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Desde que altura debes de lanzar una canica de 50g para que adquiera una energia de 100j

AnnZ [28]

Answer:

204 m

Explanation:

When the marble is dropped from a certain height, its gravitational potential energy converts into kinetic energy. So the kinetic energy gained is equal to the variation of gravitational potential energy:

$\Delta U=mg\Delta h$

where

m is the mass of the marble

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h$ is the change in height

In this problem, we have

m = 50 g = 0.05 kg

$\Delta U = 100 J$

Solving the formula for $\Delta h$ , we find the necessary height from which the marble should be dropped:

$\Delta h = \frac{\Delta U}{mg}=\frac{100}{(0.05)(9.8)}=204 m$

7 0

3 years ago

A man (weighing 763 N) stands on a long railroad flatcar (weighing 3513 N) as it rolls at 19.8 m/s in the positive direction of

Burka [1]

Answer:

0.8m/s

Explanation:

Weight of mas,F=763 N

Mass of man= $\frac{F}{g}=\frac{763}{9.8}=77.86 kg$

By using $g=9.8m/s^2$

Weight of flatcar=F'=3513 N

Mass of flatcar= $\frac{3513}{9.8}=358.5 Kg$

Total mass of the system=Mass of man+mass of flatcar=77.86+358.5=436.36 kg

Velocity of system=19.8m/s

Let v be the velocity of flatcar with respect to ground

Velocity of man relative to the flatcar= $-4.68m/s$

Final velocity of man with respect to ground=v-4.68

By using law of conservation of momentum

Initial momentum=Momentum of car+momentum of flatcar

$436.36(19.8)=77.86(v-4.68)+358.5v$

$8639.928=77.86v-364.3848+358.5v$

$8639.928+364.3848=436.36 v$

$9004.3128=436.36v$

$v=\frac{9004.3128}{436.36}$

$v=20.6 m/s$

Initial speed of flatcar=Speed of system

Increase in speed=Final speed-initial speed=20.6-19.8=0.8m/s

4 0

2 years ago

A bag of cement weighing 325 N hangs in equilibrium from three wires. Two of the wires make angles of theta1=60.0 degrees and th

Murljashka [212]

The sketch of the system is: two strings, 1 and 2, are attached to the ceiling and to a third string, 3.The third string holds the bag of cement.

The free body diagram of the weight with the string 3, drives to the tension T3 = weihgt => T3 = 325 N

The other free body diagram is around the joint of the three strings.

In this case, you can do the horizontal forces equilibrium equation as:

T1* cos(60) - T2*cos(40) = 0

And the vertical forces equilibrium equation:

Ti sin(60) + T2 sin(40) = T3 = 325 N

Then you have two equations with two unknown variables, T1 and T2

0.5 T1 - 0.766 T2 = 0

0.866 T1 + 0.643T2 = 325

When you solve it you get, T1 = 252.8 N and T2 = 165 N

Answer: T1 = 252.8 N, T2 = 165N, and T3 = 325N

7 0

3 years ago

A source at rest emits light of wavelength 500 nm. When it is moving at 0.90c toward an observer, the observer detects light of

Vesna [10]

Answer:

The observer detects light of wavelength is 115 nm.

(b) is correct option

Explanation:

Given that,

Wavelength of source = 500 nm

Velocity = 0.90 c

We need to calculate the wavelength of observer

Using Doppler effect

$\lambda_{o}=\sqrt{\dfrac{1-\beta}{1+\beta}}\lambda_{s}$

Where, $\beta=\dfrac{c}{v}$

$\lambda_{o}=\sqrt{\dfrac{c-0.90c}{c+0.90c}}\times500\times10^{-9}$

$\lambda_{o}=115\ nm$

Hence, The observer detects light of wavelength is 115 nm.

8 0

3 years ago

5. An acrobat, starting from rest, swings freely on a trapeze of

34kurt

The energy conservation and trigonometry we can find the results for the questions about the movement of the acrobat are;

a) The maximum speed is v = 4.89 m / s

b) The maximum height is h = 1.22 m

The energy conservation is one of the most fundamental principles of physics, stable that if there are no friction forces the mechanistic energy remains constant. Mechanical energy is the sum of the kinetic energy plus the potential energies.

Em = K + U

Let's write the energy in two points.

Starting point. Highest part of the oscillation

Em₀ = U = m g h

Final point. Lower part of the movement

$Em_f$ = K = ½ m v²

Energy is conserved.

Emo = $Em_f$

m g h = ½ m v²

v² = 2 gh

Let's use trigonometry to find the height, see attached.

h = L - L cos θ

h = L (1- cos θ)

They indicate that the initial angle is tea = 48º and the length is L = 3.7 m, let's calculate.

h = 3.7 (1- cos 48)

h = 1.22 m

this is the maximum height of the movement.

Let's calculate the velocity.

$v= \sqrt{2 \ 9.8 \ 1.22}$

v = 4.89 m / s

In conclusion using the conservation of energy and trigonometry we can find the results for the questions about the movement of the acrobat are;

a) The maximum speed is v = 4.89 m / s

b) The maximum height is h = 1.22 m

Learn more here: brainly.com/question/13010190

5 0

3 years ago

I have to write something here, so like hello and please help​

I have to write something here, so like hello and please help