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2 years ago

A beaker weighs 0.4N when empty and1.4N when filled with water what does ot weigh when filled with brine of density 1.2 g/cm3

Physics

1 answer:

PtichkaEL [24]2 years ago

7 0

Answer:2.47

Explanation:

So, the beaker weighs 1.40N when filled with water, brine of density weighs about 1.7N, you add the density + water. Have a good day!

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Which is likely to be more common in our Galaxy: white dwarfs or black holes? Why?

Nookie1986 [14]

Answer:

White dwarfs are likely to be much more common. The number of stars decreases with increasing mass, and only the most massive stars are likely to complete their lives as black holes. There are many more stars of the masses appropriate for evolution to a white dwarf.

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2 years ago

How is the electrostatic force affected when the magnitude of a charge is doubled?

BaLLatris [955]

The magnitude of the electrostatic force between two charges is given by:
$F=k_e \frac{q_1 q_2}{r^2}$
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We can see that the magnitude of the force is directly proportional to the charges. This means that when one of the charges is doubled, the magnitude of the electrostatic force will double as well, so the correct answer is
A) <span>The magnitude of the electrostatic force doubles</span>

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2 years ago

A weather balloon is inflated to a volume of 26.8 LL at a pressure of 744 mmHgmmHg and a temperature of 31.2 ∘C∘C. The balloon r

elena-s [515]

Answer:

43.96 L

Explanation:

We are given that

$V_1=26.8 L$

$P_1=744mm Hg$

$T_1=31.2^{\circ} C=31.2+273=304.2K$

$P_2=385mmHg$

$T_2=-14.8^{\circ}=273-14.8=258.2K$

We know that

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=\frac{P_1V_1T_2}{P_2T_1}$

Substitute the values

$V_2=\frac{744\times 26.8\times 258.2}{385\times 304.2}$

$V_2=43.96 L$

Hence, the volume of balloon at -14.8 degree Celsius=43.96 L

5 0

2 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

2 years ago

At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the ti

34kurt

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r = 0.89 x 0. 5 = 0.445 m

Angular velocity

$\omega =\frac{72}{0.445}=161.8rad/s$

Frequency

$f=\frac{2\pi}{\omega}=\frac{2\times \pi}{161.8}=0.0388rev/s=2.33rev/min$

Revolutions per minute = 2.33

b) Centripetal acceleration

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

Here linear velocity = 72 m/s

Radius, r = 0.445 m

Substituting

$a=\frac{72^2}{0.445}=11649.44m/s^2$

Centripetal acceleration = 11649.44m/s²

6 0

2 years ago