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anygoal [31]
4 years ago
9

In the graph, during which time period does the particle undergo the greatest displacement?

Physics
2 answers:
marusya05 [52]4 years ago
7 0
The best answer is bc and i'm doing plato too xD lol

stepan [7]4 years ago
4 0
D = Vt

AB
Area under it.
A = (5*4)/2 = 20/2 = 10
A = 5*4 = 20
Total = 10+20 = 30

BC
Area under it.
A = (16-4) * 5 = 12 * 5 = 60

CD
Area under it.
A = ((14-5)*(19-16))/2 = (9*3)/2 = 27/2 = 13.5
A = 5 *(19-16) = 5*3 = 15
Total = 13.5 + 15 = 28.5

BC has the greatest displacement
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find the pressure exerted by a force of 8000 newtons on an area of 25m square..give your answer in Newtown/m2​
baherus [9]

Answer:

Pressure = 320 N/m²

Explanation:

Given the following data;

Force = 8000 Newton

Area = 25 m²

To find the pressure, we would use the following formula;

Pressure = force/area

Substituting into the formula, we have;

Pressure = 8000/25

Pressure = 320 N/m²

5 0
3 years ago
According to the article, in which TWO ways is lava similar to volcanic ash?
allochka39001 [22]
I don't see the article but i personally know that

-they both originate from volcanoes
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7 0
4 years ago
An ideal monatomic gas at a pressure of 2.0×105N/m2 and a temperature of 300 K undergoes a quasi-static isobaric expansion from
liraira [26]

Answer:

a) W= 400 J

b) T_{2}=600 K

c) n=0.16 mol

d) \Delta E_{int}=598 J

e) Q=998 J

Explanation:

a) Let's recall the definition of work.

W=\int^{Vf}_{V0}PdV

Because the system is a quasi-static isobaric expansion, P is constant here, therefore:

W=P\int^{Vf}_{V0}dV=P(V_{f}-V_{0})

W=2.0\cdot 10^{5}(4.0\cdot 10^{-3}-2.0\cdot 10^{-3})= 400 [Nm^{2}]

b) Using the ideal gas equation we have:

\frac{P_{1}V_{1}}{T_{1}}=nR (1)

and \frac{P_{2}V_{2}}{T_{2}}=nR (2)

We can note that n times R is a constant in (1) and (2), so we can equal those equations.

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}} (3)

Let's solve T₂ for (3), let's recall that P₁ = P₂, so they canceled out

T_{2}=T_{1}\cdot V_{2}/V_{1}

T_{2}=300\cdot 4x10^{3}/2x10^{3}=600 K

c) Using the equation of ideal gas we have:

\frac{P_{1}V_{1}}{RT_{1}}=n

n=\frac{P_{1}V_{1}}{RT_{1}}=0.16 mol

d) We can write the internal energy as a function of Cv, and as we know the Cv is 1.5R for a monoatomic gas.

\Delta E_{int}=n1.5R\Delta T

\Delta E_{int}=0.16\cdot 1.5\cdot 8.3 \cdot (600-300)

\Delta E_{int}=598 J

e) Using the first law of thermodynamic, we have:

\Delta E_{int}=Q-W

Finally,

Q=\Delta E_{int}+W=598+400=998 J

Have a nice day!

3 0
3 years ago
In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
SpyIntel [72]

Answer:

A) F_g = 26284.48 N

B) v = 7404.18 m/s

C) E = 19.19 × 10^(10) J

Explanation:

We are given;

Mass of satellite; m = 3500 kg

Mass of the earth; M = 6 x 10²⁴ Kg

Earth circular orbit radius; R = 7.3 x 10⁶ m

A) Formula for the gravitational force is;

F_g = GmM/r²

Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Plugging in the relevant values, we have;

F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²

F_g = 26284.48 N

B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.

Thus;

GmM/r² = mv²/r

Making v th subject, we have;

v = √(GM/r)

Plugging in the relevant values;

v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))

v = 7404.18 m/s

C) From the energy principle, the minimum amount of work is given by;

E = (GmM/r) - ½mv²

Plugging in the relevant values;

E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)

E = 19.19 × 10^(10) J

5 0
3 years ago
Cicadas produce a sound that has a frequency of 123Hz. What is the wavelength of this sound in air at room temperature? (2.79m)
Llana [10]

Answer:

2.72 m

Explanation:

Sound travels at a certain speed and has a frequency and wavelength. The relationship between the speed of sound (V), its frequency (f), and wavelength (λ) is the same as for all waves, it is given by the equation:

V = fλ

The speed of sound in air (V) is 334 m/s

Frequency of sound (f) = 123 Hz

Wavelength = λ

V = fλ

λ = V/f = 334/123

λ = 2.72 m

6 0
4 years ago
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