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An unstable atomic nucleus of mass 1.82 10-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.18 10-27 kg, moves in the y direction with a speed of 6.00 106 m/s. Another particle, of mass 8.50 10-27 kg, moves in the x direction with a speed of 4.00 106 m/s. (a) Find the velocity of the third particle.

Answers

Answer 1

Answer:

Explanation:

Using Conservation of momentum (total final momentum of system is)

m1•v1f + m2•v2 f + m3•v3 f=0

and it must be zero to equal the original momentum( since the original body is at rest).

Given that

original mass M=1.82×10^-26

First disintegrate mass m1=5.18×10^-27kg

In y direction V1f=6×10^6 I'm/s

Second disintegrate mass m2=8.5×10^-27kg

In x direction V2f=4×10^6 im/s

Then the third disintegrate will be

m3=M-m1-m2

m3=1.82×10^-26-5.18×10^-27-8.5×10^-27

m3=4.52×10^-27

And the velocity is unknown

Now using the formula above

m1•v1f + m2•v2 f + m3•v3 f=0

m3•V3f= - m1•v1f - m2•v2 f

4.52E-27V3f=-5.18E-27×6E6j - 8.5E-27×4E6 i

Divide thorough by 4.52E-27

V3f= - 6.88×10^6j - 7.52×10^6i

V3f= - 7.52×10^6i - 6.88×10^6j

The final velocity of the third mass disintegrate is 6.88×10^6j - 7.52×10^6i m/s

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A basketball has a mass of 575 g. Moving to the right and heading downward at an angle of 31° to the vertical, it hits the floor with a speed of 4 m/s and bounces up with nearly the same speed, again moving to the right at an angle of 31° to the vertical. What was the momentum change ? (Take the axis to be to the right and the axis to be up. Express your answer in vector form.)

Answers

Answer:

Taking the x axis to the right and the y axis to be up, the total change of momentum is $\Delta \vec{ p} = 3.9429 (kg \ m)/(s) \hat{j}$

Explanation:

The momentum $\vec{p}$ is given by:

$\vec{p} = m \ \vec{v}$

where m is the mass and $\vec{v}$ is the velocity. Now, taking the suffix i for the initial condition, and the suffix f for the final condition, the change in momentum will be:

$\Delta \vec{ p} = \vec{p}_f - \vec{p}_i$

$\Delta \vec{ p} = m \ \vec{v}_f - m \ \vec{v}_i$

$\Delta \vec{ p} = m (\ \vec{v}_f - \ \vec{v}_i)$

As we know the mass of the ball, we just need to find the initial and final velocity.

Knowing the magnitude and direction of a vector, we can obtain the Cartesian components with the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ is the magnitude of the vector and θ is the angle measured from the x axis.

Taking the x axis to the right and the y axis to be up, the initial velocity will be:

$\vec{v}_i = 4 (m)/(s) ( \ cos ( - (90 \ °- 31 \°)) , sin( - (90 \ ° - 31\°) ) ) =$

where minus sign appears cause the ball is going downward, and we subtracted the 31 ° as it was measured from the y axis

So, the initial velocity is

$\vec{v}_i = 4 (m)/(s) ( \ cos ( - 59 \°) , sin( - 59 \°)) =$

$\vec{v}_i = ( \ 2.0601 \ (m)/(s) , - 3.4286 (m)/(s)) =$

The final velocity is

$\vec{v}_i = 4 (m)/(s) ( \ cos ( 90 \ °- 31 \°) , sin( 90 \ ° - 31\°)) =$

$\vec{v}_i = 4 (m)/(s) ( \ cos ( 59 \°) , sin( 59 \°)) =$

$\vec{v}_i = ( \ 2.0601 \ (m)/(s) , 3.4286 (m)/(s)) =$

So, the change in momentum will be

$\Delta \vec{ p} = m (\ \vec{v}_f - \ \vec{v}_i)$

$\Delta \vec{ p} = 0.575 \ kg (\ ( \ 2.0601 \ (m)/(s) , 3.4286 (m)/(s) - ( \ 2.0601 \ (m)/(s) , - 3.4286 (m)/(s)))$

$\Delta \vec{ p} = 0.575 \ kg (\ ( \ 2.0601 \ (m)/(s) - \ 2.0601 \ (m)/(s), 3.4286 (m)/(s) + 3.4286 (m)/(s))$

$\Delta \vec{ p} = 0.575 \ kg (\ 0 , 2 * 3.4286 (m)/(s) )$

$\Delta \vec{ p} = 0.575 \ kg * 2 * 3.4286 (m)/(s) \hat{j}$

$\Delta \vec{ p} = 3.9429 (kg \ m)/(s) \hat{j}$

A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the opposite direction with a speed of 22 m/s. If the ball is in contact with the bat for 0.0600 s, determine the magnitude of the average force exerted on the bat.

Answers

Answer:

42.67N

Explanation:

Step one:

Given

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

Required

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

Use the worked example above to help you solve this problem. A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 15.0 m/s. The point of release is h = 46.0 m above the ground.(a) How long does it take for the ball to hit the ground?
(b) Find the ball's speed at impact.
(c) Find the horizontal range of the ball.

Answers

Answer:

Explanation:

A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. She slows steadily, then continues on at 5.8 m/s . What is her acceleration on the rough ice?

Answers

A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. Her acceleration on the rough ice is -3.65 m/s².

v² - u² = 2 a ∆x, where u and v are initial and final velocities, respectively; a is acceleration.

and ∆x is the distance traveled (because the skater moves in only one direction).

Thus, (5.8 m/s)² - (8.8 m/s)² = 2 a (6.0 m)

a = ((5.8 m/s)² - (8.8 m/s)²) / (12 m)

a = -3.65 m/s².

Thus, A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. Her acceleration on the rough ice is -3.65 m/s².

Learn more about Acceleration, refer to the link:

brainly.com/question/2303856

#SPJ3

Recall that

v² - u² = 2 a ∆x

where u and v are initial and final velocities, respectively; a is acceleration; and ∆x is the distance traveled (because the skater moves in only one direction).

So we have

(5.8 m/s)² - (8.8 m/s)² = 2 a (6.0 m)

a = ((5.8 m/s)² - (8.8 m/s)²) / (12 m)

a = -3.65 m/s²

20 examples of scalar quantity

Answers

Answer:

Length

Time

Mass

Temperature

Energy

Direct Current (DC)

Frequency

Volume

Speed

Amount of substance

Luminous Intensity

Density

Concentration

Refractive Index

Work

Pressure

Power

Charge

Electric Potential

Entropy

A boy and a girl are pulling a heavy crate at the same time with 7 units of firce each. What is the net force acts on the ibject? Is the object balanced or unbalanced?

Answers

Answer:

Net force= 14 units

The object is unbalanced

Explanation:

The net force refers to the sum of all forces applied to an object. However, the direction of force applied determine the net force. In this question, a boy and girl is pulling a heavy crate at the same time.

This means that the force is in the same direction, hence, the net force will be:

F(N) = 7 + 7 = 14 unit

However, since the pull is occuring at the same direction. This means that the object has a net force, therefore, will move in a particular direction. This means that the OBJECT IS UNBALANCED

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