(PLEASE HELP) Which of the following describes an element?

An explosion in a rigid pipe shoots three balls out of its ends. A 6 gam ball comes out the right end. A 4 gram ball comes out t

wariber [46]

Answer:

The third ball emerges from the right side.

Explanation:

This is a conservation of Momentum problem

In an explosion or collision, the momentum is always conserved.

Momentum before explosion = Momentum after explosion

Since the rigid pipe was initially at rest,

Momentum before explosion = 0 kgm/s

- Taking the right end as the positive direction for the velocity of the balls

- And calling the speed of the 6 g ball after explosion v

- This means the velocity of the 4 g ball has to be -2v

- Mass of the third ball = m

- Let the velocuty of the third ball be V

Momentum after collision = (6)(v) + (4)(-2v) + (m)(V)

Momentum before explosion = Momentum after explosion

0 = (6)(v) + (4)(-2v) + (m)(V)

6v - 8v + mV = 0

mV - 2v = 0

mV = 2v

V = (2/m) v

Note that since we have established that the sign on m and v at both positive, the sign of the velocity of the third ball is also positive.

Hence, the velocity of the third ball according to our convention is to the right.

Hope this Helps!!!

5 0

4 years ago

Read 2 more answers

A skier reaches the end of a 0.9 km track in 1.5 hours, at what speed does it travel?

Wewaii [24]

Speed = Distance/Time

S = D/T

D = 0.9Km, T = 1.5hours,

S = 0.9/1.5

S = 0.6 m/s

hope it helps!

7 0

4 years ago

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How does friction affect basketball ad what is an example?

kaheart [24]

It affects basketball in a way called traction. The friction between the players' basketball shoes and the hard wooden floor is traction, or friction. Without it, players would slide everywhere and wouldn't be able to play the game.

5 0

4 years ago

Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer sec

Alex

Answer:

ωM = 1.7 rad/s

vM = 3.4 m/s

ωA = 1.7 rad/s

vA = 1.7 m/s

Explanation:

Calculation of the angular speed (ω)

We use the following formula for circular motion:

ω = θ / t Formula (1)

Where:

ω: angular speed (rad/s)

θ:angle that the particle travels (rad)

t: time interval (s)

Equivalence

1 revolution = 2π rad

Data

θ= 1 revolution = 2π rad

t = 3.7 s

We replace data in the formula (1):

ω = (2π rad) / (3.7s )= 1.7 rad/s

Because the angular speed depends only on the number of turns in a time interval then Mary's angular speed (ωM) is equal to Alex's angular speed (ωA).

ωM = ωA = 1.7 rad/s

Calculation of the tangential speed (v)

We use the following formula for circular motion:

v = ω*r Formula (2)

Where:

v : tangential speed (m/s)

ω: angular speed ( rad/s)

r : radius of the circular path (m)

Data

ωM = ωM = ω= 1.7 rad/s

rM = 2m : Mary's circular path radius

rA = 1m : Alex's circular path radius

We apply the formula (2) to calculate v:

vM = ω* rM = (1.7) (2) = 3.4 m/s

vA = ω* rA = (1.7) (1) = 1.7 m/s

7 0

4 years ago

Jack and Jill are maneuvering a 3100 kg boat near a dock. Initially the boat's position is < 2, 0, 3 > m and its speed is

liraira [26]

Answer:

(a) work done by Jack is -2200J

(b) work done by Jill is 0

Explanation:

Given;

Initially boat's position = < 2, 0, 3 > m

Final boat's position = < 6, 0, 1 > m

change in position = < 4, 0, -2 >

Force exerted by Jack = < -440, 0, 220 > N

Force exerted by Jill = < 170, 0, 340 > N

Part (a) How much work does Jack do?

work done = force x distance

Jack's work = < -440, 0, 220 > N x < 4, 0, -2 > m

= [(-440*4) + (0*0) + (220*-2)]

= -1760 J - 440 J

= -2200J

Part (b) How much work does Jill do?

Jill's work = < 170, 0, 340 > N x < 4, 0, -2> m

= [(170*4) + (0*0) + (340*-2)]

= 680 J - 680 J

= 0

7 0

4 years ago