Balls and rods set up to represent a solar system

What is the maximum mass that can hang without sinking from a 20-cm diameter Styrofoam sphere in water? Assume the volume of the

alex41 [277]

Answer:

the maximum mass that can hang without sinking is 2.93 kg

Explanation:

Given: details:

sphere diameter d = 20 cm

so, radius r = 10 cm = 0.10 m

density of the Styrofoam sphere D = 300 kg/m3

sphere volume $V = \frac{4}{3} \pi r^3$

$=\frac{4}{3} \pi 0.10^3$

=4.18*10^{-3} m^3

we know that

$Density = \frac{Mass}{Volume}$

mass M = Density * Volume

= (300)(4.18*10^{-3} m3)

=1.25 kg

mass of the water displace = volume *density of water

= 4.18*10^{-3} m3 * 1000

= 4.18 kg

The difference between the mass of water and mass of styrofoam is the amount of mass that the sphere can support

=4.18 kg -1.25 kg

= 2.93 kg

3 0

4 years ago

A proton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5

svlad2 [7]

The magnetic force experienced by the proton is given by
$F=qvB \sin \theta$
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and $\theta$ the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so $\sin \theta=1$ and we can ignore it in the formula.

For Netwon's second law, the force is also equal to the proton mass times its acceleration:
$F=ma$

So we have
$ma=qvB$
from which we can find the magnitude of the field:
$B= \frac{ma}{qv}= \frac{(1.67 \cdot 10^{-27}kg)(5\cdot 10^{13}m/s^2)}{(1-6 \cdot 10^{-19}C)(3.7 \cdot 10^7 m/s)}=0.014 T$

4 0

3 years ago

The potential difference between the plates of a capacitor is 145 V. Midway between the plates, a proton and an electron are rel

aniked [119]

Answer:

= 2.52 x 10^ 6 m/s

Explanation:

The force that acts on charged particles between capacitor plates =

F = (q) (Δv) ÷ d

Here, d = distance between the two plates

q = charge of the charged particle

Δv = voltage

Normally, the force that makes both proton and electron released from rest, giving the charge acceleration is F=m X a. where m= mass and a = acceleration

Poting this equation with the first one, we have:

m X a = (q) (Δv) ÷ d

So, the acceleration of a proton when moving towards a negatively charged plate is

a = (q) (Δv) ÷ (d) (m) {proton}

Likewise, the acceleration of an electron when moving towards a positively charged plate is

a = (q) (Δv) ÷ (d) (m) {electron}

Dividing the proton acceleration formula by the electron acceleration formula we have:

a (proton) / a (electron) = m (proton) / m(electron)

inserting equation of motion to get distance, s

s = ut + 1/2 at^2

recall that electron travel distance, d/2

d/2 = 1/2 at^2

making t the subject of the formula

we have, t =√(d ÷ a(electron))

The distance of proton:

d/2 = ut + 1/2 at^2 [proton}

put d/2 = ut + 1/2 at^2 [proton} into t =√(d ÷ a(electron))

Initial speed, ui = √(d ÷ a(electron)) = (d/2) - (1/2) x (d) (a(proton) + a(electron))

since acceleration wasn't given in the question, lets use mass(elect

ron) ÷ mass(proton) rather than use (a(proton) + a(electron))

Therefore, intial speed= 1/2√((e X Δv) ÷ m(electron)) (1- m(electron)/ m(proton))

Note, e = 1.60 x 10^-19

m(electron) = 9.11 X 10^-31

m(proton) = 1.67 X 10^-27

Input these values into the formula above, initial speed, UI =

= 2.52 x 10^ 6 m/s

7 0

3 years ago

A circuit is built based on the circuit diagram shown.

julsineya [31]

The current in the 50 Ω resistor is A) 1.2 A

4 0

3 years ago

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Which observation would be evidence that heat was transferred by radiation

Masteriza [31]

Radiation is a type of heat transfer wherein there is no need for medium or media through which the heat will flow. Consequently, the radiation waves are able to travel through vacuum. The best observation as evidence to conclude that heat is indeed transferred by radiation is the increase of temperature of the receiving body.

8 0

3 years ago

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