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Rasek [7]
3 years ago
11

Newtons third law of motion describes how forces act in pairs true or false HELPPPP!!!!

Physics
2 answers:
marin [14]3 years ago
8 0

Answer:

True

Explanation:

Hope this helps! Add Brainliest

Elanso [62]3 years ago
5 0

Answer:

true

Explanation:

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Imagine tying a string to a ball and twirling it around you. How is this similar to the moon orbiting the Earth? In this example
Tema [17]
Its similar to the moon orbiting the earth because lets say that the sing is moon and the ball is earth has the "moon" orbits around the "earth" the string ends up tying around the ball till its no more


i think i hope this example helps you somehow srry that i dont know more then that :/
4 0
3 years ago
Read 2 more answers
A truck with 0.410 m radius tires travels at 25.0 m/s. What is the angular velocity of the rotating tires in radians per second?
Eduardwww [97]

relation between linear velocity and angular velocity is given as

v = R\omega

here

v = linear speed

R = radius

\omega = angular speed

now plug in all data in the equation

25.0 = 0.410 \omega

\omega  = \frac{25}{0.410}

\omega = 60.9 rad/s

so rotating speed is 60.9 rad/s

6 0
3 years ago
An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su
s2008m [1.1K]

Answer: A. 23.59 minutes.

              B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

  • mass of ice, m= 0.555 kg
  • temperature of ice, T= -16.6°C
  • rate of heat transfer, q=820 J.min^{-1}
  • specific heat of ice, c_{i}= 2100 J.kg^{-1}.K^{-1}
  • latent heat of fusion of ice, L_{i}=334\times10^{3}J.kg^{-1}

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

Q=mc\Delta T

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

Q= 0.555\times2100\times16.6

Q= 19347.3 J

Now, we require 19347.3 joules of heat to bring the ice to 0°C  and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

t=\frac{Q}{q}

=\frac{19347.3}{820}

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent  heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, Q_{L}= mL_{i}

Q_{L}= 0.555\times334\times10^{3}= 185370 J

<u>Now  the time required for supply of 185370 J:</u>

t=\frac{Q_{L}}{q}

t=\frac{185370}{820}

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0
3 years ago
How much heat is absorbed by a 450g gold block as energy from the sun causes its temperature to change from 27°C to 32°C? (Speci
kotegsom [21]

Answer:

Q = 1057.5 [cal]

Explanation:

In order to solve this problem, we must use the following equation of thermal energy.

Q=m*C_{p}*(T_{final}-T_{initial})

where:

Q = heat energy [cal]

Cp = specific heat = 0.47 [cal/g*°C]

T_final = final temperature = 32 [°C]

T_initial = initial temperature = 27 [°C]

m = mass of the substance = 450 [g]

Now replacing:

Q=450*0.47*(32-27)\\Q=1057.5[cal]

5 0
3 years ago
On a sunny day, a student poured a cup of water on the sidewalk to make a puddle. When he returned later, the puddle was gone. T
ivanzaharov [21]

Answer:

The students conclusion is wrong.

Explanation:

The students conclusion is wrong because liquids can not soak into concrete. The water evaporated because it was a hot sunny day. The water turned into a gas called water vapor.

8 0
3 years ago
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