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Mrac [35]
3 years ago
11

If an object has a volume of 2.5 mL and a mass of 10 g, what is the density of the object?

Physics
2 answers:
STALIN [3.7K]3 years ago
8 0
<span>The answer is 4                                                                                                                                                                                                                                 

done









</span>
skad [1K]3 years ago
4 0
We know, density = mass /volume 
Here, m = 10 g
v = 2.5 mL = 2.5 cm³

Substitute their values, 
density = 10/2.5
<span>d = 4 g/cm³
</span>
In short, Your Answer would be <span>4 g/cm³</span>

Hope this helps!
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A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary
Snezhnost [94]

Explanation:

Let N_p\ and\ N_s are the number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p}{N_s}=\dfrac{1}{3}

A resistor R connected to the secondary dissipates a power P_s=100\ W

For a transformer, \dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}

V_s=(\dfrac{N_s}{N_p})V_p

V_s=3V_p...............(1)

The power dissipated through the secondary coil is :

P_s=\dfrac{V_s^2}{R}

100\ W=\dfrac{V_s^2}{R}

V_p^2=\dfrac{100R}{9}.............(2)

Let N_p'\ and\ N_s' are the new number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p'}{N_s'}=\dfrac{1}{24}

New voltage is :

V_s'=(\dfrac{N_s'}{N_p'})V_p'

V_s'=24V_p...............(3)

So, new power dissipated is P_s'

P_s'=\dfrac{V_s'^2}{R}

P_s'=\dfrac{(24V_p)^2}{R}

P_s'=24^2\times \dfrac{(V_p)^2}{R}

P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}

P_s'=6400\ Watts

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0
3 years ago
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