Explanation:
When a radioactive substance decays then the fast moving electrons emitted by it is known as beta ray. Basically, a number of beta particles are ejected by a beta ray.
Symbol of a beta particle is 
. A beta ray is a natural decay of a radioactive element. As we know that opposite charges get attracted towards each other. So, a beta ray gets attracted towards a positively charged plate.
Therefore, we can conclude that following are the characterizes a beta ray:
- a product of natural radioactive decay.
- is attracted to the positively charged plate in an electric field.
- is composed of electrons.
Answer:
1 second
Explanation:
The impulse exerted on the boy is equal to its change of momentum:
where
F = 500 N is the push on the boy
is the contact time
m = 25 kg is the mass of the boy
is the change in velocity of the boy
Solving the formula for the contact time, we find
The magnitude of the electric field at the third vertex of the triangle is determined as zero.
<h3>Electric field at the third vertex of the triangle </h3>
The electric field at the third vertex of the equilateral triangle due to the other charges placed on the first and second vertices is calculated as follows;
E = E(13) + E(23)
E = (kq₁)/r² + (kq₂)/r²
where;
- q1 is positive charge
- q2 is negative charge
E = (kq₁)/r² - (kq₂)/r²
E = 0
Thus, the magnitude of the electric field at the third vertex of the triangle is determined as zero.
Learn more about electric field here: brainly.com/question/14372859
#SPJ1
Hey there Kendrell!
Yes, this is very true, when the car slows down, our bodies will tend to lean forward a little bit, and this is actually due to the "motion of inertia".
Inertia allows for this to happen, this is why in this case, we have this case.
Hope this helps.
~Jurgen
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
