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masya89 [10]
3 years ago
5

Two particles, each of mass 7.0 kg, are a distance 3.0 m apart. To bring a third particle, with mass 21 kg, from far away to a r

esting point midway between the two particles, an external agent must do work equal to
Physics
1 answer:
garri49 [273]3 years ago
8 0

Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J

Explanation:

Given that;

Mass M1 = 7.0 kg

r = 3.0/2 m = 1.5 m

Mass M2 = 21 kg

we know that G = 6.67 × 10⁻¹¹ N.m²/kg²

work done by an external agent W = -2GM2M1 / r

so we substitute

W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5

W = -1.96098 × 10⁻⁸ / 1.5

W = -1.3 × 10⁻⁸ J

Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J

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B. Write true or false for the following statements.
Makovka662 [10]
1. T
2. F
3. T
4. F
5. T
I’m pretty sure.
7 0
3 years ago
A 100 Ω resistor is connected in series with a 47 µF capacitor and a source whose maximum voltage is 5 V, operating at 100.0 Hz.
Pani-rosa [81]

Answer:

X_c=-33.86275385\Omega

|Z|=105.5778675\Omega

I=0.04735841062A

\phi=20.78612878\°

Explanation:

The electrical reactance is defined as:

X_c=-\frac{1}{2\pi fC}

Where:

f=Frequency\\C=Capacitance

So, replacing the data provided by the problem:

X_c=\frac{1}{2\pi *100*(47*10^{-6} )} =-33.86275385\Omega

Now, the impedance can be calculated as:

Z=R+jX_c

Where:

R=Resistance\\X_c= Capacitive\hspace{3}reactance

Replacing the data:

Z=100-j33.86275385

In order to find the magnitude of the impedance we can use the next equation:

|Z|=\sqrt{(R^2)+(X_c^2)}=\sqrt{(100)^2+(-33.86275385)^2} =105.5778675\Omega

We can use Ohm's law to find the current:

V=I*Z\\I=\frac{V}{Z}

Therefore the current is:

I=\frac{5}{100-j33.86275385}=0.04485638113+0.01518960593j

And its magnitude is:

|I|=\sqrt{(0.04485638113)^2+(0.01518960593)^2} =0.04735841062\Omega

Finally the phase angle of the current is given by:

\phi=arctan(\frac{0.01518960593}{0.04485638113})=20.78612878\°

5 0
3 years ago
The Clean Air Act emphasizes that one way to prevent and reduce air pollution is to involve public participation true or false
Anna35 [415]
True hope this helped
7 0
3 years ago
Read 2 more answers
A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is  T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2  m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is  mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

5 0
3 years ago
A constant magnetic flux through a closed loop of wire induces an emf in that loop. True or false?
miskamm [114]

Answer: False

Explanation: In order to explain this problem we have to use the Faraday law, which say

dФm/dt=-ε  it means that the variation of the magnetic field flux with time is equal to the emf ( electromotive force). In our case the magnetic flux is constant then there is not a emf induced in a wire closed loop.

6 0
3 years ago
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