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pentagon [3]
3 years ago
8

A car maintains a constant speed v as it traverses the hill and valley as shown below. Both the hill and valley have a radius of

curvature R. At which point, A, B, or C, is the normal force acting on the car (i) the largest, (ii) smallest? Explain. (iii) Where would the driver feel heaviest and (iv) lightest? Explain. (v) How fast can the car go without losing contact with the road at A?
Physics
1 answer:
Zigmanuir [339]3 years ago
8 0

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)

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A 12.6-kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume a radial dis
Zinaida [17]

Answer:

a)27.3N

b)150.78N

Explanation:

Having in mind the conservation of energy, as the monkey goes up (gaining potential gravitational energy) the kinetic energy must be reduced, so reducing the velocity of the monkey. So the maximum velocity will be at this lower point with a velocity of 1.36m/s

From this velocity and the radius we can calculate the angular velocity for the monkey center of mass:

w=\frac{1.36m/s}{0.853m} =1.594 /s

with this we can calculate the centripetal force magnitude:

F=m*w^{2} *r=12.6kg*(1.594/s)^{2} *0.853m=27.3N

On the mokey center of mass we have two opposite forces acting, the tension of the arm and the weight, in order for the monkey to continue swinging the resolt of this two forces must be equal to the centripetal force:

T-P=F=>T=F+P

T=27.3N+12.6kg*9.8m/s^{2}=150.78N

8 0
3 years ago
A car is making a 50 mi trip. It travels the first half of the total distance 25.0 mi at 7.00 mph and the last half of the total
worty [1.4K]

Answer:

a) The total time of the trip is 4.05 h.

b) The average speed of the car is 12.35 mi/h.

c) The total time of the trip is 1.69 h.

Explanation:

Hi there!

a) The equation of traveled distance for a car traveling at constant speed is the following:

x= v · t

Where:

x = traveled distance.

v = velocity.

t = time.

Solving the equation for t, we can find the time it takes to travel a given distance "x" at a velocity "v":

x/v = t

So, the time it takes the car to travel the first half of the distance will be:

t1 = 25.0 mi / 7.00 mi/h

And for the second half of the distance:

t2= 25.0 mi / 52.00 mi / h

The total time will be:

total time = t1 + t2 = 25.0 mi / 7.00 mi/h + 25.0 mi / 52.00 mi / h

total time = 4.05 h

The total time of the trip is 4.05 h.

b) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it takes to travel that distance (t). In this case, the traveled distance is 50 mi and the time is 4.05 h. Then:

a.s = d/t

a.s = 50 mi / 4.05 h

a.s = 12.35 mi/h

The average speed of the car is 12.35 mi/h

c) Let's write the equations of traveled distance for both halves of the trip:

For the first half, you traveled a distance d1 in a time t1 at 7.00 mph:

7.00 mi/h = d1/t1

Solving for d1:

7.00 mi/h · t1 = d1

For the second half, you traveled a distance d2 in a time t2 at 52.00 mph.

52.00 mi/h = d2/t2

52.00 mi/h · t2 = d2

We know that d1 + d2 = 50 mi and that t1 and t2 are equal to t/2 where t is the total time:

d1 + d2 = 50 mi

52.00 mi/h · t/2 + 7.00 mi/h · t/2 = 50 mi

Solving for t:

29.5 mi/h · t = 50 mi

t = 50 mi / 29.5 mi/h

t = 1.69 h

The total time of the trip is 1.69 h.

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