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MAVERICK [17]
3 years ago
7

A planet orbits a start with the path shown below.

Physics
1 answer:
kvv77 [185]3 years ago
4 0

PART a)

As we know that gravitational potential energy is given by the formula

U = -\frac{Gm_1m_2}{r}

here we can see that gravitational potential energy inversely varies with the distance

so here when distance from the sun is minimum then magnitude of gravitational potential energy is maximum while since it is given with negative sign so its overall value is minimum at that position

So gravitational potential energy is minimum at the nearest point and maximum at the farthest point

PART b)

Since we know that sum of kinetic energy and potential energy is constant here

so the points of minimum potential energy is the point where kinetic energy is maximum which means speed is maximum

So here speed is maximum at the nearest point

Part C)

since gravitational potential energy inversely varies with distance so it's graph will be like hyperbolic graph with distance

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klio [65]

Lets se

And

\\ \rm\Rrightarrow T=2\pi\sqrt{\dfrac{m}{k}}

\\ \rm\Rrightarrow \sqrt{k}T=2\pi\sqrt{m}

So

\\ \rm\Rrightarrow k\propto m

If spring constant is doubled mass must be doubled

8 0
2 years ago
Angular velocity in the z direction of a flywheel is w(t)=A + Bt2 The numerical values of the constants are A=2.75 and B=1.50. W
Ivanshal [37]

Answer:

α(0) = 0 rad/s²

α(5) = 15 rad/s²

Explanation:

The angular velocity of the flywheel is given as follows:

w(t) = A + B t²

where, A and B are constants.

Now, for the angular acceleration, we must take derivative of angular velocity with respect to time:

Angular Acceleration = α (t) = dw/dt

α(t) = (d/dt)(A + B t²)

α(t) = 2 B t

where,

B = 1.5

<u>AT t = 0 s</u>

α(0) = 2(1.5)(0)

<u>α(0) = 0 rad/s²</u>

<u></u>

<u>AT t = 5 s</u>

α(5) = 2(1.5)(5)

<u>α(5) = 15 rad/s²</u>

6 0
3 years ago
What is the wavelength of an AM radio wave in a vacuum if its frequency is 810 kHz?
dalvyx [7]
Answer- 370.114 meters
6 0
2 years ago
g Which of the following wavefunctions are: a) square-integrable on the interval provided (1 point); b) valid wavefunction satis
Goshia [24]

Answer:

The answer is given in the attachment

Explanation:

5 0
3 years ago
The main water line enters a house on the first floor. The line has a gauge pressure of 2.10 x 105 Pa. (a) A faucet on the secon
Eva8 [605]

To answer this question is necessary to apply the concepts related to Bernoulli's equation. The Bernoulli-related concept describes the behavior of a liquid moving along a streamline. Pressure can be defined as the proportional ratio between height, density and gravity:

P = h\rho g

Where,

h = Height

\rho = Density

g = Gravity

Our values are

\rho = 1000kg/m^3 \rightarow density of water at normal conditions

h = 7.3m

g = 9.8m/s^2

PART A) Replacing these values to find the total pressure difference we have to

P_1 = h_1 \rho g

P_1 = (7.3)(1000)(9.8)

P_1 = 71540Pa

In this way the pressure change would be subject to

\Delta = P_2-P_1

\Delta = 2.1*10^5Pa- 0.7154*10^5Pa

\Delta = 138460Pa

\Delta = 0.135Mpa

PART B) Considering the pressure gauge of the group as the ideal so that at a height H the water cannot flow even if it is open we have to,

P_2 = H\rho g

2.1*10^5 = H (1000)(9.8)

H = 21.42m

Therefore the high which could a faucet be before no water would flow from it is 21.42m

5 0
3 years ago
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