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Colt1911 [192]
3 years ago
5

Spaceship 1 and spaceship 2 have equal masses of 200 kg. Spaceship 1 has a speed of 0 m/s, and spaceship 2 has a speed of 10 m/s

. What is the magnitude of their combined momentum
Physics
1 answer:
m_a_m_a [10]3 years ago
3 0

Answer:

2000 kg m/s

Explanation:

The momentum of an object is a vector quantity whose magnitude is given by

p=mv

where

m is the mass of the object

v is the velocity of the object

and its direction is the same as the velocity.

In this problem, we have:

- Spaceship 1 has

m = 200 kg (mass)

v = 0 m/s (zero velocity)

So its momentum is

p_1 =(200)(0)=0

- Spaceship 2 has

m = 200 kg (mass)

v = 10 m/s (velocity)

So its momentum is

p_2=(200)(10)=2000 kg m/s

Therefore, the combined momentum of the two spaceships is

p=p_1+p_2=0+2000=2000 kg m/s

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(1.6 x 10⁻¹⁹ C) x (300 N/C) = <em>4.8 x 10⁻¹⁷ Newton</em>

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4 years ago
How much do the plates move annually?
Phantasy [73]

Answer:

UP TO four inches but usually a little slower than that

Explanation:

6 0
3 years ago
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A parallel-plate capacitor has plates of area A. The plates are initially separated by a distance d, but this distance can be va
ohaa [14]

Answer:

d' = d /2

Explanation:

Given that

Distance = d

Voltage =V

We know that energy in capacitor given as

U=\dfrac{1}{2}CV^2

C=\dfrac{\varepsilon _oA}{d}

U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2

If energy become double U' = 2 U then d'

U'=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2\times \dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2 d ' = d

d' = d /2

So the distance between plates will be half on initial distance.

7 0
3 years ago
3. Which vehicle has more momentum? *
AysviL [449]

Answer:

the one with v = 25 m/s

Explanation:

Momentum = m * v

 if they both have the same mass (15000 kg) , then the one with the higher v has more momentum...I think A= 25 m/s

5 0
2 years ago
Two small frogs simultaneously leap straight up from a lily pad. frog a leaps with an initial velocity of 0.551 m/s, while frog
Setler [38]
Define
g = 9.8 m/s², acceleration due to gravity, positive downward.

Assume that wind resistance may be neglected.

Frog A:
u = 0.551 m/s, launch velocity, upward.
When the frog lands back on the pad, its vertical position is zero, and its vertical velocity will be 0.551 m/s downward.
If the time of flight is t, then
(0.551 m/s)*(t s) - 0.5*(9.8 m/s²)*(t s)² = 0
0.551t - 4.9t² = 0
t = 0, or t = 0.1124 s
t = 0 corresponds to launch, and t = 0.1124 s corresponds to landing.

Frog B:
Launch velocity is 1.75 m/s
When t = 0.1124 s, the position of the frog is
s = (1.75 m/s)(0.1124 s) - 0.5*(9.8 m/s²)*(0.1124 s)²
   = 0.135 m
The velocity of frog B is
v = (1.75 m/s) - (9.8 m/s²)*(0.1124 s)
   = 0.6485 m/s

Answer:
When frog A lands on the ground,
Frog B is 0.135 m above ground and its velocity is 0.649 m/s upward.

5 0
3 years ago
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