Ask question

Ask question

All categories

3 years ago

11

An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at

rest. After the collision, the 8.00 kg object moves south at 3.90 m/s. (a) What is the velocity of the 10.0 kg object after the collision

1 answer:

andrew-mc [135]3 years ago

6 0

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

You might be interested in

Just need the answer

marin [14]

Answer:

1. 1, 2, 4 all show some form of refraction as the bending of a light ray when passing from one media to another.

Explanation:

Number 4 is the most accurate as it also shows some light being reflected and the bending of the refracted light ray in the correct direction for going from a medium of low refractive index (air) into a higher refractive index material (crown glass)

8 0

2 years ago

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )

Also at equilibrium the moment about the point where the string holding the bar (where ( $m_1 +m_2$ ) and $m_3$ are hanged ) is attached is zero

So basically

$(m_1 + m_2 ) * 20 = m_3 * 30$

$(0.42 + 0.47) * 20 = 30 * m_3$

Making $m_3$ subject

$m_3 = \frac{(0.42 + 0.47) * 20 }{30 }$

$m_3 = 0.59 kg$

3 0

3 years ago

an object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object

emmainna [20.7K]

Answer:

The acceleration of the object is $-30\ m/s^2$

Explanation:

Given:

Initial velocity of object $v_i$ = 200 feet/second

Final velocity of object $v_f$ = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration $a$ is given by:

$a=\frac{v_f-v_i}{t}$

where $v_f$ represents final velocity, $v_i$ represents initial velocity and $t$ is time of travel.

Plugging in values to evaluate acceleration.

$a=\frac{50-200}{5}$

$a=\frac{-150}{5}$

$a=-30\ m/s^2$

The acceleration of the object is $-30\ m/s^2$ (Answer). The negative sign shows the object is slowing down.

4 0

3 years ago

Determine the slope of end a of the cantilevered beam. E = 200 gpa and i = 65. 0(106) mm4

DENIUS [597]

For E = 200 gpa and i = 65. 0(106) mm4, the slope of end a of the cantilevered beam is mathematically given as

A=0.0048rads

<h3>What is the slope of end a of the cantilevered beam?</h3>

Generally, the equation for the is mathematically given as

$A=\frac{PL^2}{2EI}+\frac{ML}{EI}$

Therefore

A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}

A=0.00288+0.00192=0.0048rads

A=0.0048rads

In conclusion, the slope is

A=0.0048rads

Read more about Graph

brainly.com/question/14375099

5 0

2 years ago

A wire 1 mm in diameter is connected to one end of a wire of the same material 2 mm in diameter of twice the length. A voltage s

miskamm [114]

Answer:

T = 2 T₀

Explanation:

To answer this question let's write the expression for electrical conductivity

σ = n e2 τ / m*

The relationship with resistivity is

ρ = 1 /σ

Whereby the resistance

R = ρ L / A = 1 /σ L / A

We see that there is no explicit relationship between time and resistance, there is only a dependence on the life time (τ) that depends on the properties of the material, not on its diameter or length.

As also the average velocity or electron velocity of electrons is constant, the time to cross 2 mm in length is twice as long as the time to cross a mm in length

T = 2 T₀

8 0

3 years ago