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Ann [662]
3 years ago
11

Whats the answer for the second question ???

Physics
1 answer:
Rama09 [41]3 years ago
6 0
We don't know the question you're talking about, hun.
Add a picture or paste it into the caption :)
You might be interested in
In which one(s) of the following situations will there be an INCREASE in Kinetic Energy? Group of answer choices A block slides
Maksim231197 [3]

Answer:

The Kinetic Energy INCREASE in the following situations:

a)A block slides down a frictionless incline:

e)A merry go round rotates faster due to the push by a person.

Explanation:

The energy kinetics is proportional to the square of the velocity:

E_k=1/2*mv^2

We study the cases:

a)A block slides down a frictionless incline:

The gravity made a positive work and the box get a extra velocity, then the kinetics energy INCREASE

b)A box is pulled across a rough floor at constant speed:

The magnitude of the velocity does not change, so the kinetics energy does not change as well

c)A stone at the end of a string is whirled in a horizontal circle at constant speed:

The magnitude of the velocity does not change, so the kinetics energy does not change as well

d)A projectile approaches its maximum height:

If the projectile approaches its maximum height, its velocity approaches to zero. Then the kinetics Energy DECREASE

e)A merry go round rotates faster due to the push by a person.

Thanks to the push, the magnitude of the velocity INCREASE, so the kinetics energy INCREASE as well

6 0
3 years ago
Female scientist who came to america to study starts at harvard
krok68 [10]
Is that a question? If it is not what its the question?
7 0
3 years ago
Para el siguiente conjunto de medidas, calcule EL ERROR RELATIVO PORCENTUAL: 1.34 m, 1.35 m, 1.37 m y 1.36 m
ivanzaharov [21]

Answer:

Ver explicacion abajo

Explanation:

En este caso para poder calcular el error relativo porcentual, es necesario calcular primero el error absoluto, que se calcula de la siguiente forma:

Error absoluto = Resultado exacto - aproximación

Sin embargo, no tenemos el resultado exacto de las medidas, pero podriamos conocerlo tomando el promedio de estas medidas y este es el que tomaremos como el verdadero resultado de las medidas:

Promedio de medidas = 1.34 + 1.35 + 1.37 + 1.36 / 4

Promedio de medidas = 1.355 m

Ya que tenemos el promedio, podemos calcular el error absoluto de cada medida y luego el error relativo porcentual:

Ea1 = 1.355 - 1.34 = 0.015

Ea2 = 1.355 - 1.35 = 0.005

Ea3 = 1.37 - 1.355 = 0.015

Ea4 = 1.36 - 1.355 = 0.005

Ya que tenemos los 4 errores absolutos, es posible calcular el porcentual:

%error relativo = (Error absoluto / resultado exacto) * 100

Aplicando la expresión con cada uno de los valores tenemos:

%Er1 = (0.015/1.34) * 100 = 1.12%

%Er2 = (0.005/1.35) * 100 = 0.37%

%Er3 = (0.015/1.37) * 100 = 1.09%

%Er4 = (0.005/1.36) * 100 = 0.37%

Espero que te sirva.

4 0
3 years ago
Can someone please explain number 8?
guajiro [1.7K]

Answer: Ax=(Vx-Vox)/(T)

Vx=Vox+Ax*T

Solving for Ax in terms of Vx, Vox, T

Vx-Vox=Ax*t

Ax=(Vx-Vox)/(T)

This is saying the acceleration in the x-direction can be found by taking the difference between the finial and initial Velocity in x-direction and dividing it by the Total Time.

Any questions please feel free to ask. Thanks

8 0
3 years ago
An object is projected from the ground with an upward speed of ų m/s has a speed of 23m/s when it is at a height of 5m above the
vovikov84 [41]

Answer:

25.08m/s

Explanation:

mgh1 + 0.5mv1² = mgh2 + 0.5mv2²

h1 = 0m

v1 = u

h2 = 5m

v2 = 23m/s

putting the values into the formula above;

m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)

0 + 0.5mu² = 50m + 264.5m

0.5mu² = 314.5m

dividing through by m

0.5u² = 314.5

u² = 629

u = <u>2</u><u>5</u><u>.</u><u>0</u><u>8</u><u>m</u><u>/</u><u>s</u>

<u>Theref</u><u>ore</u><u>,</u><u> </u><u>the</u><u> </u><u>init</u><u>ial</u><u> </u><u>speed</u><u> </u><u>"</u><u>u</u><u>"</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>m</u><u>/</u><u>s</u>

6 0
3 years ago
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