A particle leaves the origin with an initial velocity v⃗ =(2.40 m/s)xˆv→=(2.40 m/s)x^ , and moves with constant acceleration a⃗

Answer:

W = -510.98 J

Explanation:

Force = 43 N, 61° SW

Displacement = 12 m, 22° NE

The work done is calculated using:

W = F*d*cos(A)

where A is the angle between the applied force and displacement.

The angle A between the force and displacement is determined as A = 61 + 90 + 22 = 172°

Hence, W = 43 * 12 * cos(172)

This results in W = -510.98 J

The negative result indicates that the work is done contrary to the direction of the force applied.

To determine the average net force, we can calculate acceleration using:

x = 0.5*a*t^2

v = a*t

where x=3.6m and v=185 m/s.

Thus,

t=v/a and therefore x = 0.5*a*(v/a)^2 = 0.5 * (v^2)/a

which gives us a= (0.5*v^2)/x

Since we have the known values of v and x, we can compute a by substituting these numbers.

The average net force is then given as:

F = m*a,

with m=7.5kg.

Let us denote as the highest acceleration of the truck by itself. The force generated by the engine to propel the truck in this scenario is

where m is the mass of the truck alone.

Upon attaching a bus that has double the mass of the truck, the total mass of the entire system (truck+bus) becomes (m+2m)=3m. In this situation, the force generated by the engine is

where a2 represents the new acceleration. 
Since the engine remains unchanged, the force generated stays the same, allowing us to set the force equations equal to each other:

and solving for a2 gives us:

mass₃<mass₁=mass₅<mass₂=mass₄

Explanation:

Data points:-

1. mass:  m      speed: v

2. mass: 4 m   speed: v

3. mass: 2 m   speed: ¼ v

4. mass: 4 m   speed: v

5. mass: 4 m   speed: ½ v

We know that the formula for Kinetic energy (KE) is ½ mv²

Where m represents the mass of the object

           v represents the object's velocity

The KE of Body 1(mass₁) = ½*m*v²             = mv²/2

KE of Body 2(mass₂) = ½*4m*v²         = 2mv²

KE of Body 3(mass₃) = ½*2m*(1/4v)²  =  mv²/16

KE of Body 4(mass₄) = ½*4m*v ²        =  2mv ²

KE of Body 5(mass₅) = ½*4m*(1/2v)²  =   mv²/2

Answer:

7.166 hours = 430 minutes.

Explanation:

As both trains are approaching each other on the same track, their relative speed is the sum of their individual speeds. Hence, the time until they intersect (and inevitably collide) is determined by how long it takes for speeds of 65 mph and 55 mph to cover the total distance of 860 miles. One train will cover part of the distance, while the other will cover the remainder. To calculate the required time, we can apply the formula:

1 hour ---> 120 miles

X ----> 860 miles; hence X = (860 miles * 1 hour)/120 miles = 43/6 hours = 7.16666 hours. To convert this into minutes, recall that 1 hour equals 60 minutes; therefore, 43/6 hours * 60 minutes/hour = 430 minutes.