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Anna35 [415]
2 years ago
5

1) Si un mango cae a una velocidad de 75m/s y tarda 26 seg. en caer. ¿ Cuál habrá sido la velocidad con qué el mango llegó al su

elo?
Physics
1 answer:
Lyrx [107]2 years ago
8 0

Answer:

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

Explanation:

El mango experimenta un movimiento de caída libre, es decir, un movimiento uniformemente acelerado debido a la gravedad terrestre, despreciando los efectos de la viscosidad del aire y la rotación planetaria. Entonces, la velocidad final del mango, es decir, la velocidad con la que llega al suelo, se puede determinar mediante la siguiente fórmula cinemática:

v = v_{o}+g\cdot t (1)

Donde:

v_{o} - Velocidad inicial, en metros por segundo.

v - Velocidad final, en metros por segundo.

g - Aceleración gravitacional, en metros por segundo al cuadrado.

t - Tiempo, en segundos.

Si sabemos que v_{o} = -75\,\frac{m}{s}, g = -9.807\,\frac{m}{s^{2}} y t = 26\,s, entonces la velocidad final del mango es:

v = v_{o}+g\cdot t

v = -75\,\frac{m}{s}+\left(-9.807\,\frac{m}{s} \right)\cdot (26\,s)

v = -329.982\,\frac{m}{s}

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

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Sam is playing football. She kicks the ball with an average force of 75 N.
damaskus [11]

Answer:

22.5J

Explanation:

Here the force is given. Also, the displacement is given as 30cm.

First we should check if all the values are in their standard form.

Here 30cm should be converted to metre by dividing it with 100.

Which would give us 0.3m

Now we use the equation W=force x displacement =75 x 0.3=22.5J

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3 0
3 years ago
A proton and an electron traveling with the same velocity enter a uniform electric field. compared to the acceleration of the pr
kenny6666 [7]
The acceleration of the electron is larger than the acceleration of the proton.

The reason for this is that the mass of the electron is smaller (about 1000 times smaller) than the mass of the proton. The two particles have same charge (e), so they experience the same force under the same electric field E:
F=eE 
However, according to Newton's second law, the force is the product between the mass particle, m, and its acceleration, a:
F=ma 
which can be rewritten as
a= \frac{F}{m}
we said that the force exerted on the two particles, F, is the same, while the mass of the electron is smaller: therefore, from the last formula we see that the acceleration of the electron will be larger than that of the proton.

6 0
2 years ago
A rock with a mass of 0.3 kg falls from the top of a cliff. If it takes the rock 2.5 s to reach the ground, what was the impulse
kaheart [24]
Impulse=force x time
force=mass x acceleration due to gravity
force=
300 \times 10 = 3000
impulse =3000 x 2.5= ( sorry i don't have a calculator right now so you must calculate this yourself)
I converted from kg to g because it is the standard.
Hope this helps you.
4 0
3 years ago
Larry drops a 5kg ball off of a building. The ball hits the ground 4.7s later. How tall is the building?
musickatia [10]

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 4.7 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (4.7 s) + ½ (9.8 m/s²) (4.7 s)²

Δy ≈ 110 m

8 0
3 years ago
An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.
Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

u -v= d

u=71+26=97\ cm

We need to calculate the focal length

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

put the value into the formula

\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}

\dfrac{1}{f}=-\dfrac{71}{2522}

f=-35.52\ cm

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value in to the formula

\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}

\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}

\dfrac{1}{v}=-\dfrac{124}{2755}

v=-22.21\ cm

We need to calculate the magnification

Using formula of magnification

m=\dfrac{v}{u}

m=\dfrac{22.21}{95}

m=0.233

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

4 0
3 years ago
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