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Leona [35]
3 years ago
15

A car traveling 90.0 km/h is 1500 m behind a truck traveling at 76.0 km/h. 1) How long will it take the car to catch up with the

truck?
Physics
1 answer:
sergejj [24]3 years ago
8 0
It will take about 3-5 hour
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The friction force that acts on objects that are at rest is___________
weeeeeb [17]

Answer:

static friction

Explanation:

static friction is the friction force that acts on objects at rest

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
4 years ago
Read 2 more answers
You are skydiving and accelerating at 9.8 meters per second
Yuri [45]
It would take you 5.61 seconds to reach that velocity
4 0
3 years ago
HELP ASAP!!
Andreas93 [3]
C. is the correct answer


5 0
3 years ago
Read 2 more answers
PLEASE PROVIDE AN EXPLANATION<br><br> THANK YOU!
Rus_ich [418]

Answer:

(a) 0.993 s

(b) 14.0 N/m

(c) -3.02 m/s

(d) -6.01 m/s²

Explanation:

(a) The block's position can be modeled as a cosine wave:

x(t) = A cos(ωt)

where A is the amplitude (in this case, 50 cm) and ω is the angular frequency.

At t = 0.200 s, x(t) = 15.0 cm.

15.0 cm = 50.0 cm cos((0.200 s) ω)

0.3 = cos((0.2 s) ω)

1.266 rad = (0.2 s) ω

ω = 6.33 rad/s

The period is:

T = (2π rad) (1 s / 6.33 rad)

T = 0.993

(b) For a spring-mass system, ω = √(k/m).  The mass of the block is 0.350 kg, so:

ω = √(k/m)

6.33 rad/s = √(k / 0.350 kg)

6.33 rad/s = √(k / 0.350 kg)

40.1 rad/s² = k / 0.350 kg

k = 14.0 N/m

(c) Energy is conserved:

EE₀ = EE + KE

½ kx₀² = ½ kx² + ½ mv²

kx₀² = kx² + mv²

(14.0 N/m) (0.50 m)² = (14.0 N/m) (0.15 m)² + (0.35 kg) v²

v = -3.02 m/s

Alternatively, we can take the derivative of our position equation:

v(t) = -Aω sin(ωt)

v = -(0.50 m) (6.33 rad/s) sin((6.33 rad/s) (0.2 s))

v = -3.02 m/s

(d) Sum of forces on the block:

∑F = ma

-kx = ma

a = -kx / m

a = -(14.0 N/m) (0.15 m) / (0.350 kg)

a = -6.01 m/s²

Alternatively, we can take the derivative of our velocity equation:

a(t) = -Aω² cos(ωt)

a = -(0.50 m) (6.33 rad/s)² cos((6.33 rad/s) (0.2 s))

a = -6.01 m/s²

6 0
3 years ago
Read 2 more answers
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