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Novosadov [1.4K]
4 years ago
9

Which point lies on the line with point slope equation : y+5=2(x+8)

Physics
2 answers:
erik [133]4 years ago
7 0
A. (-8, -5)
the answer is A
insens350 [35]4 years ago
6 0

Answer:

A. (-8,-5)

Explanation:

If you fill in A you get

-5 + 5 = 2(-8+ 8)

0 = 0

So this point is on the line

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I would disagree with that statement.

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Benjamin places a drop of purple food coloring in a cup of water. What can Benjamin expect to see? Explain the process that will
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3 years ago
At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m
eimsori [14]

Answer:

1) The time it takes the diver to move off the end of the diving board to the pool surface, t_w, is approximately 1.392 seconds

2) The horizontal distance from the edge of the pool to where the diver enters the water, d_w, is approximately 5.76 meters

Explanation:

1) The given parameters are;

The height of the diving board above the pool's surface, h = 9.5 m

The length by which the diving board over hangs the pool L = 2 m

The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s

Taking t_w = The time it takes the diver to move off the end of the diving board to the pool surface

Therefore, we have from the equation of free fall;

h = 1/2 × g × t_w²

Where;

g = The acceleration due to gravity = 9.81 m/s²

Substituting the values, gives;

9.5 = 1/2 × 9.81 × t_w²

t_w = √(9.5/(1/2 × 9.81)) ≈ 1.392 s

The time it takes the diver to move off the end of the diving board to the pool surface = t_w ≈ 1.392 s

2) The horizontal distance, d_w, in meters from the edge of the pool to where the diver enters the water is given as follows;

d_w = L + v₀ × t_w = 2 + 2.7× 1.392 ≈ 5.76 m

∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.

7 0
3 years ago
What is umax,c, the value of the maximum energy stored in the capacitor during one cycle?
TiliK225 [7]

1) 0.266 H

2) 0.040 J

3) 0.308\Omega

Explanation:

The diagram of the circuit is missing: find it in attachment.

1) What is L, the value of the inductance of the circuit?

For an RLC circuit, the resonant angular frequency is given by:

\omega=\frac{1}{\sqrt{LC}}

where

L is the inductance of the circuit

C is the capacitance of the circuit

In this problem, we have:

\omega=164 rad/s is the angular frequency of the generator, at which the circuit is in resonance

C=140 \mu F = 140\cdot 10^{-6}F is the capacitance

Therefore, solving for L, we find the inductance of the circuit:

L=\frac{1}{\omega^2 C}=\frac{1}{(164)^2(140\cdot 10^{-6})}=0.266 H

2) What is umax,c, the value of the maximum energy stored in the capacitor during one cycle

The maximum energy stored in a capacitor is given by the equation

U=\frac{1}{2}CV^2

where

C is the capacitance of the capacitor

V is the maximum potential difference across the capacitor

Here we have:

C=140 \mu F = 140\cdot 10^{-6}F is the capacitance

V=24 V is the maximum voltage across the capacitor, which is the emf of the generator

Substituting, we find:

U=\frac{1}{2}(140\cdot 10^{-6})(24)^2=0.040 J

3) What is R, the value of the resistance of the circuit?

Here we want to find the resistance of the circuit.

The resistance of the circuit can be found by using Ohm's Law:

V=RI

where

V is the maximum voltage

R is the resistance

I is the maximum current

Here we have:

V = 24 V is the maximum voltage provided by the generator

I = 0.78 A is the maximum current in the circuit

Solving for R, we find the resistance:

R=\frac{V}{I}=\frac{0.24}{0.78}=0.308\Omega

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