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PtichkaEL [24]
3 years ago
14

Please help!!!! I will give brainliest if correct!!!

Physics
1 answer:
scZoUnD [109]3 years ago
4 0

Answer:

The density ρ of metal block is 8.92g/cm³

So from the given density table this corresponds to copper which has density of 8.92(g/mL)

Explanation:

Oh yeah, I got the correct unit update,

Now this problem bothers on the density of substances

We know that the density of a substance is expressed as

Density ρ= mass/ volume

Given data

Mass of metal block m= 62.44g

Volume of metal block v= 7 cm³

Hence we can find the density of the metal block by plugging in our data into the expression for density

ρ of metal block = 62.44/7

ρ of metal block = 8.92g/cm³

The block is a copper block

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Remember that protons are particles with positive charge and they held together in the nucleus of the atom which is a tiny tiny region. As you know, like charges repel each other, which means that the protons exert a repulsion force.
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What is negative acceleration?
Whitepunk [10]

Answer:

An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration).

Explanation:

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What is the mass of a 3,500-N rock?
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350kg because to get Newton’s it’s mass x Gravity, earths gravity is x10 so 3500 divided by 10 is 350
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A real heat engine operates between temperatures TcTcT_c and ThThT_h. During a certain time, an amount QcQcQ_c of heat is releas
Nookie1986 [14]

The maximum amount of work performed is

W_{max}=\frac{T_H-T_C}{T_C}Q_C

Explanation:

The efficiency of a real heat engine is given by the equation:

\eta = 1-\frac{T_C}{T_H} (1)

where

T_C is the temperature of the cold reservoir

T_H is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

\eta = \frac{W_{max}}{Q_H}

where

W_{max} is the maximum work done

Q_H is the heat absorbed from the hot reservoir

Q_H can be written as

Q_H=W_{max}+Q_C

where

Q_Cis the heat released to the cold reservoir

So the previous equation can be also written as

\eta=\frac{W_{max}}{W_{max}+Q_C} (2)

By combining eq.(1) and (2) we get

1-\frac{T_C}{T_H}=\frac{W_{max}}{W_{max}+Q}

And re-arranging the equation and solving for W_{max}, we find

W_{max}=\frac{T_H-T_C}{T_C}Q_C

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8 0
3 years ago
An electric bulb is marked 40volts ,230w another bulb is marked 40w,110v
Andrej [43]

Answer:

a. The ratio of their resistance is 2783:64

b. The ratio of their energy is 4:23

c. The charge on the first bulb is 5.75 C

The charge on the second bulb is 0.\overline {36} C

Explanation:

The voltage on one of the electric bulbs, V₁ = 40  volts

The power rating of the bulb, P₁ = 230 w

The voltage on the other electric bulbs, V₂ = 110 volts

The power rating of the bulb, P₂ = 40 w

a. The power is given by the formula, P = I·V = V²/R

Therefore, R = V²/P

For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96

The resistance of the second bulb, R₂ = 110²/40

The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64

∴ The ratio of their resistance, R₂:R₁ = 2783:64

b. The energy of a bulb, E = t × P

Where;

t = The time in which the bulb is powered on

∴ The energy of the first bulb, E₁ = 230 w × t

The energy of the second bulb, E₂ = 40 w × t

The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23

∴ The ratio of their energy, E₂:E₁ = 4:23

c. The charge on a bulb, 'Q', is given by the formula, Q = I × t

Where;

I = The current flowing through the bulb

From P = I·V, we get;

I = P/V

For the first bulb, the current, I = 230 w/40 V = 5.75 amperes

The charge on the first bulb per second (t = 1) is therefore;

Q₁ = 5.75 A × 1 s = 5.75 C

The charge on the first bulb, Q₁ = 5.75 C

Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = 0.\overline {36} C

The charge on the second bulb, Q₂ = 0.\overline {36} C.

d. The question has left out parts

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