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3 years ago

5

Who was the first scientist to propose that an object could emit only certain amounts of energy?

1 answer:

Rzqust [24]3 years ago

3 0

It was Niels Bohr who proposed it

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A 133-foot antenna is on top of a tall building. From a point on the ground, the angle of elevation to the top of the antenna is

77julia77 [94]

Answer:

Explanation:

<ADC = 30.5 Degree

<BDC = 21.5 Degree

AB = 133 ft

$in \Delta BCD$

$tan 21.5= \frac{BC}{CD}$

$CD = \frac{BC}{tan21.5}$

in $\Delta ACD$

$tan 30.5 =\frac{AC}{CD}$

$CD = \frac{AB+BC}{tan30.5}$

$\frac{BC}{tan21.5} = \frac{AB}{tan30.5} +\frac{BC}{tan30.5}$

$\frac{BC}{tan21.5} - \frac{BC}{tan30.5} = 225.78$

2.5BC - 1.69BC = 225.78

BC = 281.40 ft

6 0

3 years ago

If the amount of gas in a sealed container increases, what happens to the pressure in the container

Mariulka [41]

If you don't let it cool, then the pressure increases.

4 0

3 years ago

Read 2 more answers

A commonly used unit of mass in the English system is the pound-mass, abbreviated lbm, where 1 lbm = 0.454 kg. What is the densi

Gemiola [76]

Answer: The density of glycerine will be $0.028lbm/foot^3$

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{Volume}$

Given:

Density of glycerine= $1.26\times 10^3kg/m^3$

1 lbm = 0.454 kg

1 kg = $\frac{1}{0.454}=2.2lbm$

Thus $0.454kg=\frac{2.2}{1}\times 0.454=0.99$

Also $1m^3=35.3147foot^3$

Putting in the values we get:

$Density=\frac{0.99lbm}{35.3147}=0.028lbm/foot^3$

Thus density of glycerine will be $0.028lbm/foot^3$

3 0

3 years ago

You are climbing in the High Sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of t

s344n2d4d5 [400]

Answer:

Height with sound ignored = Gravity x Time taken = 9.8 x 8.60 = 84.28 meters

Time taken by the sound = 84.28/330 = 0.255 seconds

Height with sound involved = (84.28 x 0.255) + 84.28 = 105.80 meters

a. Underestimated

5 0

3 years ago

In a 5000 m race, the athletes run 12 1/2 laps; each lap is 400 m.Kara runs the race at a constant pace and finishes in 17.9 min

Ksju [112]

Answer:

No. of laps of Hannah are 7 (approx).

Solution:

According to the question:

The total distance to be covered, D = 5000 m

The distance for each lap, x = 400 m

Time taken by Kara, $t_{K} = 17.9 min = 17.9\times 60 = 1074 s$

Time taken by Hannah, $t_{H} = 15.3 min = 15.3\times 60 = 918 s$

Now, the speed of Kara and Hannah can be calculated respectively as:

$v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s$

$v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s$

Time taken in each lap is given by:

$(v_{H} - v_{K})t = x$

$(5.45 - 4.65)\times t = 400$

$t = \frac{400}{0.8}$

t = 500 s

So, Distance covered by Hannah in 't' sec is given by:

$d_{H} = v_{H}\times t$

$d_{H} = 5.45\times 500 = 2725 m$

No. of laps taken by Hannah when she passes Kara:

$n_{H} = \frac{d_{H}}{x}$

$n_{H} = \frac{2725}{400} = 6.8$ ≈ 7 laps

3 0

3 years ago