Answer:
Initial velocity = 10 m/s
θ = 60°
This is the case of projectile motion
So the horizontal component of velocity 10 m/s = 10 cosθ
u = 10 cosθ
u = 10 cos 60°
u=5 m/s
x= 5 m
So in the horizontal direction
x = u .t
5 = 5 .t
t = 1 sec The vertical component of velocity 10 m/s = 10 sinθ
Vo= 10 sinθ
Vo= 10 sin 60°
Vo = 8.66 m/s
h=3.75 m
So height of robot = 3.75 - 0.75 m
height of robot =3 m
Answer:
TEJ as this is a thing you wont get
Answer:
1.2cm
Explanation:
V=(2ev/m)^1/2
=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2
=6.2x10^5m/s
Radius of resulting path= MV/qB
= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6
=0.012m
=1.2cm
C real,inverted and smaller than the object
Answer:
Explanation:
First, It's important to remember F = ma, and in this problem m = 13.3 kg
This can be reduced to a simple system of equations problem. Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them. So let's call them F1 and F2, with F1 arger than F2. Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.
Can you solve this system of equations seeing them like this, or do you need more help?
