Answer:
The magnitude of the acceleration is 
The direction is
i.e the negative direction of the z-axis
Explanation:
From the question we are that
The mass of the particle 
The charge on the particle is 
The velocity is 
The the magnetic field is 
The charge experienced a force which is mathematically represented as

Substituting value



Note :

Now force is also mathematically represented as

Making a the subject

Substituting values



Answer:
The mass of the cart is 5 kg
Explanation:
You divide 25 by 5 and get 5. Have a great day! :D
<em>The Equation:</em>
25/5 = 5
Answer:
As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.
(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.
(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.
(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.
(iv) At point A, the driver will feel the lightest.
(v)The car can go that much fast without losing contact with the road at A can be determined as follow:
Fn=0 - lose contact with road
Fg= mv²/r
mg=mv²/r
v=sqrt (gr)
<span>a. the amount of matter in a given volume </span>