Answer: C. -1.16 meters/second2
Explanation:
A= v/t (velocity/time)
in this case: v=7 and t=6
So, A= 7/6
A=1.16
The graph is decreasing so accelleration would be negative
A= <u>-1.16 meters/second2</u>
<u>Option C!</u> ; )
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Answer:
The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction
Explanation:
Inelastic Collision
Given data
mass of glider A m1= 0.125kg
initial velocity u1=0
final velocity v1= 0.600 m/s
mass of glider B m2= 0.375kg
initial velocity u2=0
final velocity v2=?
We know that the expression for the conservation of momentum is given as
m1u1+m2u2=m1v1+m2v2
since u1=u2=u=0m/s
u(m1+m2)=m1v1+m2v2
substituting we have
0(0.125+0.0375)=0.125*0.6+0.375*v2
0=0.075+0.375v2
0.375v2=-0.075
v2=-0.075/0.375
v2=-0.2m/s
The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction
Answer:
Part a)

Part b)

Part c)

Part d)

Part e)

Part f)

Explanation:
As we know that catapult is projected with speed 19.9 m/s
so here we have


similarly we have


Part a)
Horizontal displacement in 1.03 s



Part b)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part c)
Horizontal displacement in 1.71 s



Part d)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part e)
Horizontal displacement in 5.44 s



Part f)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Answer:
The magnetic force points in the positive z-direction, which corresponds to the upward direction.
Option 2 is correct, the force points in the upwards direction.
Explanation:
The magnetic force on any charge is given as the cross product of qv and B
F = qv × B
where q = charge on the ball thrown = +q (Since it is positively charged)
v = velocity of the charged ball = (+vî) (velocity is in the eastern direction)
B = Magnetic field = (+Bj) (Magnetic field is in the northern direction; pointing forward)
F = qv × B = (+qvî) × (Bj)
F =
| î j k |
| qv 0 0|
| 0 B 0
F = i(0 - 0) - j(0 - 0) + k(qvB - 0)
F = (qvB)k N
The force is in the z-direction.
We could also use the right hand rule; if we point the index finger east (direction of the velocity), the middle finger northwards (direction of the magnetic field), the thumb points in the upward direction (direction of the magnetic force). Hence, the magnetic force is acting upwards, in the positive z-direction too.
Hope this Helps!!!
Explanation:
c. if the vector is oriented at 0° from the X -axis.