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likoan [24]
2 years ago
5

A student is walking back up the hill from the field to the school building. His

Physics
1 answer:
myrzilka [38]2 years ago
4 0

Answer:

Time cause it's so long ago I was just

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Which of the following is true about wedges?
timofeeve [1]

Answer:

wedges are a type of inclined plane.

Explanation:

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3 years ago
Hey, I need help can someone help me out, please?
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7) f =ma = 15*15 = 225N

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2. What is a mirror? How does it work?​
Murljashka [212]

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5 0
3 years ago
You charge an initially uncharged 89.9-mf capacitor through a 30.5-ω resistor by means of a 9.00-v battery having negligible int
blsea [12.9K]
<span>1) The differential equation that models the RC circuit is :

(d/dt)V_capacitor </span>+ (V_capacitor/RC)​ = (V_source/<span>RC)​​</span>

<span>Where the time constant of the circuit is defined by the product of R*C

Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s


2)
C<span>harge of the capacitor 1.57 time constants

1.57*(2.742) = 4.3048 s

The solution of the differential equation is

</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T)  +  </span>V_capac(∞)

Since the capacitor is initially uncharged V_capac(0) = 0

And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery  V_capac(∞) = 9V 

This means,

V_capac (t) = (-9V)e ^(-t /T)  +  9V

The charge in a capacitor is defined as Q = C*V

Where C is the capacitance and V is the Voltage across

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V = -1.87V +9V

V_capac (4.3048 s) = 7.1275 V

Q (4.3048 s)  = 89.9mF*(7.1275V) = 0.6407 C

3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V

Q (∞)  = 89.9mF*(9V) = 0.8091 C
7 0
3 years ago
g a mass of 1.3 kg is pushed horizontally against a massless spring with a spring constant of 58 n/m until the spring compresses
ExtremeBDS [4]

Answer: 1.102\ J

Explanation:

Given

Mass m=1.3\ kg

Spring constant k=58\ N/m

Compression in the spring x=19.5\ cm\ or\ 0.195\ m

When the mass leaves the spring, the elastic potential energy of spring is being converted into kinetic energy of mass i.e.

\Rightarrow \dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}\cdot 58\cdot (0.195)^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}mv^2=1.102\ J

The kinetic energy of the mass is 1.102 J.

5 0
3 years ago
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