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3 years ago

HI CAN ANYONE PLS ANSWER DIS PLS!!!!!!

Physics

1 answer:

Evgesh-ka [11]3 years ago

4 0

Answer:

Volume of an object (L): blue 105.00, yellow 105.00, green 102.50, red 101.25

Density of an object (kg/L) Mass / Volume: blue 1, yellow 1, green 1.014, red .992

Sink or Float: blue sink, yellow sink, green float, red float

hope this helps

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Electronics and inhabitants of the International Space Station generate a significant amount of thermal energy that the station

katrin2010 [14]

Answer:

4462.0927 W

Explanation:

$\epsilon$ = Emissivity of the panel = 1

$\sigma$ = Stefan-Boltzmann constant = $5.67\times 10^{-8}\ W/m^2K^4$

T = Temperature = (273.15+6)

Area of the panel is given by

$A=2\times 1.8\times 3.6\\\Rightarrow A=12.96\ m^2$

The power radiated is given by

$P=\epsilon \sigma AT^4\\\Rightarrow P=1\times 5.67\times 10^{-8}\times 12.96\times (273.15+6)^4\\\Rightarrow P=4462.0927\ W$

The power radiated from each panel is 4462.0927 W

5 0

4 years ago

What are two reasons why a home computer scanner requires

alexandr1967 [171]

Answer: C and D

Explanation: a p e x

4 0

3 years ago

Read 2 more answers

During the experiment if you could double the breakaway magnetic force with all other quantities left unchanged, what is the new

sergiy2304 [10]

There are some missing information in the question.
However, since you are talking about magnetic force, I think you refer to the Lorentz force. When a particle of charge q and velocity v is immersed in a magnetic field of intensity B, the force acting on the particle is:
$F=qvBsin\theta$
where $\theta$ is the angle between the magnetic field and the direction of the particle.
Therefore, if force F is doubled, then also the velocity v must be double of its initial value:
$v=2v_0$

6 0

3 years ago

Help on matching please, struggling on it. Even just 1 or 2 would help

n200080 [17]

Answer:

7. free fall -- h. 9.8m/s^2

3. Velocity -- x. 60 km/hr west

6. Acceleration -- d. change in velocity/time

8. Centrifugal -- s. towards the centre

13. Work done --w. Force * displacement

5. Uniform circular motion --j. spin cycle in washer

18. Power -- r. kW an hour

7. g -- a. 10N

hope this helps

6 0

2 years ago

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

3 0

2 years ago