Which list of observations is the best evidence of only a chemical change occurring?

A 70.9-kg boy and a 43.2-kg girl, both wearing skates face each other at rest on a skating rink. The boy pushes the girl, sendin

Lelechka [254]

Answer:

Explanation:

Given

mass of boy $m_b=70.9\ kg$

mass of girl $m_g=43.2\ kg$

speed of girl after push $v_g=4.64\ m/s$

Suppose speed of boy after push is $v_b$

initially momentum of system is zero so final momentum is also zero because momentum is conserved

$P_i=P=f$

$0=m_b\cdot v_b+m_g\cdot v_g$

$v_b=-\frac{m_g}{m_b}\times v_g$

$v_b=-\frac{43.2}{70.9}\times 4.64$

$v_b=-2.82\ m/s$

i.e. velocity of boy is 2.82 m/s towards west

8 0

2 years ago

A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling

Setler [38]

m1=1500kg
m_2=3000kg
v_1=5m/s
v_2=7m/s

Using law of conservation of momentum

$\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3$

$\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3$

$\\ \sf\Rrightarrow 7500-21000=4500v_3$

$\\ \sf\Rrightarrow -13500=4500v_3$

$\\ \sf\Rrightarrow v_3=-3m/s$

6 0

2 years ago

Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second

shepuryov [24]

Answer:

distance between both oasis ( 1 and 2) is 27.83 Km

Explanation:

let d is the distance between oasis1 and oasis 2

from figure

OC = 25cos 30

OE = 25sin30

OE = CD

Therefore BC = 30-25sin30

distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem

in $\Delta BCO$

$OB^2 = BC^2 + OC^2$

PUTTING ALL VALUE IN ABOVE EQUATION

$d^2 = 930-25sin30)^2 + (25cos30)^2$

$d^2 = 775$

d = 27.83 Km

distance between both oasis ( 1 and 2) is 27.83 Km

3 0

2 years ago

5 description of a motion

yuradex [85]

Answer:

<h2>Displacement</h2><h2>Distance</h2><h2>Velocity</h2><h2> Acceleration</h2><h2>Speed</h2><h2> Time</h2>

Explanation:

HOPE IT HELPS

8 0

3 years ago

A ball is thrown straight up. If the launch velocity is 15 m/s, at what SPEED will the ball return to the thrower’s hand?

IgorC [24]

Answer:

Vf = 15 m/s

Explanation:

First we consider the upward motion of ball to find the height reached by the ball. Using 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = -9.8 m/s² (negative sign for upward motion)

h = height =?

Vf = Final Velocity = 0 m/s (Since, ball momentarily stops at highest point)

Vi = Initial Velocity = 15 m/s

Therefore,

2(-9.8 m/s²)h = (0 m/s)² - (15 m/s)²

h = (-225 m²/s²)/(-19.6 m/s²)

h = 11.47 m

Now, we consider downward motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = 9.8 m/s²

h = height = 11.47 m

Vf = Final Velocity = ?

Vi = Initial Velocity = 0 m/s

Therefore,

2(9.8 m/s²)(11.47 m) = Vf² - (0 m/s)²

Vf = √(224.812 m²/s²)

8 0

2 years ago