Answer:
the shooting angle ia 18.4º
Explanation:
For resolution of this exercise we use projectile launch expressions, let's see the scope
R = Vo² sin (2θ) / g
sin 2θ = g R / Vo²
sin 2θ = 9.8 75/35²
2θ = sin⁻¹ (0.6)
θ = 18.4º
To know how for the arrow the tree branch we calculate the height of the arrow at this point
X2 = 75/2 = 37.5 m
We calculate the time to reach this point since the speed is constant on the X axis
X = Vox t
t2 = X2 / Vox = X2 / (Vo cosθ)
t2 = 37.5 / (35 cos 18.4)
t2 = 1.13 s
With this time we calculate the height at this point
Y = Voy t - ½ g t²
Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²
Y = 6.23 m
With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch
B: Energy lose
i say this because in order to change they lose energy.
Answer:
Invisible UV energy reacts with emulsion sensitizer and hardens the stencil so it won't dissolve with water and rinse down the drain
Explanation:
<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q</h2>
Explanation:
Specific heat capacity
It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .
It is given as :
Heat absorbed = mass of substance x specific heat capacity x rise in temperature
or ,
Q= m x c x t
In above question , it is given :
For Q
mass of Q = m
Temperature changed =T₂/2
Heat supplied = x
Q= mc t
or
X=m x C₁ X T₁
or, X =m x C₁ x T₂/2
or, C₁=X x 2 /m x T₂ (equation 1 )
For another quantity : P
mass of P =m/2
Temperature= T₂
Heat supplied is same that is : X
so, X= m/2 x C₂ x T₂
or, C₂=2X/m. T₂ (equation 2 )
Now taking ratio of C₂ to c₁, We have
C₂/C₁= 2X /m.T₂ /2X /m.T₂
so, C₂/C₁= 1/1
so, the ratio is 1: 1
D.to bring light together
