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neonofarm [45]
4 years ago
7

Models can be used to obtain information about scientific processes and objects that may be difficult to study in real life. Whi

ch of the following scientific processes or objects would best be studied by using a model? A. the effects of an antibiotic on a disease-causing bacterium B. the formation of a solar system C. the chemical composition of a protein D. the growth pattern of a pea plant
Physics
1 answer:
Snezhnost [94]4 years ago
5 0
The answer is B.
________
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Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus
tino4ka555 [31]

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

3 0
3 years ago
What % of an object’s mass is above the water line if the object’s density is 0.82g/ml?
Sladkaya [172]

To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:

\rho_w =1 g/mL \rightarrow \text{Water Density}

\rho_o = 0.82g/mL \rightarrow \text{Object density}

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore

\sum F= 0

F_w-F_o = 0

F_w = F_o

m_w g = m_og

\rho_w V_w g = \rho_o V_o g

The value of gravity is canceled because it is a constant

\frac{V_w}{V_o} = \frac{\rho_o}{\rho_w}

\frac{V_w}{V_o} = \frac{0.82}{1}

\frac{V_w}{V_o} = 0.82

The portion of the object that is submerged corresponds to 82%, while the portion that is visible, above the water level will be 18%

4 0
3 years ago
True Or False;
tino4ka555 [31]
False: because atoms are base on the elements on the periodic table.
3 0
3 years ago
¿Cuál es la frecuencia de una ola con una velocidad de 14 m / s y una longitud de onda de 20 metros?
Margaret [11]

Responder:

<h2>0.7Hertz </h2>

Explicación:

Usando la fórmula para calcular la velocidad de onda que se expresa como se muestra.

Velocidad de una onda = frecuencia * longitud de onda

v = fλ

Dada la velocidad de onda = 14 m / sy longitud de onda = 20 metros

frecuencia f = v / λ

f = 14/20

f = 0.7Hertz

La frecuencia de la onda es de 0.7 Hertz.

7 0
3 years ago
1. A rocket is fired vertically from the launch pad with a
zavuch27 [327]

Answer:

Max height= 36000 meters

Total Time = 120 seconds

Explanation:

0 = U - at

U = at

U= 20*60

U= 1200 m/s

MAX altitude would be

(U²Sin²tita)/2g

Max height= 1200² *( SIN90)²/(2*20)

Time of FLIGHT

2 * 1200/20

2400/20

120 sec onds

7 0
4 years ago
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