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KengaRu [80]
2 years ago
7

un objeto de 10kg cae desde una altura y tarda 2 segundos en caer al suelo. cual era su velocidad al llegar al suelo? de que alt

ura cayo?
Physics
1 answer:
const2013 [10]2 years ago
8 0

Explanation:

hru3he. djehe isbs she house siege dbsus d beuh

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What happens to the particles of the liquid inside a thermometer when the thermometer is heated?
aev [14]
The particles of the liquid inside a thermometer speed up and spread apart when the thermometer is heated. In short, the particles expand from one another when they're heated, and become condensed and compact when chilled.

5 0
3 years ago
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How do fossils provide evidence of continental drift?
Alex777 [14]
C explanation: There are many examples of fossils found on separate continents and nowhere else, suggesting the continents were once joined. If Continental Drift had not occurred, the alternative explanations would be: They swam to the other continent/s in breeding pairs to establish a second population. ...
8 0
2 years ago
A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f
Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram.  There are three forces acting on the car.  Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank.  If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

4 0
2 years ago
In the data table , distance is measured in meters and time is in seconds. Calculate the mans average velocity using the equatio
Artist 52 [7]

Answer:

3.626 m/s

Explanation:

v=d/t

1. -0.02/0 = 0 m/s

2. 0.86/0.2 = 4.3 m/s

3. 1.71/0.4 = 4.275 m/s

4. 2.54/0.6 = 4.23 m/s

5. 3.32/0.8 = 4.15 m/s

6. 4.08/1.0 = 4.08 m/s

7. 4.79/1.2 = 3.99 m/s

8. 5.48/1.4 = 3.91 m/s

9. 6.15/1.6 = 3.84 m/s

10. 6.76/1.8 = 3.76 m/s

11. 7.37/2.0 = 3.66 m/s

12. 7.92/2.2 = 3.6 m/s

13. 8.45/2.4 = 3.52 m/s

14. 8.96/2.6 = 3.45 m/s

the mean of these numbers is 3.626

his average velocity ks 3.626 m/s

6 0
2 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.60×106 N, one at an angle 13.0 west of north, an
Juliette [100K]

Answer:W=1.93\times 10^9 J      

Explanation:

Given

Force F=1.6\times 10^{6} N

one at an angle of 13^{\circ} East of North and another at 13^{\circ} West of North

Net Force is in North Direction

F_{net}=2F\cos 13

Forces in horizontal direction will cancel out each other

thus Work done will be by north direction forces  

W=2F\cdot \cos 30\cdot s

here s=0.7 km

W=2\times 1.6\times 10^{6}\cdot \cos 30\cdot 700

W=1.93\times 10^9 J                  

3 0
3 years ago
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