3 years ago

Is ice cream a solid -or is it actually a solid?

Physics

2 answers:

blondinia [14]3 years ago

6 0

It is a solid when is frozen and a liquid when it melts

zaharov [31]3 years ago

6 0

It has to be a liquid turned into a solid because if heat accurs then it turns back into a liquid

You might be interested in

A 60-kg block initially at rest on a frictionless horizontal surface is acted upon by a force of 3.0 N for a distance of 7.0 m.

Schach [20]

Answer:

Explanation:

Kinetic energy generated = work done by force = force x displacement

= 3 x 7 = 21 J

5 0

3 years ago

What tool of dream analysis says that feelings and urges are hidden inside an event? condensation symbolism projection displacem

Brilliant_brown [7]

Answer:

Condensation

Explanation:

4 0

3 years ago

Read 2 more answers

A solid cylinder of mass 10 kg is pivoted about a frictionless axis thought the center O. A rope wrapped around the outer radius

taurus [48]

Torque acting dowward = 6 x 0.5 = 3 Nm

Torque acting to the right = 5 x 1 = 5 Nm

5 - 3 = 2 Nm

inertia = 1/2 mr^2

0.5 x 10 x 1^2 = 5 kg-m^2
2/5 = alpha = 0.4 rad /s^2

Hope this helps

5 0

3 years ago

Read 2 more answers

What is a good activity for strengthening your stomach muscles?

koban [17]

Answer:

sit-ups

Explanation:

3 0

3 years ago

Read 2 more answers

Two point charges are on the y-axis. A 3.0 µC charge is located at y = 1.15 cm, and a -2.28 µC charge is located at y = -2.00 cm

Ghella [55]

Answer:

Total electric potential, $V=1.32\times 10^6\ volts$

Explanation:

It is given that,

First charge, $q_1=3\ \mu C=3\times 10^{-6}\ C$

Second charge, $q_2=-2.28\ \mu C=-2.28\times 10^{-6}\ C$

Distance of first charge from origin, $r_1=1.15\ cm=0.0115\ m$

Distance of second charge from origin, $r_2=2\ cm=0.02\ m$

We need to find the total electric potential at the origin. The electric potential at the origin is given by :

$V=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}$

$V=k(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2})$

$V=9\times 10^9(\dfrac{3\times 10^{-6}}{0.0115}+\dfrac{-2.28\times 10^{-6}}{0.02})$

V = 1321826.08 V

$V=1.32\times 10^6\ volts$

So, the total electric potential at the origin is $1.32\times 10^6\ volts$ . Hence, this is the required solution.

3 0

3 years ago