Answer:
Stationary
20N
Explanation:
From the graph, we see that the body traveling is on a fixed position. Therefore, it is a stationary body.
The graph given is a position - time curve.
This curve depict a body changing position with given time.
Since the line of the curve is on a single position, the body is not changing position with the passage of time therefore, it is a stationary object.
B. 20N
From Newton's third law of motion we understand that "action and reaction force are equal but oppositely directed".
Since the person is exerting a force of 20N on the balance.
So, the reaction force by the balance is 20N upward.
Answer:
The maximum speed will be 26.475 m/sec
Explanation:
We have given mass of the toy m = 0.50 kg
radius of the light string r = 1 m
Tension on the string T = 350 N
We have to find the maximum speed without breaking the string
For without breaking the string tension must be equal to the centripetal force
So 
So 
v = 26.475 m /sec
So the maximum speed will be 26.475 m/sec
In an alpha decay, an atom emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons: this means that during this kind of decay, the original atom loses 2 protons and 2 neutrons from its nucleus.
This also means that the atomic number Z of the element (the atomic number is the number of protons in the nucleus) decreases by 2 units in the process, while the mass number A (the mass number is the sum of the number of protons and neutrons) decreases by 4 units.
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
