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myrzilka [38]
3 years ago
7

Consider a person sliding down a water slide at constant velocity. what are the forces acting on the person as they slide? are t

he forces balanced or unbalanced? explain.

Physics
2 answers:
Andru [333]3 years ago
4 0

Explanation :

When a person is sliding with constant velocity, the forces that are acting on him are :

(1) Weight, W = mg where g is acceleration due to gravity.

(2) Friction forces acting opposite to the direction of motion.

(3) Normal force.

The forces acting on the person is balanced because the person is sliding with constant velocity.

From figure it is clear that, friction force is balanced by w\ sin\theta and normal is balanced by w\ cos\theta.

Umnica [9.8K]3 years ago
4 0

Answer:

It is D.

Explanation:

Gravity and friction are acting on the person. The forces are balanced. There is no net force because the person is moving at a constant velocity. In this example, there is no net force when a mass moves at constant velocity. Although friction is acting on the person, there is no change in velocity and friction is not a net force in this case. Friction is only a net force if it changes the velocity of a mass. If friction slowed the person down, then there is a net force.

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If there were no friction between the girl and the slide on the hill, what would most likely happen to her motion
aivan3 [116]

Answer: she would move faster and more efficiently due to no friction

Explanation:

7 0
3 years ago
In fig. 2-27, a red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel t
lbvjy [14]

Answer:

a) V₀ = 13.5m/s

b) a = 2.1 m/s²

Explanation:

1) Convert velocities to m/s

i) 20 km/h × 1000m/kg × 1h/3600s = 5.556m/s

ii) 40km/h = 11.111m/s

2) Red car, time elapsed to reach position x = 44.5m

Constant velocity ⇒ x = V×t ⇒ t = x / V

⇒ t₁ = 44.5m / 5.556m/s = 8s

3) Red car, time elapsed to reach position x = 77.6m

t₂ = 76.6m / 11.111/s = 6.9s

4) Green car, distance run at t₁ = 8s, x = 44.5m

i) uniform acceleration equation d = V₀t + at² / 2

ii) d = 220m - x = 220m - 44.5m = 175.5m = V₀ (8) + a (8)² /2

175.5 = 8V₀ + 32a ↔ equation (1)

5) Green car, distance run at t₂ = 6.9s, x = 76.6m

i) d = 220m - x = 220m - 76.6m = 143.4

ii) 143.4 = V₀t₂ + at₂² / 2

143.4 = V₀ (6.9) + a(6.9)² / 2

143.4 = 6.9V₀ + 23.8a ↔ equation (2)

6) Solve the system of equations:

175.5 = 8V₀ + 32a ↔ equation (1)

143.4 = 6.9V₀ + 23.8a ↔ equation (2)

V₀ = 13.5m/s

a = 2.1 m/s²

5 0
3 years ago
Read 2 more answers
Two masses are connected by a string which passes over a pulley with negligible mass and friction. One mass hangs vertically and
lora16 [44]
The tension in the string and the acceleration must be equal for both masses. (See the free body diagrams)





8 0
3 years ago
After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 68.3 m horizontally from the end o
inessss [21]

Answer:

a)52.58 m/s

b)56.13°

Explanation:

assume the upward direction as positive

x-component of the velocity = 29.3×cos33.6°=24.40 m/s (remain constant)

y-component of the velocity which is -29.3sin33.6°= -16.21 m/s

time of flight = 68.3/24.40= 2.7991 seconds

now, we can obtain final velocity in y-direction

v_f_y= v_i_y-gt

v_f_y=-16.21-(-9.8)×2.7991

=43.66 m/s

v_0=\sqrt{29.3^2+43.66^2}

=52.58 m/s

for direction

tan^{-1}\frac{43.66}{29.3}

56.13° from the horizontal

3 0
3 years ago
Suppose you take a trip that covers 490 km and takes 1 hours to make. Your average speed is
Vsevolod [243]

Answer:

B

Explanation:

490km/h

490km/1h

490000/3600

note: 1h= (60×60)=3600

hope it help??

8 0
3 years ago
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