On a time line.

Suzette had prepared the graph below to add to her lab

Charra [1.4K]

Answer:

A title

Explanation:

Because this is middle school.

4 0

2 years ago

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A 4kg book sits on a table and your pet hamster puts his front paws on the book and pushes down with a force of 3N. What is the

Natali [406]

There are three forces acting on the book.
1. Force due to gravity
2. Force exerted downward by the hamster
3. Normal Force in reaction to the downward forces

Since the book is not moving, the net force is zero. The summation of all forces must be zero. Then we could find the normal force which is unknown (denoted as x).

∑F = -(4 kg)(9.81 m/s2) - 3 N + x =0
∑F = -39.24 N - 3N + x =0
x = 42 N

Therefore, the normal force is 42 N.

4 0

3 years ago

Two wires A and B of the same material, having radii in the ratio 1 : 2 and carry currents in the ratio 4 : 1. The ratio of drif

Brums [2.3K]

uhhhh idk cheif...

thats a big oof right there.

ged is always an option

3 0

3 years ago

A light bulb emits light uniformly in all directions. The average emitted power is 150.0 W. At a distance of 5 m from the bulb,

beks73 [17]

Answer:

a) 0.477 W/m²

b) 13.407 N/C

c) 18.96 N/C

Explanation:

P = Power = 150 W

r = Distance = 5 m

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

a) Average intensity

$\bar{S}=\frac{P}{A}\\\Rightarrow \bar{S}=\frac{150}{4\pi r^2}\\\Rightarrow \bar{S}=\frac{150}{4\pi 5^2}\\\Rightarrow \bar{S}=0.477\ W/m^2$

∴ Average intensity is 0.477 W/m²

b) Rms value

$\bar{S}=c\epsilon_0E_{rms}^2\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{S}}{c\epsilon_0}}\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{0.477}}{3\times 10^8\times 8.854\times 10^{-12}}}\\\Rightarrow E_{rms}=13.407\ N/C$

∴ Rms value of the electric field is 13.407 N/C

c) Peak value

$E_0=\sqrt 2E_{rms}\\\Rightarrow E_0=\sqrt 2\times 13.407\\\Rightarrow E_0=18.96\ N/C$

∴ Peak value of the electric field is 18.96 N/C

8 0

2 years ago

Monochromatic light from a distant source is incident on a slit 0.705 mm wide. On a screen 2.13 m away, the distance from the ce

vampirchik [111]

Answer:

492.183 nm

Explanation:

x = Distance from the central maximum to the first minimum = 1.35 mm

l = Distance of screen = 2.13 m

d = Distance of gap = 0.705 mm

m = Order = 1

We have the relation

$tan\theta=\dfrac{x}{l}\\\Rightarrow \theta=tan^{-1}\dfrac{1.35\times 10^{-3}}{2.13}\\\Rightarrow \theta=0.04^{\circ}$

$dsin\theta=m\lambda\\\Rightarrow \lambda=\dfrac{dsin\theta}{m}\\\Rightarrow \lambda=\dfrac{0.705\times 10^{-3}\times sin0.04}{1}\\\Rightarrow \lambda=4.92183\times 10^{-7}=492.183\ nm$

The wavelength of the light is 492.185 nm

5 0

2 years ago

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