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2 years ago

An eclipse of the moon is caused by the

1 answer:

6 0

An eclipse is a phenomenon of an astronomical object being obscured by something. In this case, it is the Moon that is being obscured from sunlight by Earth's shadow. Answer is D.

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In the diagram, q1= -2.60*10^-9 C and

Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

$E_{net}=90.37\: N/c$

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the magnitude of the net electric field will be the E1 + E2.

Let's find first E1 and E2.

The electric field equation is given by:

$|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}$

Where:

k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
q1 is the first charge
d1 is the distance from q1 to P

$|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}$

$|E_{1}|=80.84\: N/C$

And E2 will be:

$|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}$

$|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}$

$|E_{2}|=40.39\: N/C$

Finally, we need to use the Pythagoras theorem to find the magnitude of the net electric field.

$E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}$

$E_{net}=\sqrt{80.84^{2}+40.39^{2}}$

$E_{net}=90.37\: N/c$

I hope it helps you!

7 0

2 years ago

Calculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperat

tigry1 [53]

Answer:

Q = 62383.44 Joules

Explanation:

Given that,

Mass of water, m = 710 gm

Initial temperature of water, $T_i=4^{\circ} C$

Final temperature of water, $T_f=25^{\circ} C$

The specific heat capacity of liquid water is, $c=4.184\ J/g\ ^oC$

Heat transferred is given by :

$Q=mc(T_f-T_i)$

$Q=710\times 4.184\times (25-4)$

Q = 62383.44 Joules

So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.

3 0

3 years ago

A student sits on a pivoted stool while holding a pair of weights. The stool is free to rotate about a vertical axis with neglig

Blababa [14]

Answer:

Please Mark As Brainliest!!

a) 4.99 rad/sec b) 6.24 rad/sec c) 7.03 J

Explanation:

a) If the student completes one turn in 1.26 sec, this is called the period of the movement.

If we take into account that the angle rotated during one turn is 2π rads, by definition of angular velocity, we can get this value as follows:

ω = Δθ / Δt = 2*π rad / 1.26 seg = 4.99 rad/sec.

b) As no external torques are acting on the system, the total angular momentum must be conserved, so we can write the following equation:

Li = Lf ⇒ I₁ * ω₁ = I₂* ω₂

So, we can solve for ω₂, as follows:

ω₂ = (I₁ * ω₁) / I₂ = 6.24 rad/sec

c) Appying the work-energy theorem, we know that the work done by the student, must be equal to the change in the kinetic energy, which in this case is only rotational, so we can write:

W = 1/2 I₂* ω₂² - 1/2 I₁ ω₁²

W =1/2 ((2.25 kg.m² * (6.24)²) (rad/sec)² - (1.8 kg.m²* (4.99)²) (rad/sec)²)

W = 7.03 J

4 0

2 years ago

At what speed does a falling hailstone travel? Does the speed depend on the distance that the hailstone falls?

Anton [14]

Answer:The speed if hailstone dependly largely on its size. A hailstone with a diameter of 0.39 inches,falls wit a speed of 20mph while a hailstone with 3.1 inches in diameter falls at a speed of 110mph.

No speed does not depend on the distance that the hailstone falls.

Explanation: There are other factors that affect the speed of the falling hailstone apart from its size.They are:

1. Friction between the air and the hailstone

2. Wind condition( windy or moist air)

3. The rate at which it melts falling.

7 0

2 years ago

A rock dropped on the moon will increase it's speed from 0 m/s to 8.15 m/s in about 5 seconds what is the acceleration of the ro

Lunna [17]

Using the formula:

a = (Vf - Vi) / t

Our initial velocity is 0 m/s, and our final velocity is 8.15 m/s, with a time period of 5 seconds:

a = (8.15 - 0.0) / 5

a = 1.63 m/s^2

If you know the acceleration due to gravity on the Moon, you can confirm this answer. The recorded gravitational acceleration on the Moon is 1.62 m/s^2.

5 0

3 years ago