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Gnesinka [82]
3 years ago
5

Which letter (A, B, or C) shows where you should apply the most effort to lift the stone?

Physics
1 answer:
sergij07 [2.7K]3 years ago
8 0
A becuz its at da it dont got no wa
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Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem
statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

x=x_o+v_1t    (1)

xo = 10 km

v1 = 55km/h

car 2:

x'=v_2t    (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}

t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s

The position in which both cars coincides is:

x=(55km/h)(0.33h)=18.33km

6 0
3 years ago
Four forces act on a hot-air balloon, as shown
dolphi86 [110]

The magnitude of the resultant force on the balloon is 374.13 N.

The given forces from the image;

  • <em>Upward force = 514 N</em>
  • <em>Downward force = 267 N</em>
  • <em>Eastward force = 678 N</em>
  • <em>Westward force = 397 N</em>

The net vertical force on the balloon is calculated as follows;

F_y = 514 \ N \ \ - \ \ 267 \ N\\\\F_y = 247 \ N

The net horizontal force on the balloon is calculated as follows;

F_x = 678 \ N \ - \ 397 \ N\\\\F_x = 281 \ N

The magnitude of the resultant force on the balloon is calculated as follows;

F = \sqrt{F_y^2 + F_x^2} \\\\F = \sqrt{(247)^2 + (281)^2} \\\\F= 374.13 \ N

Thus, the magnitude of the resultant force on the balloon is 374.13 N.

Learn more here:brainly.com/question/4404327

5 0
2 years ago
2nd question!!!!!!!!!!!!!!!!!!!
DochEvi [55]

Answer:

D.

Explanation:

6 0
2 years ago
A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2 , and 1.85 ms (1 ms = 10−3
ankoles [38]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 =  u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

negative sign indicate that the ball bounce in opposite directon

4 0
3 years ago
What happens to myosin and actin as sarcomeres relax?
Leokris [45]
The myosin heads pull on the actin, bringing them closer together

4 0
3 years ago
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