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xenn [34]
3 years ago
8

4.-Una atleta patea un balon (masa de 650gr y angulo de 45 grados) arriba de un edificio con una velcidad 21 m/s, el edificio mi

de 10 metros de altura. Si el balon cae fuera del edificio, calcula la velocidad (m/s) que este alcanza un instante antes de tocar el suelo
Physics
1 answer:
marta [7]3 years ago
3 0

Answer:

v = 25 m/s

Explanation:

½mv₀² + mgh = ½mv₁²

v₁² = v₀² + 2gh

v₁ = √(21² + 2(9.8)(10))

v₁ = 25.238858..

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A thin rod (length = 2.97 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.
nlexa [21]

Answer:

a)  w = 2.57 rad / s , b)   α = 3.3 rad / s²

Explanation:

a) Let's use the conservation of mechanical energy, we will write it in two points the highest and when touching the ground

Initial. Higher

       Em₀ = U = m g h

Final. Touching the ground

       Em_{f} = K = ½ I w²

How energy is conserved

       Em₀ = Em_{f}

       mg h = ½ I w2

The moment of specific object inertia

        I = m L²

We replace

       m g h = ½ (mL²) w²

       w² = 2g h / L²

The height of the object is the length of the bar

        h = L

        w = √ 2g / L

       w = √ (2 9.8 / 2.97)

       w = 2.57 rad / s

b) the angular acceleration can be found from Newton's second rotational law

       τ = I α

       W L = I α

       mg L = (m L²) α

       α = g / L

       α = 9.8 / 2.97

       α = 3.3 rad / s²

3 0
3 years ago
A 1502.7 kg car is traveling at 33.1 m/s when
Bond [772]

By definition of average acceleration,

<em>a</em> = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²

Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:

∑ <em>F</em> = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N

If you just want the magnitude, drop the negative sign.

5 0
3 years ago
Two resistances, R1 and R2, are connected in series across a 9-V battery. The current increases by 0.450 A when R2 is removed, l
Rina8888 [55]

Answer:

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

Explanation:

Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.

When the current increases by 0.450 A wen only R1 is in the circuit, the current is

9/R1 + R2 + 0.450 A = 9/R1       (1)

When the current increases by 0.225 A when only R2 is in the circuit, the current is

9/R1 + R2 + 0.225 A = 9/R2       (2)

equation (1) - (2) equals

9(1/R1 - 1/R2) = 0.450 A - 0.225

9(1/R1 - 1/R2) = 0.125

(1/R1 - 1/R2) = 0.125 A/9 = 0.0138

1/R1 = 0.0138 + 1/R2

R1 = R2/(1 + 0.0138R2)     (3)

From (1)

9/R1 - 9/R1 + R2 = 0.450 A

9R2/[R1(R1 + R2)] = 0.450 A

R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5

R2/[R1(R1 + R2)] = 0.5    (4)

From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,

(1 + 0.0138R2) = 0.5(R1 + R2)

0.5R1 + 0.5R2 = 1 + 0.0138R2

0.5R1 = 1 + 0.0138R2 - 0.5R2

0.5R1 = 1 - 0.4862R2        (5)

Substituting (3) into (5) we have

0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2

R2 = (1 + 0.0138R2)(1 - 0.4862R2)

R2 = 1 - 0.4724R2 - 0.0067R2²

Collecting like terms, we have

0.0067R2² + 0.4724R2 + R2 - 1 = 0

0.0067R2² + 1.4724R2 - 1 = 0

Using the quadratic formula,

R_{2} = \frac{-1.4724 +/-\sqrt{(1.4724)^{2} - 4 X 0.0067 X -1} }{2 X 0.0067}  \\= \frac{-1.4724 +/-\sqrt{2.1680 + 0.0268} }{0.0268}\\= \frac{-1.4724 +/-\sqrt{2.1948} }{0.0268}\\= \frac{-1.4724 +/- 1.4815 }{0.0268}\\= \frac{-1.4724 + 1.4815 }{0.0268} or \frac{-1.4724 - 1.4815 }{0.0268}\\= \frac{0.0091 }{0.0268} or \frac{-2.9539}{0.0268}\\= 0.340 or -110.22

We choose the positive answer.

So R2 = 0.340 Ω

From (5)

R1 = 0.5 - 0.9931R2

   = 0.5 - 0.9931 × 0.340

   = 0.5 - 0.338

   = 0.162 Ω

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

5 0
2 years ago
A stuffed toy with a mass of 0.900 kilograms sits on the edge of a bed at a height of 0.830 if the toy falls off the bed what wi
Olenka [21]
Mechanical energy (ME) is the sum of potential energy (PE) and kinetic energy (KE). When the toy falls, energy is converted from PE to KE, but by conservation of energy, ME (and therefore PE+KE) will remain the same.

Therefore, ME at 0.500 m is the same as ME at 0.830 m (the starting point). It's easier to calculate ME at the starting point because its just PE we need to worry about (but if we wanted to we could calculate the instantaneous PE and KE at 0.500 m too and add them to get the same answer).

At the start:

ME = PE = mgh
ME = 0.900 (9.8) (0.830)
ME = 7.32 J
8 0
2 years ago
What is meant by free fall
Firlakuza [10]
Downward movement under the force of gravity only.
5 0
3 years ago
Read 2 more answers
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