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lyudmila [28]
3 years ago
8

Se deja caer un objeto desde la cornisa de una casa la cual tarda en llegar al suelo 645/500 segundos. ¿Cúal es la altura de la

casa?
Physics
1 answer:
salantis [7]3 years ago
4 0

8.15m. La altura de la casa de la cual se deja caer un objeto que tarda en caer \frac{645}{500}s es de 8.15m.

La clave para resolver este problema es mediante caída libre, la velocidad inicial del objeto es 0 por lo que podemos calcular la altura h de la casa mediante la relación h= \frac{gt^{2}}{2}.

Conocemos la gravedad g=9.8\frac{m}{s^{2}} y el tiempo en que tarda en caer el objeto t=\frac{645}{500}s, sustituyendo los valores tenemos que:

h=\frac{(9.8\frac{m}{s^{2}})(\frac{645}{500}s)^{2}}{2}=8.15m

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How many joules of work are done on an object when a force of 10 N pushes it 5 m?
zhenek [66]

Answer:

option C

Explanation:

given,                            

Force on the object = 10 N

distance of push = 5 m

Work done = ?              

we know,              

work done is equal to Force into displacement.

W = F . s            

W = 10 x 5              

W = 50 J                

Work done by the object when 10 N force is applied is equal to 50 J

Hence, the correct answer is option C

5 0
3 years ago
A variable that is not changed.
AleksAgata [21]

Answer:

A controlled variable does not change during a experiment

Explanation:

it's c

5 0
3 years ago
The y-component of the force F which a person exerts on the handle of the box wrench is known to be 86 lb. Determine the x-compo
emmainna [20.7K]

Answer:

x-component of force is  38.18 lb where as magnitude of Force is 93.16

Explanation:

Fy of the force F exerted on the handle of the box wrench = 86 lb

Considering the triangle in Fig 1

magnitude of perpendicular = P =  12

magnitude of base = B = 5

using Pythagoras theorem

                        H= \sqrt{P^{2} + B^{2}}

                 H= \sqrt{12^{2} + 5^{2}}

            H=13\\\implies cos \theta = \frac{5}{13} \\\implies sin \theta = \frac{12}{13}\\

y-component of force is given given as:

                              86 = Fsin\theta\\F = 86 (\frac{13}{12})\\F = 93.16 lb\\F_{x} =F cos \theta\\F_{x} = 93.16 (\frac{5}{13})\\F_{x} =38.18 lb.

5 0
3 years ago
*WILL MARK BRAINLIEST FOR RIGHT ANSWER* How much current must be applied across a 60 Ω light bulb filament in order for it to co
sp2606 [1]

Answer: The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

Explanation:

You can find the current in amperes using ohms and watts from this formula:

I = \sqrt{\frac{P}{R} }

Where P represents power in watts, R represents resistance in ohms, and I represents current in amperes.

You can then substitute 60 and 55 into the equation to find I:

I = \sqrt{\frac{55}{60} } \\I = \frac{\sqrt{55} }{\sqrt{60} }

Then, simplify the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} }

Rationalize the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} } * \frac{\sqrt{15} }{\sqrt{15} } = \frac{\sqrt{825} }{30}

Simplify the numerator by finding its factors:

I = \frac{5\sqrt{33} }{30} = \frac{\sqrt{33} }{6}

The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

8 0
3 years ago
1. What is the heat energy when 114.32g of water ( c = 4.18 J/g °C) at 14.85°C is raised to
timama [110]
Dnt listen to the file stuff
6 0
3 years ago
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