Answer:
the willing power of student is more important then the material. If a student is willing to make something and express his/her identity then there are more than enough substances he/she can make. For example:- a wooden box can turn into a cute playing car, piggybank or a pencil holder according to his her wish.
Their inferences are based on evidence that they collect during their investigations. Readers learn that scientists gather and interpret evidence and draw conclusions based on this evidence. ... Once scientists have gathered evidence, they use it to make inferences about the things they are investigating.
Explanation:
Given that,
Force, F = 70 N
Angle, ![\theta=50^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D50%5E%7B%5Ccirc%7D)
Acceleration of the boy, a = 1.8 m/s²
(a) Net force is equal to the product of mass and acceleration. So,
![m=\dfrac{F}{a}\\\\m=\dfrac{70}{1.8}\\\\m=38.89\ kg](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7BF%7D%7Ba%7D%5C%5C%5C%5Cm%3D%5Cdfrac%7B70%7D%7B1.8%7D%5C%5C%5C%5Cm%3D38.89%5C%20kg)
(2) Normal force = upward force
So,
![F=mg\cos\theta\\\\F=38.89\times 9.81\times \cos(50)\\F=245.23\ N](https://tex.z-dn.net/?f=F%3Dmg%5Ccos%5Ctheta%5C%5C%5C%5CF%3D38.89%5Ctimes%209.81%5Ctimes%20%5Ccos%2850%29%5C%5CF%3D245.23%5C%20N)
Hence, mass is 38.89 kg and the normal force is 245.23 N.
Answer:
I = 1.06886 N s
Explanation:
The expression for momentum is
I = F t = Δp
therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor
Let's find the components of the initial velocity
sin 28.2 = v_y / v
cos 28.2= vₓ / v
v_y = v sin 282
vₓ = v cos 28.2
v_y = 42.8 sin 28.2 = 20.225 m / s
vₓ = 42.8 cos 28.2 = 37.72 m / s
since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making
θ = -28.2
v_y = -20.55 m / s
v_x = 37.72 m / s
X axis
Iₓ = Δpₓ = ![p_{fx} - p_{ox}](https://tex.z-dn.net/?f=p_%7Bfx%7D%20-%20p_%7Box%7D)
since the ball moves in the x-axis without changing the velocity, the change in moment must be zero
Δpₓ = m
- m v₀ₓ = 0
v_{fx} = v₀ₓ
therefore
Iₓ = 0
Y axis
I_y = Δp_y = p_{fy} -p_{oy}
when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards
v_{fy} = - v_{oy}
Δp_y = 2 m v_{oy}
Δp_y = 2 0.0260 (20.55)
= 1.0686 N s
the total impulse is
I = Iₓ i ^ + I_y j ^
I = 1.06886 j^ N s