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faust18 [17]
4 years ago
11

Lucy is cruising through space in her new spaceship. As she coasts along, a tiny spacebug drifts into her path and bounces off t

he window. Consider several statements concerning this scenario. Evaluate each statement according to the law of momentum conservation and match it to the appropriate category.
Physics
1 answer:
Elden [556K]4 years ago
4 0

Your question is not complete, please let me assume this to be your complete question:

Lucy is cruising through space in her new spaceship. As she coasts along, a spacebug drifts into her path and bounces off the window. Consider several statements concerning this scenario. Evaluate each statement according to the law of momentum conservation and match it to the appropriate category.

Below are several statements concerning this scenario. Evaluate each statement and decide if it is true, false, or undetermined by the principle of momentum conservation.

i) The total change in momentum for this interaction is zero.

ii) The change in the space bug's momentum is greater than the change in the spaceship's momentum.

iii) If the space bug had stuck to the spaceship instead of bouncing off, momentum would not have been conserved for this interaction.

Answer:

I is true

II is false

III is undetermined

Explanation:

STATEMENT I : This statement is TRUE because because they was no change in velocity and mass of the spaceship and spacebug, after the collision, as the bug bounced outside the window and the spaceship retains it's velocity. Therefore the total momentum in the system before the collision, is equal to the total momentum of the system after the collision.

Where;

momentum = Mass × velocity

M1V1 = M2V2

Therefore;

M2V2 - M1V1 = 0

STATEMENT II : This statement is FALSE, because the change in the momentum of the spaceship and bug are equal, as the spaceship and bug remains in constant motion after collision. The collision did not have any effect in their velocity nor mass.

Ps = Pb

Ps = Momentum of spaceship

Pb = Momentum of spacebug

STATEMENT III : The statement is UNDETERMINE, because it will depend on the momentum we are considering. If the spaceship is still in constant motion, that means moment of the spaceship is conserved, while that of the spacebug is not conserved.

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After a surface interval my pressure group is D. I want to do a repetitive dive to 38 feet for 32 minutes. What will be my new p
ExtremeBDS [4]

Answer:

New pressure group is N

Explanation:

The Repetitive Dive table has to be checked, the residual nitrogen for the maximum depth of 38 feet as a "D" diver, then add that to the actual dive time to get the total nitrogen time. The total nitrogen time and the maximum depth of 38 feet would put you on a pressure group N.

5 0
3 years ago
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A 2.0 x 10^3-kilogram car travels at a constant speed of 12 meters per second around a circular curve of radius 30. meters. What
Vikentia [17]

The magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

<h3>Circular motion</h3>

From the question, we are to determine the magnitude of the centripetal acceleration.

Centripetal acceleration can be calculated by using the formula

a_{c} =\frac{v^{2} }{r}

Where a_{c} is the centripetal acceleration

v is the velocity

and r is the radius

From the given information

v = 12 \ m/s

and r = 30 \ m

Therefore,

a_{c} =\frac{12^{2} }{30}

a_{c} =\frac{144 }{30}

a_{c} = 4.8\ m/s^{2}

Hence, the magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

Learn more on circular motion here: brainly.com/question/20905151

4 0
2 years ago
A car of mass 500kg travelling at 60m/s has it speed reduced to 40m/s by a constant breaking force over a distance of 200m. find
uranmaximum [27]

Answer:

Ek1 = 900000 [J]

Ek1 = 400000 [J]

Explanation:

In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:

E_{k1}=\frac{1}{2}*m*v1^{2}

where:

m = mass = 500 [kg]

v1 = 60 [m/s]

So we have:

Ek1 = 0.5*500*(60^2)

Ek1 = 900000 [J]

and:

Ek2 = 0.5*500*(40^2)

Ek2 = 400000 [J]

6 0
3 years ago
A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next
zalisa [80]

Complete Question

A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?

Answer:

V=1.4*10^5m/s

Explanation:

From the question we are told that:

Electric field B=1.5*10N/C

Distance d=2 x 10^{-3}

At negative plate

Generally the equation for Velocity is mathematically given by

V^2=2as

Therefore

V^2=\frac{2*e_0E*d}{m}

V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}

V=\sqrt{19.2*10^9}

V=1.4*10^5m/s

5 0
3 years ago
In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of
aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0
3 years ago
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