvec%20r%7C%3D339.82%5C%20m)
![\theta=-6.67^o](https://tex.z-dn.net/?f=%5Ctheta%3D-6.67%5Eo)
Explanation:
<u>Displacement
</u>
It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:
![\vec r=\vec r_1+\vec r_2](https://tex.z-dn.net/?f=%5Cvec%20r%3D%5Cvec%20r_1%2B%5Cvec%20r_2)
Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with
![x=rcos\theta](https://tex.z-dn.net/?f=x%3Drcos%5Ctheta)
![y=rsin\theta](https://tex.z-dn.net/?f=y%3Drsin%5Ctheta)
And the vector is expressed as
![\vec z==](https://tex.z-dn.net/?f=%5Cvec%20z%3D%3Cx%2Cy%3E%3D%3Crcos%5Ctheta%2Crsin%5Ctheta%3E)
The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is
![\vec r_1==\ km=\ m](https://tex.z-dn.net/?f=%5Cvec%20r_1%3D%3C0.198cos0%5Eo%2C0.198sin0%5Eo%3E%3D%3C0.198%2C0%3E%5C%20km%3D%3C198%2C0%3E%5C%20m)
The second move is (145 m , -15.8°). The angle is negative because it points South of East. The second displacement is
![\vec r_2==\ m](https://tex.z-dn.net/?f=%5Cvec%20r_2%3D%3C145cos%28-15.8%5Eo%29%2C145sin%28-15.8%5Eo%29%3E%3D%3C139.52%2C-39.48%3E%5C%20m)
The total displacement is
![\vec r=\ m+\ m](https://tex.z-dn.net/?f=%5Cvec%20r%3D%3C198%2C0%3E%5C%20m%2B%3C139.52%2C-39.48%3E%5C%20m)
![\vec r=\ m](https://tex.z-dn.net/?f=%5Cvec%20r%3D%3C337.52%2C-39.48%3E%5C%20m)
In (magnitude,angle) form:
![|\vec r|=\sqrt{337.52^2+(-39.48)^2}=339.82\ m](https://tex.z-dn.net/?f=%7C%5Cvec%20r%7C%3D%5Csqrt%7B337.52%5E2%2B%28-39.48%29%5E2%7D%3D339.82%5C%20m)
![\boxed{|\vec r|=339.82\ m}](https://tex.z-dn.net/?f=%5Cboxed%7B%7C%5Cvec%20r%7C%3D339.82%5C%20m%7D)
![\displaystyle tan\theta=\frac{-39.48}{337.52}=-0.1169](https://tex.z-dn.net/?f=%5Cdisplaystyle%20tan%5Ctheta%3D%5Cfrac%7B-39.48%7D%7B337.52%7D%3D-0.1169)
![\boxed{\theta=-6.67^o}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctheta%3D-6.67%5Eo%7D)