Answer:
2.43J
Explanation:
Given parameters:
Mass of the arrow = 0.155kg
Velocity = 31.4m /s
Unknown:
Kinetic energy when it leaves the bow = ?
Solution:
The kinetic energy of a body is the energy in motion of the body;
it can be derived using the expression below:
K.E = 
m v²
m is the mass
v is the velocity
Solve for K.E;
K.E = x 0.155 x 31.4 = 2.43J
Answer:5101.35v
Explanation:
Radius of gold nucleus=7.3×10-15m and a charge of +79e
Q= 79e
e=1.6×10^-19
q= +2e
The nucleus is considered as the point charge where the potential energy between the charges are
U = 1/(4×3.142×Eo)×(qQ)/r
Where r is distance between the charges and the nucleus
r=R+d
V=U/q
U= 1/(4×3.142×Eo)×Q/r
V= 1/(4×3.142×Eo)×Q/(r+d)
9.0×10^9 ×(79×10^-19)/(7.3×10^-15)+(1.5×10^-14)
V= 9.0×10^9 ×(1.264×10^-17)/(2.23×10^-14)
V= 9×10^9×(5.67×10^-14)= 5101.35v
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
Answer:
D.amplifying sound vibrations from the eardrum
this is correct
Answer:
true
Explanation:
Yes, it is true.
As the wattage is more than the prescribed wattage, it becomes overheated.
