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zvonat [6]
3 years ago
6

It takes 56.5 kilojoules of energy to raise the temperature of 150 milliliters of water from 5°C to 95°C. If you

Physics
1 answer:
Debora [2.8K]3 years ago
7 0
You'd get an extra 40/60 of the energy, or 2/3. Multiply 5/3 by the required energy to get the actual consumption.
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What fraction of all the electrons in a 25 mg water
mihalych1998 [28]

Answer:

9.11\times 10^{-15}.

Explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of  -40 nC is removed from the water droplet.

The charge on one electron, \rm e=-1.6\times 10^{-19}\ C.

Let the N number of electrons have charge -40 nC, such that,

\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.  

Now, mass of one electron = \rm 9.11\times 10^{-31}\ kg.

Therefore, mass of N electrons = \rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is 25\ \rm mg = 25\times 10^{-6}\ kg.

Then,

\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.

It is the required fraction of mass of the droplet.

3 0
3 years ago
An 8.75 kg point mass and a 14.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be
marysya [2.9K]

Answer

given,

mass of the = m₁ = 8.75 Kg

another mass of the object = m₂ = 14 Kg

distance between them = 50 cm

R₁ = 17 cm

R₂ = 50 -17 = 33 cm

a) Force applied due to the Mass 8.75  in +ve x- direction

F_1 = \dfrac{GM_1 m}{R_1^2}

F_1 = \dfrac{6.67\times 10^{-11} \times 8.75\ m}{0.17^2}

F_1 = 2.019\times 10^{8}\ m

Force applied due to mass 14 Kg in -ve x-direction

F_2 = \dfrac{GM_2 m}{R_2^2}

F_2 = \dfrac{6.67\times 10^{-11} \times 14\ m}{0.33^2}

F_2 = 0.857\times 10^{8}\ m

net force

F = F₁ + F₂

F = 2.019\times 10^{8}\ m - 0.857\times 10^{8}\ m

F = 1.162\times 10^{8}\ m

Using newton second law

a = \dfrac{F}{m}

a = \dfrac{ 1.162\times 10^{8}\ m}{m}

a =1.162\times 10^{8} \ m/s^2

b) As the acceleration of mass comes out to be  +ve hence, the direction will be toward the mass of 8.75 Kg

6 0
3 years ago
When the effect of aerodynamic drag is included, the y‐acceleration of a baseball moving vertically upward is au = −g − kv2, whi
alexira [117]

Answer:

Explanation:

The detailed steps and appropriate integration and differentiation is as shown in the attached files.

au = −g − kv2

ad = −g + kv2

3 0
3 years ago
Our entire ______ _____<br> is built in the backbone of metals.
Aliun [14]

Answer:

modern world

Explanation:

6 0
3 years ago
A uniform ladder of length 10.8 m is leaning against a vertical frictionless wall. The weight of the ladder is 323 N, and it mak
love history [14]

Answer:

0.3625

Explanation:

From the given information:

Consider the equilibrium conditions;

On the ladder, net torque= 0

Thus,

\tau_{net} = 0; and

-fL \s in \theta +m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta = 0

However, by rearrangement;

fL \s in \theta =m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta  \\ \\  \mu(m_L + m) gL \ sin \theta = (323  \ N) ( 10.8 \ meters) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos  \  66.46^0

\mu= \dfrac{ (323  \ N) ( 10.8 \ m) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos  \  66.46^0}{\Big [(323 \ N)+(734 \ N) \Big] (10.8 \ m)}

\mathbf{\mu= 0.3625 }

8 0
3 years ago
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