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Brut [27]
3 years ago
12

Disturbed by speeding cars outside his workplace, Nobel laureate Arthur Holly Compton designed a speed bump (called the "Holly h

ump") and had it installed. Suppose a 1 800-kg car passes over a hump in a roadway that follows the arc of a circle of radius 18.8 m.
a. If the car travels at 33.0 km/h what force does the road exert on the car as the car passes the highest point of the hump?
b. What is the maximum speed the car can have without losing contact with the road as it passes this highest point?
Physics
1 answer:
Bezzdna [24]3 years ago
3 0
:<span>  </span><span>30.50 km/h = 30.50^3 m / 3600s = 8.47 m/s 

At the top of the circle the centripetal force (mv²/R) comes from the car's weight (mg) 

So, the net downward force from the car (Fn) = (weight - centripetal force) .. and by reaction this is the upward force provided by the road .. 

Fn = mg - mv²/R 
Fn = m(g - v²/R) .. .. 1800kg (9.80 - 8.47²/20.20) .. .. .. ►Fn = 11 247 N (upwards) 
(b) 
When the car's speed is such that all the weight is needed for the centripetal force .. then the net downward force (Fn), and the reaction from the road, becomes zero. 

ie .. mg = mv²/R .. .. v² = Rg .. .. 20.20m x 9.80 = 198.0(m/s)² 

►v = √198 = 14.0 m/s</span>
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3 years ago
You run a race with your friend. At first you each have the same kinetic energy, but then you find that she is beating you. When
Blizzard [7]

Answer:

52.49 Kg

Explanation:

Let m1 and v1 denote your mass and velocity respectively

Let m2 and v2 denote your friends mass and velocity respectively

Kinetic energy is given by

KE= 0.5mv^{2}  

Since your kinetic energies are the same hence

0.5m1(v1)^{2}=0.5m2(v2)^{2}

m1(v1)^{2}=m2(v2)^{2} and making m2 the subject then  

m2=\frac { m1(v1)^{2}}{(v2)^{2}}

Since v2 is v1+0.28v1=1.28v1

Substituting m1 for 86 Kg

m2=\frac { 86 Kg(v1)^{2}}{(1.28v1)^{2}}= 52.49023\approx 52.49 Kg

3 0
3 years ago
A softball is fouled off with a vertical velocity of 20 m/s and a horizontal velocity of 15 m/s. what is the resultant velocity
raketka [301]
25 m/s is the answer
8 0
3 years ago
ball A of mass 5.0 kg moving at 20 meters per second collides with ball B of unknown mass moving at 10. meters per second in the
antiseptic1488 [7]

Answer:The mass of ball B is 10 kg.

Explanation;

Mass of ball A = M_A=5 kg

Velocity of the ball A before collision:U_A=20 m/s

Velocity of ball A after collision=V_A=10 m/s

Mass of ball B= M_B

Velocity of the ball B before collision:U_B=10 m/s

Velocity of ball B after collision=V_B=15 m/s

M_AV_A+M_BV_B=M_AU_A+M_BU_B

5 kg\times 10 m/s+M_B\times 15=5 kg\times 20m/s+M_B\times 10m/s

M_B=10kg

The mass of ball B is 10 kg.

8 0
2 years ago
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0
3 years ago
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