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3 years ago

You are asked to determine the density of a liquid. What lab equipment is most helpful to find density of the liquid?

Physics

2 answers:

crimeas [40]3 years ago

7 0

The answer is D, the graduated cylinder and a triple beam-balance

AnnyKZ [126]3 years ago

6 0

D. The graduated cylinder is used to find the volume, and triple beam balance is used to find the mass.

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If railroad tracks were built over the boundary of two plates what would happen to the railroad tracks as the plates moved

Oliga [24]

The railroad tracks will move with the plate boundaries

6 0

3 years ago

Una bola de 1 kg gira alrededor de un circulovrtical en el extremo de un cuerda. El otro extremo de la cuerda esta fijo en el ce

Vladimir79 [104]

Answer:

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

Explanation:

Puesto que la bola gira en un círculo vertical, existe claramente una diferencia entre las tensiones debido a la influencia de la gravedad y la tensión que resulta de la aceleración centrípeta experimentada por la masa. La máxima tensión ocurre cuando la bola se encuentra en el nadir (o la sima) del trayecto circular, la cual se describe por la Segunda Ley de Newton:

$T_{max} - m\cdot g = m\cdot \frac{v^{2}}{L}$

En cambio, la mínima tensión aparece cuando la bola se encuentra en el cénit (o la cima) del trayecto circular, descrita por la misma ley de Newton:

$T_{min} + m\cdot g = m\cdot \frac{v^{2}}{L}$

Donde:

$T_{min}$ , $T_{max}$ - Tensiones mínima y máxima, medidas en newtons.

$m$ - Masa de la bola, medida en kilogramos.

$g$ - Constante gravitacional, medida en metros por segundo al cuadrado.

$L$ - Distancia con respecto al eje de rotación, medida en metros.

$v$ - Rapidez tangencial, medido en metros por segundo.

Se elimina la aceleración centrípeta de ambas expresiones por igualación:

$T_{min} + m\cdot g = T_{max} - m\cdot g$

Ahora, la diferencia entre las tensiones máxima y mínima es:

$T_{max} - T_{min} = 2\cdot m \cdot g$

Si $m = 1\,kg$ y $g = 9.807\,\frac{m}{s^{2}}$ , entonces:

$T_{max} - T_{min} = 2\cdot (1\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$T_{max}-T_{min} = 19.614\,N$

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

5 0

3 years ago

A dog leaps horizontally off a 70 m cliff with a speed of 6 m/s, how far from the base will the dog land?

iris [78.8K]

Answer:

$\Delta x=22.67786838m$

Explanation:

Let's use projectile motion equations. First of all we need to find the travel time. So we are going to use the next equation:

$y-y_0=v_o*sin(\theta)*t-\frac{1}{2}*t^2$ (1)

Where:

$y=Final\hspace{3}position\hspace{3}at\hspace{3}y-axis$

$y_o=Initial\hspace{3}position\hspace{3}at\hspace{3}y-axis$

$v_o=initial\hspace{3}velocity$

$t=travel\hspace{3}time$

$g=gravity\hspace{3}constant$

$\theta=Initial\hspace{3}launch\hspace{3}angle$

In this case:

$\theta=0$

Because the dog jumps horizontally

Let's asume the gravity constant as:

$g=9.8$

$y=0$

Because when the dog reach the base the height is 0

$y_o=70$

$v_o=6$

Now let's replace the data in (1)

$y_o-\frac{1}{2} *(9.8)*t^2+70$

Isolating t:

$t=\pm\sqrt{\frac{2*70}{9.8} } =3.77964473$

Finally let's find the horizontal displacement using this equation:

$\Delta x=v_o*cos(\theta)*t$

Replacing the data:

$\Delta x=6*1*3.77964473=22.67786838m$

8 0

2 years ago

A circular conical reservoir has depth 20 feet and radius of the top 10 feet. water is leaking out so that the surface is fallin

levacccp [35]

As per given condition we know that vertex angle of the cone is given as

$tan\theta = \frac{R}{H}$

so here we can say that vertex angle will remain constant

so here

$\frac{r}{y} = \frac{R}{H}$

$\frac{r}{8} = \frac{10}{20}$

$r = 4 feet$

now for the volume we can say

$V = \frac{1}{3}\pi r^2 y$

also we can say

$r = \frac{y}{2}$

so here we will have

$V = \frac{1}{3}\pi (\frac{y}{2})^2y = \frac{1}{12}\pi y^3$

now for volume flow rate

$Q = \frac{dV}{dt} = \frac{3}{12}\pi y^2\frac{dy}{dt}$

$Q = \frac{1}{4}\pi y^2 v_y$

now plug in all data

$Q = \frac{1}{4} \pi (8)^2 (12) ft^3/h$

$Q = 603.2 ft^3/h$

4 0

2 years ago

A boat traveling at a velocity of 20 m/s leaves island heading east . The boat slows to rest at a rate of 1.5 m/s2 How far away

ioda

The boat's initial velocity is:

$v_0 = 20 m/s$

While the boat's acceleration is

$a=-1.5 m/s^2$

with a negative sign, since the boat is slowing down, so it is a deceleration. The distance traveled by the boat until it comes to a stop can be found by using the following equation:

$v_f^2 -v_0^2 = 2aS$

where vf=0 is the final velocity of the boat and S is the distance covered. Re-arranging the formula, we can find S:

$S=\frac{-v_0^2}{2a}=\frac{-(20 m/s)^2}{2(-1.5 m/s^2)}=133.3 m$

7 0

2 years ago