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myrzilka [38]
3 years ago
10

A 30.0 g mass of iron at 24.5°C is heated to 45.0°C. The theoretical specific

Physics
1 answer:
diamong [38]3 years ago
8 0

Answer:

276.135 J

Explanation:

Given that:

mass of Fe = 30.0 g

initial temperature = 24.5°C

final temperature = 45.0°C

specific heat of Fe = 0.449 J/g°C

We can determine the thermal energy added by using the formula;

Q = mcΔT

Q = 30.0g × 0.449 J/g°C × (45.0 - 24.5)°C

Q = 276.135 J

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Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)
SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c²  =>  c = 0.707 m

Charge to the right:  In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x 10^{9} N m²/C²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.707)²] cos 45

e = 225000√2 N/C

X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below:  For y direction

e = kq/r²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

4 0
3 years ago
How far away is mars?
Elanso [62]

Answer:

about: 110.14 million mi

Explanation:

the distance to Mars from Earth is 140 million miles (225 million kilometers).But, distance to Mars from Earth is constantly changing.

Hope that was helpful.Thank you!!!

3 0
3 years ago
Read 2 more answers
A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting
ololo11 [35]

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

T=\dfrac{38}{8}\\\\T=4.75\ s

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

k = spring stiffness constant

So,

k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m

When the cord is in air,

mg=kx

x = the extension in the cord

x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

5 0
3 years ago
A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless, horizontal s
mafiozo [28]

Answer:

block velocity   v = 0.09186 = 9.18 10⁻² m/s  and speed bollet   v₀ = 11.5 m / s

Explanation:

We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.

Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)

Before the crash

     p₀ = m v₀ + 0

After the crash

   p_{f} = (m + M) v

    p₀ = p_{f}

    m v₀ = (m + M) v                    (1)

Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring

Initial

    Em₀ = K = ½ m v2

Final

    E m_{f}= Ke = ½ k x2

   Emo = E m_{f}

   ½ m v² = ½ k x²

   v² = k/m  x²

Let's look for the spring constant (k), with Hook's law

   F = -k x

   k = -F / x

   k = - 0.75 / -0.25

   k = 3 N / m

Let's calculate the speed

  v = √(k/m)   x

  v = √ (3/8.00)   0.15

  v = 0.09186 = 9.18 10⁻² m/s

This is the spped of  the  block  plus bullet rsystem right after the crash

We substitute calculate in equation  (1)

   m v₀ = (m + M) v

  v₀ = v (m + M) / m

  v₀ = 0.09186 (0.008 + 0.992) /0.008

  v₀ = 11.5 m / s

6 0
3 years ago
A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal ener
Elina [12.6K]

Answer:

Explanation:

From the given information:

The initial PE (PE)_i = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = P.E_f -P.E_i

ΔP.E = 0 - (PE)_i

ΔP.E = -P.E_i

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

\Delta P.E + \Delta K.E + \Delta U = 0

\Delta U = -\Delta P.E - \Delta K.E

this can be re-written as:

\Delta U =- (-\Delta P.E_i) - \Delta K.E

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

\Delta U =(70\%) \Delta P.E_i-0

\Delta U =(0.70) (490.5)

ΔU = 343.35  J

Thus, the magnitude of the increase is = 343.35 J

7 0
3 years ago
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