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3 years ago

A water heater warms 35 l of water from a temperature of 22.7 c to a temperature of 83.7

1 answer:

8 0

The amount of energy needed to increase the temperature of a substance by $\Delta T$ is given by
$Q=m C_s \Delta T$
where
m is the mass of the substance
$C_s$ is its specific heat capacity
$\Delta T$ is the increase in temperature

The water volume is $V=35 L= 35 dm^3 = 0.035 m^3$ , since its density is $d=1000 kg/m^3$ , the mass of this sample of water is
$m=dV=(1000 kg/m^3)(0.035 m^3)=35 kg$

The water specific heat capacity is $C_s = 4.18 kJ/kg ^{\circ}C$

and the increase in temperature is $\Delta T=83.7 ^{\circ}C-22.7 ^{\circ}C=61^{\circ}C$

Therefore, the amount of energy needed is
$Q=mC_s \Delta T=(35 kg)(4.18 kJ/kg ^{\circ}C)(61^{\circ}C)=8924 kJ = 8.92 \cdot 10^6 J$

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Atoms of the same element that differ only in the number of neutrons are known as.

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Answer:

They are known as isotopes

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2 years ago

The train 'A' travelled a distance is f 120 km in 3 hours , Whereas another train 'B' travelled a distance of 180 km in 4 hours

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Answer:

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Explanation:

A train travelled at 40 km/hr. (120/3)

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3 years ago

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-Plz Help Meh-<br> Calculate the acceleration of a car if the force is 450N and the mass is 1300kg.

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Answer:

Explanation:

F=Ma

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2 years ago

A force of 5 N produces an acceleration of 2 m/s2 on the object. What is the mass of the object?

hichkok12 [17]

Hello!

A force of 5 N produces an acceleration of 2 m/s2 on the object. What is the mass of the object ?

Data:

F (force) = 5 N

m (mass) = ?

a (acceleration) = 2 m/s²

Solving:

$F = m*a$

$5 = m*2$

$2\:m = 5$

$m = \dfrac{5}{2}$

$\boxed{\boxed{m = 2.5\:kg}}\end{array}}\qquad\quad\checkmark$

Answer:

2.5 kg

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

7 0

2 years ago

At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is

n200080 [17]

a) $8.93\cdot 10^{-30} N$ , downward

b) $1.76\cdot 10^{-17} N$ , upward

c) $1.20\cdot 10^{-17} N$ , downward

Explanation:

The gravitational force of an object (also known as weight of the object) is the attractive force with which the object is pulled towards the Earth's centre.

For an object near the Earth's surface, the magnitude of the gravitational force is given by the equation

$F=mg$

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

$g=9.8 m/s^2$ is the acceleration due to gravity

Therefore, the gravitational force on the electron is:

$F=(9.11\cdot 10^{-31})(9.8)=8.93\cdot 10^{-30} N$

And the direction is downward, towards the Earth's centre.

The electric force on a charged particle is the force produced by the presence of an electric field.

In particular, this force is:

- Repulsive (away from the source of the field) if the charge has the same sign of the charge source of the field

- Attractive (towards the source of the field) if the charge has opposite sign to the charge source of the field

The magnitude of the electric force is given by:

$F=qE$

where

q is the charge of the particle

E is the strength of the electric field

In this problem:

$q=1.6\cdot 10^{-19} C$ is the charge of the electron

$E=110 N/C$ is the strength of the electric field

Substituting,

$F=(1.6\cdot 10^{-19})(110)=1.76\cdot 10^{-17} N$

And since the electron has negative charge, the direction of the force is opposite to that of the electric field, so upward.

When a charged particle is moving in a magnetic field, it experiences a force which is perpendicular to both the direction of motion of the charge and to the direction of the magnetic field.

The magnitude of this force is given by (if the charge moves perpendicular to the magnetic field)

$F=qvB$

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

In this problem:

$q=1.6\cdot 10^{-19} C$ is the charge of the electron

$v=1.50\cdot 10^6 m/s$ is the velocity

$B=50.0\mu T = 50.0 \cdot 10^{-6} T$ is the strength of the magnetic field

Substituting,

$F=(1.6\cdot 10^{-19})(1.50\cdot 10^6)(50.0\cdot 10^{-6})=1.20\cdot 10^{-17} N$

And the direction can be determined using the right-hand rule:

- Index finger: direction of velocity (eastward)

- Middle finger: direction of magnetic field (northward)

- Thumb: direction of force (upward) --> however the electron has negative charge, so the direction of the force is reversed --> downward

5 0

2 years ago