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uysha [10]
3 years ago
6

Which of these changes will increase the energy loss of a building?

Physics
1 answer:
fenix001 [56]3 years ago
6 0

1. The velocity decreases, and the kinetic energy decreases.

2. An increase in temperature difference between the inside and outside of the building.

3. The total kinetic energy remains the same.

4. 76,761 J

5. The energy loss must increase.

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An electron enters a region of space containing a uniform 0.0000193-T magnetic field. Its speed is 121 m/s and it enters perpend
Mandarinka [93]

Answer

Given,

Magnetic field, B = 0.0000193 T

speed, v = 121 m/s

mass of electron, m = 9.11 x 10⁻³¹ Kg

charge of electron, q = 1.6 x 10⁻¹⁹ C

radius of the electron path, r = ?

r = \dfrac{mv}{qB}

r = \dfrac{9.31\times 10^{-31}\times 121}{1.6\times 10^{-19}\times 0.0000193}

r = 3.64 x 10⁻⁵ m

We know frequency is inverse of time period

d = v t

2 \pi r = v \times t

t = \dfrac{2 \pi r}{v}

t = \dfrac{2 \pi \times 3.64\times 10^{-5}}{121}

t = 1.889 x 10⁻⁶ s.

now, frequency

f = \dfrac{1}{t}

f = \dfrac{1}{1.889\times 10^{-6}}

f = 529380\ Hz

3 0
3 years ago
you are pushing a friend in a wagon.you push with a force of 40N. your friend and wagon together have a mass of 80kg. ignoring f
hram777 [196]

Answer:

a= 0.5 m/s^2

Explanation:

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3 years ago
Andre is studying snails and how fast they travel. He observes three groups of snails containing four snails each. Here are the
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The material that the snails are moving on affects their speed. On a smooth material, (petri dish) the snails moved the fastest. On a rough material (sandpaper) the snails moved the slowest. On dirt, a compromise between smooth and rough, the snails moved at a medium pace. The material and possibly friction affect this.
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Resistance training should be performed every day for maximum conditioning benefits.
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7 0
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Dos cargas puntuales q1 = −50μC y q2 = +30μC se encuentran
alina1380 [7]

Answer:

Datos:

q1 = -50 μC = 50*10^{-6}

q2 = +30 μC = 30*10^{-6}

F = 10 N

a) x si la <em>F = 10N</em>

Aplicando la Ley de Coulomb:

x = \sqrt\frac{k0 * q1 *q2}{F} = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{10} = 1,162m

b) x si la <em>F = 20 N</em>

x=<em> </em>\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{20}<em> </em>= 0,822m

c)x si la <em>F = 50 N</em>

x = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{50} = 0,520m

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3 years ago
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