A student measures out 5.0 g of magnesium and burns it. After the reaction, the student

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

m is the mass of object

$m=\dfrac{W}{g}$

$m=\dfrac{2.45}{9.8}$

m = 0.25 kg

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}$

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0

2 years ago

If I bought 59 kids off the black market in a secret door in the stalls and I sell 21 kids then ask for 81 more then 12 off the

larisa [96]

You will be left with 106 kids

<h3>Meaning of word problem</h3>

A word problem can be defined as a mathematical problem that is written in word or written in a sentence format.

In a word problem, the student is expected to decode the sentence into a mathematical expression before solving

In conclusion, You will be left with 106 kids

Learn more about word problems: brainly.com/question/13818690

#SPJ1

5 0

2 years ago

Assume a reaction takes 7.5 moles of anhydrous calcium chloride and energy transfer occurs, which we record as 21.2 joules of en

OLga [1]

Answer: 2.83 J/mol

Explanation:

Heat of solution, sometimes interchangeably called enthalpy of solution, is said to be the energy released or absorbed when the solute dissolves in the solvent. A solute is that which can dissolve in a solvent, to form a solution

Given

No of moles of CaCl = 7.5 mol

Total energy used = 21.2 J

Heat of solution = q/n where

q = total energy

n = number of moles

Heat of solution = 21.2 / 7.5

Heat of solution = 2.83 J/mol

8 0

3 years ago

Read 2 more answers

Un trineo de 20 kg descansa en la cima de una pendiente de 80 m de longitud y 30° de inclinación. Si µ = 0.2, ¿cuál es la veloci

Mariulka [41]

Answer:

v= 26.70 m/seg

Explanation: Ver anexo ( diagrama de cuerpo libre)

De acuerdo a la segunda ley de Newton

∑ F = m*a

∑ Fx = m* a(x) ∑ Fy = m* a(y)

También sabemos que el coeficiente de roce dinámico es:

μ = 0.2 = F(r)/N siendo N la fuerza normal.

Si descomponemos la fuerza P = mg = 20Kg* 9.8m/seg²

P = 196 [N] en sus componentes sobre los ejes x y y tenemos

Py = P* cos30 = 196* √3/2 = 98*√3

Px = P* sen30 = 196*1/2 = 98

La sumatoria sobre el eje y es :

∑ F(y) = m*a Py - N = 0 98*√3 = N ( no hay movimiento en la dirección y)

∑ F(x) = m*a P(x) - Fr = m*a

Fr = μ *N = 0.2* 98*√3

Fr = 19.6*√3 [N]

98 - 19.6*√3 = m*a

98 - 33.52 = m*a

a = (98 - 33.52 ) / 20

a = 3.22 m/seg²

Para calcular la velocidad del trineo al pié del plano, sabemos que al pié del plano el trineo ha recorrido 80 m, y que de cinemática

v² = v₀² + 2*a*d ( se pueden chequear unidades para ver la consistencia de la ecuación v y v₀ vienen dados en m/seg entonces v² y v₀² vienen en m²/seg², el producto de a (m/seg²) por la distancia d (m) resulta en m²/seg² entonces es consistente la relación

v² = 0 + 2*3.22*80 ( la velocidad inicial es cero)

v² = 515.2 m²/seg²

v = √515.2 m/seg

v= 26.70 m/seg

6 0

2 years ago