Bromine vapour is heavier than air. even so it's spreads upwards in the experiment above. Why?

Total these measurements. Your answer should indicate the proper accuracy. Be sure to include the units in your answer. (Remembe

Mila [183]

Your answer is 8. You add 2 + 1 + 5.3 to get 8.3. You round down to 8 because of the sig fig rules.

5 0

3 years ago

Use the data provided to calculate the gravitational potential energy of each cylinder mass. Round your answers to the nearest t

Goryan [66]

Answer: 14.7kJ, 29.4kJ, 44.1kJ

Explanation:

<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field. </em>

<em />

In the case of the Earth, in which the gravitational field is considered constant, the value of the gravitational potential energy $U_{p}$ will be:

$U_{p}=mgh$

Where $m$ is the mass of the object, $g=9.8m/s^{2}$ the acceleration due gravity and $h=500m$ the height of the object.

Knowing this, let's begin with the calculaations:

For m=3kg

$U_{p}=(3kg)(9.8m/s^{2})(500m)$

$U_{p}=14700J=14.7kJ$

For m=6kg

$U_{p}=(6kg)(9.8m/s^{2})(500m)$

$U_{p}=29400J=29.4kJ$

For m=9kg

$U_{p}=(9kg)(9.8m/s^{2})(500m)$

$U_{p}=44100J=44.1kJ$

6 0

3 years ago

The force of the added water produces a torque on the dam. In a simple model, if the torque due to the water were enough to caus

Fittoniya [83]

Answer:

The appropriate response is " $\tau=\frac{1}{6} PgLh^3$ ". A further explanation is described below.

Explanation:

The torque ( $\tau$ ) produced by the force on the dam will be:

⇒ $d \tau=XdF$

On applying integration both sides, we get

⇒ $\tau = \int_{0}^{a}x pgL(h-x)dx$

⇒ $= pgL\int_{0}^{h}(h-x)dx$

⇒ $=pgL[\frac{h^3}{2} -\frac{h^3}{3} ]$

⇒ $=\frac{1}{6} PgLh^3$

8 0

2 years ago

The barrel of a rifle has a length of 0.89 m. A bullet leaves the muzzle of a rifle with a speed of 620 m/s. What is the acceler

guapka [62]

Answer:

215955.06 m/s^2

Explanation:

length of barrel, s = 0.89 m

initial velocity of the bullet, u = 0 m/s

Final velocity of the bullet, v = 620 m/s

Let a be the acceleration of the bullet in the barrel

Use third equation of motion, we get

$v^{2}=u^{2}+ 2as$

$620^{2}=0^{2}+ 2\times a \times 0.89$

a = 215955.06 m/s^2

Thus, the acceleration of the bullet inside the barrel is 215955.06 m/s^2.

6 0

3 years ago

A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa

Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

$Q=P\tan\phi$ ...(I)

We need to calculate the $\phi$

Using formula of $\phi$

$\phi=\cos^{-1}(Power\ factor)$

Put the value into the formula

$\phi=\cos^{-1}(0.6)$

$\phi=53.13^{\circ}$

Put the value of Φ in equation (I)

$Q=600\tan(53.13)$

$Q=799.99\ kVAR$

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

$Q'=600\tan(0.95)$

$Q'=9.94\ KVAR$

We need to calculate the difference between Q and Q'

$Q''=Q-Q'$

Put the value into the formula

$Q''=799.99-9.94$

$Q''=790.05\ KVAR$

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0

3 years ago