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3 years ago

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An object is thrown directly upwards from the ground at a velocity of 9ms. Recalling that the acceleration due to gravity is −gm

s2, find the time at which the object reaches its maximum height. What is this maximum height? Your answers should be in terms of g.

1 answer:

olga55 [171]3 years ago

6 0

Answer:

Explanation:

Given

Object is thrown with a velocity of $u=9\ m/s$

Acceleration due to gravity is -g (i.e. acting downward)

Vertical distance traveled by object is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

at maximum height final velocity is zero

$0-(9)^2=2\times (-g)\times (s)$

$s=\frac{81}{2g}=\frac{40.5}{g}\ m$

time taken to reach maximum height

using

v=u+at

0=9-gt

$t=\frac{9}{g}\ s$

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observe the figure given carefully volume of water in each vessel is shown arrange them in order of decreasing pressure at the b

mihalych1998 [28]

Answer:

See the explanation below

Explanation:

The pressure is defined as the product of the density of the liquid by the gravitational acceleration by the height, and can be easily calculated by means of the following equation.

$P=Ro*g*h$

where:

Ro = density of the fluid [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = elevation [m]

In this way we can understand that the greater pressure is achieved by means of the height of the liquid, that is, as long as the fluid has more height, greater pressure will be achieved at the bottom.

Therefore in order of decreasing will be

The largest pressure with the largest height of the liquid, container B. The next is obtained with container D, the next with container A and the lowest pressure with container C.

The pressure decreases as we go from the container B - D - A - C

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3 years ago

a) ¿En qué posición es mínima la magnitud de la fuerza sobre la masa de un sistema masa-resorte? 1) x 0, 2) x A o 3) x A. ¿Por q

Tatiana [17]

Answer:

a) the correct answer is 1 , b) x=0 F=0, a=0

x= 0.050 F= -7.5 N, a= -15 m/s²

x= 0.150 F= 22.5 N, a=- 45 m/s²

Explanation:

a) In a mass - spring system the force is given by the Hooke force,

F = - k x

Analyzing this equation we see that the outside is proportional to the elongation from the equilibrium position, therefore the force is zero when the spring is in its equilibrium position

the correct answer is 1

b) we assume that the given values are from the equilibrium position of the spring.

Let's calculate the force

x = 0

F = 0

x = 0.050

F = - 150 0.050

F = - 7.50 N

x = 0.150

F = - 150 0.150

F = - 22.5 N

let's use Newton's second law to find the acceleration

F = m a

a = F / m

x = 0 m

a = 0

x = 0.050 m

a = -7.50 / 0.50

a = - 15 m / s²

x = 0.150 m

a = - 22.5 / 0.50

a = - 45 m/s²

TRASLATE

a) En un sistema masa – resorte la fuerza es dada por la fuerza de Hoke,

F= - k x

analizando esta ecuación vemos que la fuera es proporcional a la elongación desde la posición de equilibrio, por lo tanto la fuerza es cero cuando el resorte esta en su posición de equilibrio

la respuesta correcta es 1

b)suponemos que los valores dados son desde la posición de equilibrio del resorte.

Calculemos la fuerza

x=0

F= 0

x=0.050

F = - 150 0.050

F= - 7.50 N

x= 0.150

F= - 150 0.150

F= - 22.5 N

usemos la segunda ley de Newton para encontrar la aceleración

F = m a

a = F/m

x =0 m

a = 0

x= 0.050 m

a = -7.50/ 0.50

a =- 15 m/s²

x= 0.150 m

a= - 22.5 / 0.50

a= - 45 m/s²

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3 years ago

What was one main point of Dalton's atomic theory

goblinko [34]

All matter is composed of atoms, indestructible building blocks.

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3 years ago

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A ball is kicked from a location 7, 0, −8 (on the ground) with initial velocity −11, 19, −5 m/s. The ball's speed is low enough

Alja [10]

Answer:

$\vec{v} =$

Explanation:

given,

location of the ball ⟨7,0,−8⟩

initial velocity of the ball ⟨-11,19,−5⟩

time = 0.4 s

speed of the ball = ?

using Momentum Principle

change in momentum = Force x time

$m \vec{v} - m \vec{u}= \vec{F}\times \Delta t$

$\vec{v} =\vec{u} + \dfrac{\vec{F}}{m}\times \Delta t$

Net force acting in this case will be equal to force due to gravity because air resistance is negligible.

F_net = F_g = ⟨0 ,-9.8 m , 0⟩

now,

$\vec{v} = + \dfrac{}{m}\times (0.4-0)$

$\vec{v} = +$

$\vec{v} =$

hence, the velocity of the ball 0.4 s after being kicked is equal to $\vec{v} =$

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3 years ago

A stone is tied to a string and whirled at constant speed in a horizontal circle. The speed is then doubled without changing the

Tanya [424]

Answer:

Afterward the magnitude of the acceleration of the stone is four times as great

Explanation:

A stone tied to a string and whirled at a constant speed in a horizontal circle will have a centripetal acceleration given by

$a = \frac{v^{2} }{r}$

Where $a$ is the centripetal acceleration

$v$ is the speed

and $r$ is the radius of the circle

Here, the radius of the circle is the length of the string.

Now, from the question

The speed is then doubled without changing the length of the string,

Let the new speed be $v_{2}$ , that is

$v_{2} = 2v$

and without changing the length of the string means radius $r$ is constant.

To determine the magnitude of the acceleration of the stone afterwards,

Let the new acceleration be $a_{2}$ .

Then we can write that

$a_{2} = \frac{v_{2}^{2} }{r}$

From

$a = \frac{v^{2} }{r}$

$v = \sqrt{ar}$

Recall that

$v_{2} = 2v$

∴ $v_{2} = 2\sqrt{ar}$

Now, we will put the value of $v_{2}$ into

$a_{2} = \frac{v_{2}^{2} }{r}$

Then,

$a_{2} = \frac{(2\sqrt{ar}) ^{2} }{r}$

$a_{2} = \frac{4ar }{r}$

$a_{2} = 4a$

The new magnitude of the acceleration of the stone is four times the initial acceleration.

Hence,

Afterward the magnitude of the acceleration of the stone is four times as great

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3 years ago