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Alexeev081 [22]

3 years ago

11

. Determinar la magnitud de la fuerza que recibe un cuerpo de 45 kg, la cual le produce una aceleración cuya magnitud es de 5 m/

s2.

1 answer:

PSYCHO15rus [73]3 years ago

6 0

Answer:

225 N

Explanation:

Según la segunda ley de Newton;

F = ma

F = fuerza

m = masa

a = aceleración

Por eso;

F = 45 kg * 5 m / s ^ 2

F = 225 N

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In Fig. 21-25, particle 1 of charge &1.0 mC and particle 2 of charge $3.0 mC are held at separation L ! 10.0 cm on an x axis

Kaylis [27]

Answer:

Since the particle 1 and 2 are on the x-axis, the 3rd particle should also be on the x-axis in order the net force on it to be zero.

Let's denote the distance between particles 1 and 3 as x. Therefore the distance between particles 2 and 3 is (0.1 - x), since the distance between 1 and 2 is 0.1 m.

Coulomb's Law states the force between charges as

$F_{1-3} = \frac{1}{4\pi \epsilon_0}\frac{q_1q_3}{x^2}$

$F_{2-3} = \frac{1}{4\pi \epsilon_0}\frac{q_2 q_3}{(0.1 -x)^2}$

The question asks that $F_{1-3} = F_{2-3}$ , so

$\frac{1}{4\pi \epsilon_0}\frac{1\times 10^{-3} \times q_3}{x^2} = \frac{1}{4\pi \epsilon_0}\frac{3\times 10^{-3}\times q_3}{(0.1 - x)^2}\\\frac{1}{x^2} = \frac{3}{(0.1 - x)^2}\\3x^2 = 0.01 -0.2x + x^2\\2x^2 + 0.2x - 0.01 = 0\\x_1 = 0.036m\\x_2 = -0.136m$

We will take the positive root:

$x = 0.036~m$ away from the first particle.

Explanation:

Since Fig. 21-25 is not given in the question, the exact locations are not known. However, the location of the third particle is found to be 0.036 m away from the first particle and the third particle is located between the particles 1 and 2.

8 0

3 years ago

At launch a rocket ship weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 km/h; at the end

SpyIntel [72]

Answer:

$5.590278\ m/s^2$

$7.74038461538\ m/s^2$

178.888896 m

12790.56 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{161}{3.6}-0}{8}\\\Rightarrow a=5.590278\ m/s^2$

The acceleration is $5.590278\ m/s^2$

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{1610}{3.6}-\dfrac{161}{3.6}}{60-8}\\\Rightarrow a=7.74038461538\ m/s^2$

The acceleration is $7.74038461538\ m/s^2$

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 5.590278\times 8^2\\\Rightarrow s=178.888896\ m$

Distance traveled in the first 8 seconds is 178.888896 m

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=\dfrac{161}{3.6}\times 52+\dfrac{1}{2}\times 7.74038461538\times 52^2\\\Rightarrow s=12790.56\ m$

Distance traveled during 8-60 second interval is 12790.56 m

6 0

3 years ago

Can someone help me with these physics questions?

wariber [46]

hi! Glad to help =)

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3 years ago

How many electrons make up a charge of - 30.0 μc?

Ymorist [56]

Each electron has a charge of

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Therefore, the number of electrons contained in this total charge will be given by the total charge divided by the charge of a single electron:

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3 0

3 years ago

Read 2 more answers