Expand each vector into their component forms:
![\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N](https://tex.z-dn.net/?f=%5Cvec%20A%3D%284.5%5C%2C%5Cmathrm%20N%29%28%5Ccos%5Ctheta_A%5C%2C%5Cvec%5Cimath%2B%5Csin%5Ctheta_A%5C%2C%5Cvec%5Cjmath%29%3D%282.58%5C%2C%5Cvec%5Cimath%2B3.69%5C%2C%5Cvec%5Cjmath%29%5C%2C%5Cmathrm%20N)
Similarly,
![\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N](https://tex.z-dn.net/?f=%5Cvec%20B%3D%28-1.23%5C%2C%5Cvec%5Cimath%2B0.860%5C%2C%5Cvec%5Cjmath%29%5C%2C%5Cmathrm%20N)
![\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N](https://tex.z-dn.net/?f=%5Cvec%20C%3D%28-3.44%5C%2C%5Cvec%5Cimath-4.91%5C%2C%5Cvec%5Cjmath%29%5C%2C%5Cmathrm%20N)
Then assuming the resultant vector
is the sum of these three vectors, we have
![\vec R=\vec A+\vec B+\vec C](https://tex.z-dn.net/?f=%5Cvec%20R%3D%5Cvec%20A%2B%5Cvec%20B%2B%5Cvec%20C)
![\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N](https://tex.z-dn.net/?f=%5Cvec%20R%3D%28-2.09%5C%2C%5Cvec%5Cimath-0.368%5C%2C%5Cvec%5Cjmath%29%5C%2C%5Cmathrm%20N)
and so
has magnitude
![\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N](https://tex.z-dn.net/?f=%5C%7C%5Cvec%20R%5C%7C%3D%5Csqrt%7B%28-2.09%29%5E2%2B%28-0.368%29%5E2%7D%5C%2C%5Cmathrm%20N%5Capprox2.12%5C%2C%5Cmathrm%20N)
and direction
such that
![\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ](https://tex.z-dn.net/?f=%5Ctan%5Ctheta_R%3D%5Cdfrac%7B-0.368%7D%7B-2.09%7D%5Cimplies%5Ctheta_R%3D-170%5E%5Ccirc%3D190%5E%5Ccirc)
Answer:
a) 46.5º b) 64.4º
Explanation:
To solve this problem we will use the laws of geometric optics
a) For this part we will use the law of reflection that states that the reflected and incident angle are equal
θ = 43.5º
This angle measured from the surface is
θ_r = 90 -43.5
θ_s = 46.5º
b) In this part the law of refraction must be used
n₁ sin θ₁ = n₂. Sin θ₂
sin θ₂ = n₁ / n₂ sin θ₁
The index of air refraction is n₁ = 1
The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface
θ_s = 90 - θ
θ_s = 90- 43.5
θ_s = 46.5º
sin θ₂ = 1 / 1.68 sin 46.5
sin θ₂ = 0.4318
θ₂ = 25.6º
The angle with respect to the surface is
θ₂_s = 90 - 25.6
θ₂_s = 64.4º
measured in the fourth quadrant
Answer:
A hypothesis is what you think will happen.
A conclusion is the results of an experiment summarized.
Hope this helps.
Answer:
v=32.9m/s
Explanation:
The acceleration needed to mantain a circular motion of radius r and speed v is given by the equation ![a=v^2/r](https://tex.z-dn.net/?f=a%3Dv%5E2%2Fr)
This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).
We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:
![v=\sqrt{ar} = \sqrt{12.5gr}=\sqrt{(12.5)(9.8m/s)(8.84m)}=32.9m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bar%7D%20%3D%20%5Csqrt%7B12.5gr%7D%3D%5Csqrt%7B%2812.5%29%289.8m%2Fs%29%288.84m%29%7D%3D32.9m%2Fs)
Answer:
100 cc
Explanation:
Heat released in cooling human body by t degree
= mass of the body x specific heat of the body x t
Substituting the data given
Heat released by the body
= 70 x 3480 x 1
= 243600 J
Mass of water to be evaporated
= 243600 / latent heat of vaporization of water
= 243600 / 2420000
= .1 kg
= 100 g
volume of water
= mass / density
= 100 / 1
100 cc
1 / 10 litres.