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andrey2020 [161]
2 years ago
3

A sunbather is lying on a floating raft that is stationary. She then gets up and walks to one end of the raft. Consider the sunb

ather and raft as an isolated system.
(a) What is the velocity of the center of mass of this system while she is walking?
(b) Does the raft itself move while she is walking? If so, what is the direction of the raft's velocity relative to that of the sunbather?
Physics
1 answer:
emmainna [20.7K]2 years ago
5 0

Explanation:

1.Since initial velocity of the system of the girl and raft is zero, and the there is no external force, the center of mass of system remains stationary

NO.

2 . The center of mass of the system is at rest. So when the person moves towards the end, she applies an opposite force on the boat in the opposite direction, which pushes it backwards. Hence it moves in the direction opposite to that of her movement.

Yes.

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Using the superposition method, calculate the current through R5 in Figure 8-71
Vladimir79 [104]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

r_1 = \frac{2.2 * 1}{2.2 + 1}

r_1 = 0.6875 k ohm

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}

r_2 = 0.95 k ohm

now the current flowing through the battery is

i = \frac{2}{1 + 0.95} = 1.02 mA

now this will divide into R3 and R2 so current flowing in R3 will be

i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA

now this will again divide in R4 and R5

so current in R5 will be

i_5 = \frac{R_4}{R_4 + R_5}* i_1

i_5 = 0.18 mA

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm

also after its series with R3

r_2 = 1 + 0.6875 = 1.6875 k ohm

now it is in parallel with R5 on other side

r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm

now current through the battery will be given as

i = \frac{3}{1 + 0.95} = 1.53 mA

now it is divide in r2 and R5

so current in R5 is given as

i_5 = \frac{r_2}{r_2 + R_5}*i

i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53

i_5 = 0.67 mA

now the total current in R5 will be given by super position which is

i = 0.67 + 0.18 = 0.85 mA

so there is 0.85 mA current through R5 resistance

5 0
2 years ago
A cubical box, 5.00 cm on each side, is immersed in a fluid. The gauge pressure at the top surface of the box is 594 Pa and the
Zolol [24]

Answer:

The density of the fluid is 1100 kg/m³.

Explanation:

Given that,

Height = 5.00 cm

Pressure at top =594 Pa

Pressure at bottom = 1133 Pa

We need to calculate the change in pressure

Using formula of change in pressure

\Delta P=P_{b}-P_{t}

Where, P_{b} = Pressure at bottom

P_{t} = Pressure at top

put the value into the formula

\Delta P=1133-594

\Delta P=539\ Pa

Using formula of pressure for density

\Delta P = \rho g h

\rho =\dfrac{\Delta P}{gh}

Where, \rho = density

P = pressure

h = height

Put the value in to the formula

\rho =\dfrac{539}{5.00\times10^{-2}\times9.8}

\rho =1100\ kg/m^3

Hence, The density of the fluid is 1100 kg/m³.

4 0
2 years ago
A tennis player standing 12.6m from the net hits the ball at 3.00 degrees above the horizontal. To clear the net, the ball must
mezya [45]
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction. 
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end) 
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
4 0
2 years ago
Read 2 more answers
Think about it: suppose a meteorite collided head-on with mars and becomes buried under mars's surface. what would be the elasti
Ket [755]

Answer:

Perfectly inelastic collision

Explanation:

There are two types of collision.

1. Elastic collision : When the momentum of the system and the kinetic energy of the system is conserved, the collision is said to be elastic. For example, the collision of two atoms or molecules are considered to be elastic collision.

2. Inelastic collision: When the momentum  the system is conserved but the kinetic energy is not conserved, the collision is said to be inelastic. For example, collision of a ball with the mud.

For a perfectly elastic collision, the two bodies stick together after collision.

Here, the meteorite collide with the Mars and buried inside it, the collision is said to be perfectly inelastic. here the kinetic energy of a body lost completely during the collision.

4 0
2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

Explanation:

\omega = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J

Energy used in one acceleration

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J

Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

So the number of complete accelerations is 230

8 0
2 years ago
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