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Vikki [24]
2 years ago
10

Help please help! Need help

Physics
1 answer:
Alborosie2 years ago
7 0

Answer:

M. Magnetism is a property of individual atoms.

Explanation:

when a magnet is broken into pieces the new pieces behave like the original magnet this observation shows that magnetism is the property of individual atoms.

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Two forces, one four times as large as the other, pull in the same direction on a 10kg mass and impart to it an acceleration of
notsponge [240]

Answer:

The acceleration of the mass is 2 meters per square second.

Explanation:

By Newton's second law, we know that force (F), measured in newtons, is the product of mass (m), measured in kilograms, and net acceleration (a), measured in meters per square second. That is:

F = m\cdot a (1)

The initial force applied in the mass is:

F = (10\,kg)\cdot \left(2.5\,\frac{m}{s^{2}} \right)

F = 25\,N

In addition, we know that force is directly proportional to acceleration. If the smaller force is removed, then the initial force is reduced to \frac{4}{5} of the initial force. The acceleration of the mass is:

\frac{25\,N}{20\,N} = \frac{2.5\,\frac{m}{s^{2}} }{a}

a = 2\,\frac{m}{s^{2}}

The acceleration of the mass is 2 meters per square second.

4 0
3 years ago
Y’all I have best describes the purpose of a bar graph
Sholpan [36]

so you can see all the different categories at once. both as a whole and on an individual scale.

3 0
3 years ago
What is the scientific meaning of energy?
JulijaS [17]

Answer:

it is B

Explanation:

8 0
3 years ago
What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe
Eduardwww [97]

Answer:

Final temperature, T_f=21.85^{\circ}

Explanation:

Given that,

Mass of silver ring, m = 4 g

Initial temperature, T_i=41^{\circ}C

Heat released, Q = -18 J (as heat is released)

Specific heat capacity of silver, c=0.235\ J/g\ C

To find,

Final temperature

Solution,

The expression for the specific heat is given by :

Q=mc\Delta T

Q=mc(T_f-T_i)

T_f=\dfrac{-Q}{mc}+T_i

T_f=\dfrac{-18}{4\times 0.235}+41

T_f=21.85^{\circ}

So, the final temperature of silver is 21.85 degrees Celsius.

5 0
3 years ago
A 1.0-kg block of aluminum is at a temperature of 50°C. How much thermal energy will it lose when its temperature is reduced by
Marat540 [252]

Answer:

22425 J

Explanation:

From the question,

Applying

Q = cm(t₂-t₁).................. Equation 1

Where Q = Thermal Energy, c = specific heat capacity of aluminium, m = mass of aluminium, t₂ = Final Temperature, t₁ = Initial Temperature.

Given: c = 897 J/kg.K, m = 1.0 kg, t₁ = 50 °C, t₂ = 25 °C (The final temperature is reduced by half)

Substitute these values into equation 1

Q = 897×1×(25-50)

Q = 897×(-25)

Q = -22425 J

Hence the thermal energy lost by the aluminium is 22425 J

5 0
3 years ago
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