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3 years ago

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At the moment t = 0, a 24.0-v battery is connected to a 5.00-mh coil and a 6.00-ω resistor. (a) immediately thereafter, how does

the potential difference across the resistor compare to the emf across the coil?

1 answer:

agasfer [191]3 years ago

5 0

At the moment the answer is Yeet

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The terminal velocity is not dependent on which one of the following properties? the drag coefficient 1 the force of gravity 2 c

ahrayia [7]

<h2>Answer: the falling time</h2>

Explanation:

When a body or object falls, basically two forces act on it:

1. The force of air friction, also called<em> </em><u><em>"drag force"</em></u> $D$ :

$D={C}_{d}\frac{\rho V^{2} }{2}A$ (1)

Where:

$C_ {d}$ is the drag coefficient

$\rho$ is the density of the fluid (air for example)

$V$ is the velocity

$A$ is the transversal area of the object

So, this force is proportional to the transversal area of the falling element and to the square of the velocity.

2. Its <u>weight </u>due to the gravity force $W$ :

$W=m.g$

(2)

Where:

$m$ is the mass of the object

$g$ is the acceleration due gravity

So, at the moment <u>when the drag force equals the gravity force, the object will have its terminal velocity:</u>

$D=W$ (3)

${C}_{d}\frac{\rho V^{2} }{2}A=m.g$ (4)

$V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}}$ (5) This is the terminal velocity

As we can see, there is no "falling time" in this equation.

Therefore, the terminal velocity is not dependent on the falling time.

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2 years ago

On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total

Stella [2.4K]

Explanation:

Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.

Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.

Weight of the air displaced = density of air × volume

The density of air at 1 atm pressure and 20º C is 1.2 kg/m³

the volume V = 20,000/(1.2×9.8) = 1700 m³

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* 1a Average speed

ruslelena [56]

Explanation:

$\implies v_{av} = \dfrac{total \: displacement}{total \: time}$

$\implies v_{av} = \dfrac{100}{10}$

$\implies v_{av} =10 \: {ms}^{ - 1}$

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2 years ago

Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl

lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

$m g sin\theta - F_{rope} = ma$

velocity is constant, a = 0

$m g sin\theta - F_{rope} =0$

$F_{rope} = m g sin\theta$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0$

$F_{rope}= 102.73\ N$

b) now, when acceleration, a = 0.135 m/s²

$F_{rope}-m g sin\theta = ma$

velocity is constant, a = 0.135 m/s₂

$F_{rope} = m g sin\theta+ma$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135$

$F_{rope}= 112.19\ N$

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A rock is thrown down from the top of a cliff with a velocity of 3.61 m/s (down). The cliff is 28.4 m above the ground. Determin

Katarina [22]

I’m not sure look it up in the internet!

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