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1 year ago

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A net force is applied on a 100 kg rocket which causes the rocket to acceleration at 10 m/s2. The same net force is applied on a

25 kg rocket. What is the acceleration of the second rocket?

1 answer:

ELEN [110]1 year ago

8 0

Answer:

i dont know lo

Explanation:

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A scuba diver has his lungs filled to half capacity (3 liters) when 10 m below the surface. If the diver holds his breath while

rjkz [21]

To solve this problem it is necessary to apply Boyle's law in which it is specified that

$P_1V_1 =P_2 V_2$

Where,

$P_1$ and $V_1$ are the initial pressure and volume values

$P_2$ and $V_2$ are the final pressure volume values

The final pressure here is the atmosphere, then

$P_2 = 101325 \approx 1*10^5Pa$

$h = 10m$

$\rho_w = 1000kg/m^3$

$V_1 = 3.0L$

Pressure at the water is given by,

$P_1 = P_2 -\rho gh$

$P_1 = 1*10^5 +1000*9.8*10 =198000Pa$

Using Boyle equation we have,

$V_2 = \frac{P_1V_1}{P_2}$

$V_2 = \frac{198000*3*10^5}{10^5}$

$V_2 = 5.9L$

Therefore the volume of the lungs at the surface is 5.9L

4 0

1 year ago

The first law of thermodynamics states that ___. when a process converts energy from one form to another, some energy converted

barxatty [35]

The first law of thermodynamics states that energy cannot be created or destroyed, but it can change from one form to another

7 0

1 year ago

Read 2 more answers

Determine the magnitude and sense (direction) of the current in the 500-latex: \omega ω resistor when i = 30 ma.

VARVARA [1.3K]

Complete Question:

Check the circuit in the file attached to this solution

Answer:

Total current = 0.056 A(From left to right)

Explanation:

Let the current in loop 1 be I₁ and the current in loop 2 be I₂

Applying KVL to loop 1

30 - (I₁ - I₂)500 + I₂R + 15 = 0

45 - 500I₁ - 500I₂ + RI₂ = 0

I₁ = 30mA = 0.03 A

45 - 500(0.03) - 500I₂ + RI₂ = 0

30 -500I₂ + RI₂ = 0...............(1)

Applying kvl to loop 2

-RI₂ - 15 + 10 - 400I₁ = 0

-RI₂ = 5 + 400*0.03

RI₂ = -17 ................(2)

Put equation (2) into (1)

30 -500I₂ -17 = 0

-500I₂ = 13

I₂ = -13/500

I₂ = -0.026 A

The total current in the 500 ohms resistor = I₁ - I₂ = 0.03+0.026

Total current = 0.056 A

The current will flow from left to right

5 0

1 year ago

The intensity at a distance of 6.0 m from a source that is radiating equally in all directions is 6.0 × 10-10 w/m2 . what is the

satela [25.4K]

The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:
$I= \frac{P}{A}$ (1)
where
P is the power
A is the area

In our problem, the intensity is $I=6.0 \cdot 10^{-10} W/m^2$ . At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:
$A=4 \pi r^2 = 4 \pi (6.0 m)^2 = 452.2 m^2$

And so if we re-arrange (1) we find the power emitted by the source:
$P=IA = (6.0 \cdot 10^{-10}W/m^2)(452.2 m^2)=2.7 \cdot 10^{-7} W$

3 0

2 years ago

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov

frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v $_{1}$ be the velocity of ball A along positive x axis

v $_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v $_{1}$

∴ v $_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v $_{2}$

2 = u + v $_{2}$ → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 = v $_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v $_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

= 8·85×m J

Total momentum = m×√(2² + 3·7²)

= m× 4·21 kgm/s

3 0

2 years ago