Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1
Answer:
ax = -3.29[m/s²]
ay = -1.9[m/s²]
Explanation:
We must remember that acceleration is a vector and therefore has magnitude and direction.
In this case, it is accelerating downwards, therefore for a greater understanding we will make a diagram of said vector, this diagram is attached.
![a_{x}=-3.8*cos(30) = -3.29 [m/s^{2}]\\ a_{y}=-3.8*sin(30) = -1.9 [m/s^{2}]](https://tex.z-dn.net/?f=a_%7Bx%7D%3D-3.8%2Acos%2830%29%20%3D%20-3.29%20%5Bm%2Fs%5E%7B2%7D%5D%5C%5C%20a_%7By%7D%3D-3.8%2Asin%2830%29%20%3D%20-1.9%20%5Bm%2Fs%5E%7B2%7D%5D)
Answer:
16.12
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Answer:
The strength of magnetic field is 0.2 Tesla.
Explanation:
Data from the question is
Length (L) of wire ; L=0.10 m
Current in wire ; I= 2.0 A
Force on wire ; F = 0.04 N
Angle = Right angle So, 

Now ,
We have to find the magnetic Field strength (B)
For this formula for Force on wire in magnetic field is

Further modified as

Now insert values in the formula


So, the strength of magnetic field is 0.2 Tesla.