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zhannawk [14.2K]

3 years ago

11

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

1 answer:

hichkok12 [17]3 years ago

7 0

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! =)

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A Christmas light is made to flash via the discharge of a capacitor. The effective duration of the flash is 0.25 s (which you ca

Sonbull [250]

Answer:

The correct solution is:

(a) $1.375\times 10^{-2} \ J$

(b) $4.43\times 10^{-3} \ C$

(c) $1.42\times 10^{-3} \ F$

(d) $178.57 \ \Omega$

Explanation:

The given values are:

Effective duration of the flash,

ζ = 0.25 s

Average power,

$P_{avg}=55 \ mW$

$=55\times 10^{-3} \ W$

Average voltage,

$V_{avg}=3.1 \ V$

Now,

(a)

⇒ $E=P_{avg}\times \zeta$

On substituting the values, we get

⇒ $=55\times 10^{-3}\times 0.25$

⇒ $=1.375\times 10^{-2} \ J$

(b)

⇒ $E=Q\times V_{avg}$

then,

⇒ $Q=\frac{E}{V_{avg}}$

On substituting the values, we get

⇒ $=\frac{1.375\times 10^{-2}}{3.1}$

⇒ $=4.43\times 10^{-3} \ C$

(c)

⇒ $C=\frac{Q}{V}$

⇒ $=\frac{4.43\times 10^{-3}}{3.1}$

⇒ $=1.42\times 10^{-3} \ F$

(d)

As we know,

⇒ $R=\frac{1}{4C}$

⇒ $=\frac{1}{4\times 1.42\times 10^{-3}}$

⇒ $=\frac{1000}{5.6}$

⇒ $=178.57 \ \Omega$

5 0

2 years ago

When a 20-V emf is placed across two resistors in series, a current of 2.0 A is present in each of the resistors. When the same

ehidna [41]

Answer:

7.24 ohm

Explanation:

Let R1 and R2 are resistance of two resistors.

Emf=E=20 V

Current,I=2 A

Current,I'=10 A

We have to find the magnitude of the greater of the two resistances.

In series

$R=R_1+R_2$

$V=IR$

By using the formula

$20=2(R_1+R_2)$

$R_1+R_2=\frac{20}{2}=10$ ...(1)

In parallel

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R}=\frac{R_2+R_1}{R_1R_2}$

$R=\frac{R_1R_2}{R_1+R_2}$

$20=10(\frac{R_1R_2}{R_1+R_2}$

$2=\frac{R_1R_2}{10}$

$R_1R_2=20$

$R_2=\frac{20}{R_1}$

Substitute the value

$\frac{20}{R_1}+R_1=10$

$R^2_1+20=10R_1$

$R^2_1-10R_1+20=0$

$R_1=\frac{10\pm\sqrt{(-10)^2-4(20)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$R_1=\frac{10\pm 2\sqrt 5}{2}$

$R_1=\frac{10+2\sqrt 5}{2}=5+\sqrt 5=7.24 ohm$

$R_1=\frac{10-2\sqrt 5}{2}=2.76 ohm$

Substitute the value

$R_2=\frac{20}{7.24}=2.76 ohm$

$R_2=\frac{20}{2.76}=7.24 ohm$

Hence, the magnitude of the greater of the two resistance=7.24 ohm

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