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svetlana [45]
3 years ago
10

Halley is standing outside on a cloudy day. When she hears thunder, she goes back inside so that she doesn't get caught in

Physics
1 answer:
jarptica [38.1K]3 years ago
6 0

Answer:

Inference

Explanation:

An inference involves the application of logic to progress from a premise to a conclusion or logical consequence on the basis of the evidence or known fact. Inference is a process of thought that be divided into a deduction and an induction aspect.

In the given question Halley, by standing outside was able to deduce the sound of thunder she is then able by inductive reasoning from the fact that storms are usually preceded by and accompany lightening, conclude that there is a storm coming.

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A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin
Yanka [14]

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

3 0
3 years ago
Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s
Viefleur [7K]

Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density =  0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

  =  -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level )  = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = \sqrt{CRT}  =  \sqrt{1.4*287*200.5 } = 283.8 m/s

hence it is a subsonic aircraft

4 0
3 years ago
Two parallel-plate capacitors have the same plate area, but the plate gap in capacitor 1 is twice as big as capacitor 2. If capa
-BARSIC- [3]

Answer:

Capacitance of the second capacitor = 2C

Explanation:

\texttt{Capacitance, C}=\frac{\varepsilon_0A}{d}

Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.

Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.

We have    

              \texttt{Capacitance, C}_1=\frac{\varepsilon_0A_1}{d_1}=C

Similarly for capacitor 2

               \texttt{Capacitance, C}_2=\frac{\varepsilon_0A_2}{d_2}=\frac{\varepsilon_0A_1}{\frac{d_1}{2}}=2\times \frac{\varepsilon_0A_1}{d_1}=2C

Capacitance of the second capacitor = 2C

6 0
3 years ago
(URGENT) A ball rolls off a ledge. Its velocity is 7.70m/s in a horizontal direction. It falls on the floor 1.60m below the ledg
elena-s [515]

Answer:

the horizontal distance is 4.355 meters

Explanation:

The computation of the horizontal distance while travelling in the air is shown below:

Data provided in the question is as follows

Velocity = u = 7.70 m/s

H = 1.60 m

R = horizontal direction

Based on the above information

As we know that

R = u × time

where,

Time = \sqrt{\frac{2H}{g} }

So,

= 7.70\times \sqrt{\frac{2\times 1.60}{10} }

= 4.355 meters

hence, the horizontal distance is 4.355 meters

6 0
2 years ago
Hydrogen bonds create unusual properties in water. What are they?
xeze [42]

Answer:

High boiling and melting points: Hydrogen bonds increase the amount of energy required for phase changes to occur, thereby raising the boiling and melting points.

High specific heat: Hydrogen bonds increase the amount of energy required for molecules to increase in speed, thereby raising the specific heat.

Lower density as a solid than as a liquid: Hydrogen bonds increase the volume of the solid by holding molecules apart, thereby decreasing the density

High surface tension: Hydrogen bonds produce strong intermolecular  attractions, which increase surface tension

Explanation:

7 0
2 years ago
Read 2 more answers
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