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3 years ago

- Complete the chart. What does "charge” mean in this passage?

Physics

1 answer:

Maslowich3 years ago

3 0

Answer:

demand (an amount) as a price from someone for a service rendered or goods supplied.

You might be interested in

________ Is a push or pull

victus00 [196]

Answer:

lever

Explanation:

because you can push and pull a lever

5 0

2 years ago

"A 3 kg crate slides down a ramp. The ramp is 1 m in length and inclined at an angle of 30 degrees. The crate starts from rest a

OlgaM077 [116]

Answer:

v = 2.57 m /sec

Explanation: See Annex Free Body Diagram

From free body diagram and Newton´s second law we have

There is not movements in the y axis direction

cos 30° = √3/2 sin 30° = 1/2

We have P = mg = 3 Kg * 9.8 m/sec²

P = 29.4 Kg*m/ sec² P = 29.4 [N]

Py = P * cos 30° Py = 29.4 [N] * √3/2 ⇒ Py = 25.43 [N]

Px = P * sin 30° Px = 29.4 [N] * 1/2 ⇒ Px = 14.7 [N]

∑ F = m* a ⇒ ∑ Fy = 0 ∑ Fx = m *a

∑ Fy = Fn - Py = 0 Py = P*cos30° Py = 25.43 [N]

Fn = 25.43 [N]

Fr = μk * Fn ⇒ Fr = 0.19 * 25.43 ⇒ Fr = 4.83 [N]

Now

∑ Fx = m *a mg sin30° - Fr = m*a ⇒ Px - Fr = m*a

14.7 [N] - 4.83 [N] = 3 [kg] * a ⇒ 9.87 /3 = a [m /sec²]

a = 3.29 [m/sec²]

From uniformly accelerated movement

distance = x₀ + V₀*t ± at²/2 but x₀ and V₀ = 0

Then

d = ( 1/2 )*a*t² ⇒ 1 [m] * 2 = 3.29 [m/sec²] * t²

t = 0.78 sec

And finally

v = a*t ⇒ v = 3.29 *(.78) ⇒ v = 2.57 m /sec

5 0

3 years ago

A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The

lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s

Explanation:

1) We can find the speed of the object by conservation of energy:

$E_{i} = E_{f}$

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s$

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s$

3) When the object is at the equilibrium position, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s$

I hope it helps you!

8 0

3 years ago

Unpolarized light whose intensity is 1.46 W/m2 is incident on the polarizer in the drawing. (a) What is the intensity of the lig

Serhud [2]

Answer:

0.73 W/m²

0.2522 W/m²

Explanation:

$I_0$ = Unpolarized light = 1.46 W/m²

$\theta$ = Analyzer angle = 54°

Light through first filter

$I_1=\frac{I_0}{2}\\\Rightarrow I_1=\frac{1.46}{2}\\\Rightarrow I_1=0.73\ W/m^2$

The intensity of the light leaving the polarizer is 0.73 W/m²

After passing through analyzer

$I=I_1cos^2\theta\\\Rightarrow I=0.73\times cos^2(54)\\\Rightarrow I=0.2522\ W/m^2$

The intensity of the light that reaches the photocell is 0.2522 W/m²

6 0

3 years ago

What is the medium that sounds waves travel through when a person speaks

meriva

The medium that sound travels through when a person speaks anything that has loose floating atoms for them to bounce off of so it is most commonly air.

8 0

3 years ago