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lakkis [162]
3 years ago
14

Galileo performed an experiment to measure the speed of light by timing how long it took light to travel from a lamp he was hold

ing to an assistant about 1.5 km away and back again. Why was Galileo unable to conclude that light had a finite speed?
Physics
1 answer:
krok68 [10]3 years ago
6 0

Answer:

The time it takes light to cover 1.5 km was too short to be measured by Galileo's instruments.

Explanation:

The speed of light is c=3*10^8m/s, which means the time it takes to cover a distance of 1.5 km (or 1,500m) will be

t= \dfrac{1500m}{3*10^8m/s}

t= 0.000005s

which is \dfrac{1}{200000} of a second! This time delay could in no way be measured by Galileo considering the fact that he was using his heartbeat to measure time!

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Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec
Dmitrij [34]

Answer:

(a) F_{net} = 4.19 x 10^{-5} N, and its direction is towards m_{2}.

(b) It must be placed inside a hollow shell.

Explanation:

Let, m_{1} = 165 kg, m_{2} = 465 kg, m_{3} = 60 kg, and the distance between m_{1} and m_{2} is 0.340 m.

(a) Since m_{3} is placed midway between m_{1} and m_{2}, then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = \frac{Gm_{1}m_{2}  }{r^{2} }

Where all variables have their usual meaning.

Then,

a. F_{net} = F_{23} - F_{13}

F_{13} = \frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }

     = 2.25 x 10^{-5} N

F_{23} = \frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }

     = 6.44 x 10^{-5} N

∴ F_{net} =  = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

The net force exerted by the two masses on the 60 kg object is 4.19 x 10^{-5}  N.

(ii) /F_{net}/ = /F_{23}/ - /F_{13}/

              = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

(iii) The direction of the net force is to the right i.e towards m_{2}.

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0
3 years ago
A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el
GuDViN [60]

Answer:

7.2\cdot 10^{-19} J

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

\Delta U = q \Delta V

where

q is the magnitude of the charge of the particle

\Delta V is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J

- the potential difference is

\Delta V=4.50 V

So, the change in electrical potential energy is

\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J

7 0
3 years ago
If 2.40 g of KNO3 reacts with sufficient sulfur (S8) and carbon (C), how much P-V work will the gases do against an external pre
creativ13 [48]

Answer:

-112.876J

Explanation:

In order to solve this question, we would need to incorporate Stoichiometry, which involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Here's a balanced equation for the reaction:

16KNO_3(s) + 24C(s) + S_8(s)    \to 24CO_2(g) + 8N_2(g) + 8K_2S(s)

Let us define P - V work as;

w_{pv} = - P_{external}  \triangle Volume

where  \triangle (Volume) = (V_{final} - V_{initial})

External pressure is given as  1.00atm , therefore the work solely depends on the change in volume and since the reactants are solids, none of the reactants contribute to the volume. Hence,  V_i = 0.

To find the volume of the products, we need to first find the amount of moles of the product made from  2.40_gKNO_3, using the molar mass of  KNO_3  which is 101.1032 g/mol  

2.40_gKNO_3 . {\frac{1molKNO_3}{101.1032_g}} = 0.0237molKNO_3

Now let us convert moles of  KNO_3  into moles of CO_2 and N_2  using the stoichiometric ratios from our balanced equation of the reaction.

0.0237molKNO_3 . {\frac{24molCO_2}{16molKNO_3}} = 0.0356molCO_2

0.0237molKNO_3 . {\frac{8molN_2}{16molKNO_3}} = 0.01185molN_2

K_2S is not factored into the volume calculation because it is a solid.

Now let us also convert the moles of  CO_2  and  N_2 into grams using their respective molar masses.

0.0356molCO_2 . {\frac{44.01_g}{1molCO_2}} = 1.567_gCO_2

0.01185molN_2 . {\frac{28.014_g}{1molN_2}} = 0.332_gN_2

We will now proceed to convert grams into volume using the density values provided.

1.567_gCO_2 . {\frac{1L}{1.830_g}} = 0.856LCO_2

0.332_gN_2 . {\frac{1L}{1.165_g}} = 0.285LN_2

Summing up the two volumes, we get the final volume

0.856L + 0.258L = 1.114L = V_f

Plugging everything into the w_{pv} equation, we get:

w_{pv} = -1atm(1.114L - 0L) = -1.114L.atm

Finally, let us convert L.atm into joules using the conversion rate of;

1L.atm = 101.325J\\-1.114L.atm. {\frac{101.325J}{1L.atm}} = -112.876J

7 0
3 years ago
How far does a car go in 30 seconds at a speed of 29 m/s? (The equation for<br> distance is d = st.)
Harrizon [31]

d = speed x time

distance = 29 x 30

distance = 870m

please mark as BRAINLIEST

8 0
2 years ago
ariel has 2.6 grams of zinc . what volume of the material does she have if zinc had density of 7.13 g/cm
vaieri [72.5K]

Answer:

Ap. 4.6

Explanation:

5 0
3 years ago
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