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Assoli18 [71]
3 years ago
11

What is the acceleration of a 11 kg box sliding across the floor under the action of a net force 35N

Physics
1 answer:
Fofino [41]3 years ago
8 0

Answer:

<h3>The answer is 3.18 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

where

f is the force

m is the mass

We have

a =  \frac{35}{11}  \\  = 3.181818...

We have the final answer as

<h3>3.18 m/s²</h3>

Hope this helps you

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A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou
elena55 [62]

Answer:

3.2\times 10^{-7}\ m or 0.32 μm.

Explanation:

Given:

The radiations are UV radiation.

The frequency of the radiations absorbed (f) = 9.38\times 10^{14}\ Hz

The wavelength of the radiations absorbed (λ) = ?

We know that, the speed of ultraviolet radiations is same as speed of light.

So, speed of UV radiation (v) = 3\times 10^8\ m/s

Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

v=f\lambda

Now, expressing the above equation in terms of wavelength 'λ', we have:

\lambda=\frac{v}{f}

Now, plug in the given values and solve for 'λ'. This gives,

\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m

Therefore, the wavelength of the radiations absorbed by the ozone is nearly 3.2\times 10^{-7}\ m or 0.32 μm.

7 0
3 years ago
g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal r
Naddika [18.5K]

Answer:

2577 K

Explanation:

Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.

So, T = ⁴√(P/σεA)

Since P = 60 W, we substitute the vales of the variables into T. So,

T = ⁴√(P/σεA)

= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)

= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)

= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)

= ⁴√(0.00441 × 10¹⁶K⁴)

= 0.2577 × 10⁴ K

= 2577 K

6 0
3 years ago
Help I’ll cash app you $5 if you get it right!
Thepotemich [5.8K]

Answer: B and E

Explanation:

8 0
3 years ago
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Two long parallel wires carry currents of 3.35 A and 6.99 A . The magnitude of the force per unit length acting on each wire is
Nady [450]

Answer:

244mm

Explanation:

I₁ = 3.35A

I₂ = 6.99A

μ₀ = 4π*10^-7

force per unit length (F/L) = 6.03*10⁻⁵N/m

B = (μ₀ I₁ I₂ )/ 2πr .........equation i

B = F / L ..........equation ii

equating equation i & ii,

F / L = (μ₀ I₁ I₂ )/ 2πr

Note F/L = B = F

F = (μ₀ I₁ I₂ ) / 2πr

2πr*F = (μ₀ I₁ I₂ )

r = (μ₀ I₁ I₂ ) / 2πF

r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵

r = 1.4713*10⁻⁵ / 6.03*10⁻⁵

r = 0.244m = 244mm

The distance between the wires is 244m

7 0
3 years ago
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7.
Usimov [2.4K]

Answer:

A

Explanation:

7 0
3 years ago
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