A 2.1 × 103 kg car starts from rest at the top of a driveway that is sloped at an angle of 20.0° with the horizontal. An average

Answer:

R=V/I=6/2=3 ohm

time = 5 minutes = 5*60=300 seconds

I=2 A

Energy = I²Rt=(2)²*3*300=4*900=3600 J

Answer:

Electric flux is calculated as

Explanation:

We start with the given parameters:

The electric field impacting the circular surface is

Our objective is to ascertain the electric flux passing through a circular region with a radius of 1.83 m situated in the xy-plane. The area vector is oriented in the z direction. The formula for electric flux is expressed as:

Applying properties of the dot product, we calculate the electric flux as:

Consequently, the electric flux for the circular area is . Thus, this represents the required answer.

The duration required for the seventh car to pass amounts to 13.2 seconds. The train's movement is characterized by uniform acceleration, enabling the application of suvat equations. Initially, we analyze the movement of the first car, utilizing the equation for distance s covered in time t, which corresponds to the length of one car, with u = 0 as the initial velocity and a representing acceleration, over t = 5.0 s. We can rearrange the equation reflecting L as the length of one car. This is similarly applicable for the initial seven cars, accounting for the distance of 7L and the required time t'. With constant acceleration retained, we can derive t' through substitution in the equation, leading to fundamental conclusions regarding the relationship exhibited in the graph of distance against time in uniformly accelerated motion.

<span>A centripetal force maintains an object's circular motion. When the ball is at the highest point, we can assume that the ball's speed v is such that the weight of the ball matches the required centripetal force to keep it moving in a circle. Hence, the string will not become slack. centripetal force = weight of the ball m v^2 / r = m g v^2 / r = g v^2 = g r v = sqrt { g r } v = sqrt { (9.80~m/s^2) (0.7 m) } v = 2.62 m/s Thus, the minimum speed for the ball at the top position is 2.62 m/s.</span>

Answer:

a = 18.28 ft/s²

Explanation:

the values provided are:

duration of force application, t= 10 s

Work done = 10 Btu

mass of the object = 15 lb

acceleration, a =? ft/s²

1 Btu = 778.15 ft.lbf

thus, 10 Btu = 7781.5 ft.lbf

m = 0.466 slug

So,

the work is equivalent to the change in kinetic energy

 

The acceleration of the object is therefore

 

 

         a = 18.28 ft/s²

the constant acceleration of the object is calculated to be 18.28 ft/s²