Please help true or false

A student notices that an inflated balloon gets larger when it is warmed by a lamp. Which best describes the mass of the balloon

nirvana33 [79]

It gets larger because
well let me give you an example
so today in class we looked at a lava lamp with wax inside and there was a lightbulb at the bottom.
we watched as the wax floated up because the molecules inside the wax spreads out and makes the wax less dense.
the wax floats up because (which is related to the balloon getting bigger) the wax is getting less dense and the particles get bigger which ALSO makes the wax less dense.
hope this helps and hope you can relate it to your problem! say thanks if I did help AT ALL! :)

7 0

3 years ago

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A truck pushes a pile of dirt horizontally on a frictionless road with a net force of 20\, \text N20N20, start text, N, end text

meriva

Answer:

300 Nm ; 300 J

Explanation:

Given that:

Force (F) = 20 N

Distance (d) = 15 m

The kinetic energy (Workdone) = Force * Distance

Kinetic Energy = 20N * 15m

Kinetic Energy = 300Nm

K. E = 1/2

4 0

3 years ago

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You are standing 2.5m directly in front of one of the two loudspeakers. They are 3.0m apart and both are playing a 686Hz tone in

ahrayia [7]

Answer:

distance from speaker is 17.87 m

Explanation:

given data

distance from loudspeaker = 2.5 m

distance between loudspeaker = 3.0 m

room temperature = 20c

wavelength f = 686Hz

to find out

what distances from the speaker

solution

we know sound velocity c = 331.5 + 0.6 × 20c = 343.5

so wavelength of sound λ = c / f

wavelength = 343.5 / 686 = 0.5 m

when the difference in distance of speaker destructive interference will be

d = λ/2 × (2n-1)

for n = 1, 2 3 4 ..

d = 0.5/2 × (2n-1)

d = 0.250 , 0.75 , 1.25 , 1.750............ for n = 1, 2 3 .............

so

for d = 0.250

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1)$ = 0.250

0.5 x1 = 7.6875

x1 = 15.375 m

for d = 0.75

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2)$ = 0.75

1.5 x2 = 4.6875

x2 = 3.125 m

for d = 1.250

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3)$ = 1.250

2.5 x2 = 1.1875

x3 = 0.475 m

for d = 1.750

x4 will be negative so we stop here

so the distance from speaker here is given below

distance = 2.5 + x

here x = 0.475 , 3.125 and 15.375 so

distance 1 = 2.5 + 0.475 = 2.975 m

distance 2 = 2.5 + 3.125 = 5.625 m

distance 3 = 2.5 + 15.375 = 17.875 m

final distance from speaker is 17.87 m

8 0

3 years ago

A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the

Sergio039 [100]

Answer:

W = 1.06 MJ

Explanation:

- We will use differential calculus to solve this problem.

- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

F = p*V*g

F = 1000*(2*0.5*x*8*dx)*g

F = 78480*x*dx

- Now, the work done is given by:

W = F.s

- Where, s is the distance from top of hose to the differential volume:

s = (5 - x)

- We have the work as follows:

dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

W = 78400*(2.5*x^2 - (1/3)*x^3)

W = 78400*(2.5*3^2 - (1/3)*3^3)

W = 78400*13.5 = 1058400 J

5 0

3 years ago

An isolated conducting sphere has a 10 cm radius. One wire carries a current of 1.000 002 0 A into it. Another wire carries a cu

romanna [79]

Answer:

t = 5.56 ms

Explanation:

Given:-

- The current carried in, Iin = 1.000002 C

- The current carried out, Iout = 1.00000 C

- The radius of sphere, r = 10 cm

Find:-

How long would it take for the sphere to increase in potential by 1000 V?

Solution:-

- The net charge held by the isolated conducting sphere after (t) seconds would be:

qnet = (Iin - Iout)*t

qnet = t*(1.000002 - 1.00000) = 0.000002*t

- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:

V = k*qnet / r

Where, k = 8.99*10^9 ..... Coulomb's constant

qnet = V*r / k

t = 1000*0.1 / (8.99*10^9 * 0.000002)

t = 5.56 ms

7 0

3 years ago