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3 years ago

14

You can feel the heat waves from a small electric heater on your hands and face even though the heater is on the other side of t

he room. Which is this is an example of?

1 answer:

NeX [460]3 years ago

8 0

That's an example of heat transfer by radiation.

You might be interested in

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 1.00 mm away. part a

mihalych1998 [28]

<span>We can use Coulomb's law to find the force F acting on the proton that is released. F = k x Q1 x Q2 / r^2 k = 9 x 10^9 Q1 is the charge on one proton which is 1.6 x 10^{-19} C Q2 is the same charge on the other proton r is the distance between the protons F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2 F = 2.304 x 10^{-22} N We can use the force to find the acceleration. F = ma a = F / m a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg) a = 1.38 x 10^5 m/s^2 The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>

8 0

3 years ago

Place the left charge at the 2 cm position and the right one at the 4 cm position. Vary the left and right charge to the values

Vlad1618 [11]

Answer:

Explanation:

Force between two charges can be expressed as follows

F = k q₁ q₂ / d²

q₁ and q₂ are two charges , d is distance between them , k is a constant whose value is 9 x 10⁹

distance between charges is fixed which is 4 -2 = 2 cm = 2 x 10⁻² m

force between 1μC and 4μC

= 9 x 10⁹ x 1 x 4 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 4μC and 1μC

= 9 x 10⁹ x 4 x 1 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 2μC and 2μC

= 9 x 10⁹ x 2 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 1μC and 2μC

= 9 x 10⁹ x 1 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 4.5 x 10 = 45 N

force between 1μC and 8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 18 x 10 = 180 N

force between 2μC and 8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 36 x 10 = 360 N

Left Charge Right Charge Resulting force(N)

1μC 4μC 90 N

4μC 1μC 90 N

2μC 2μC 90 N

1μC 2μC 45 N

1μC 8μC 180 N

2μC 8μC 360 N

5 0

3 years ago

If an object went from 0 m/s to 6 m/s in 1.7 seconds after a 10 N force was applied to it; what is the object's mass? No links p

Mashcka [7]

The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was

<em>a</em> = ∆<em>v</em> / ∆<em>t</em> = (6 m/s - 0) / (1.7 s) ≈ 3.52941 m/s²

By Newton's second law, the magnitude of the force <em>F</em> is proportional to the acceleration <em>a</em> according to

<em>F</em> = <em>m a</em>

where <em>m</em> is the object's mass. Solving for <em>m</em> gives

<em>m</em> = <em>F</em> / <em>a</em> = (10 N) / (3.52941 m/s²) ≈ 2.8 kg

4 0

3 years ago

When you jump straight up as high as you can, what is the order of magnitude of the maximum recoil speed that you give to the Ea

Klio2033 [76]

Answer:

5.66 × 10⁻²³ m/s

Explanation:

If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.

Since initial momentum = final momentum,

mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.

My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?

So mv₁ + M × 0 = m × 0 + MV₂

mv₁ = MV₂

V₂ = mv₁/M = 54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s

5 0

3 years ago

How many atoms of each element are in the reactants?

Gnom [1K]

6.022*10^23 atoms/mole of reactant
(this is chemistry not physics)

6 0

3 years ago