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3 years ago

Pure science is best defined as the what?

Physics

2 answers:

Lapatulllka [165]3 years ago

7 0

A (continuing search for new knowledge)

WINSTONCH [101]3 years ago

6 0

Answer:

A) continuing search for new knowledge

Explanation:

Science has always been about discovery and gaining knowledge

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Secretions from the skin, such as sweat, protect the body because they

IgorLugansk [536]

Sweat is slightly acidic which helps to protect the body.

<em>(Please mark this answer as Brainliest and leave a Thanks if I helped you!)</em>

4 0

3 years ago

Read 2 more answers

Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.

andreev551 [17]

Answer:

the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = $m_A$

then, the mass of object B = $m_B = \frac{m_A}{4}$

work done on object A = -18 J

work done on object B = -18 J

let $v_i$ be the initial speed

let $v_f$ be the final speed

For object A;

$K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2 = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ - v_i^2 )\ =- 18\\\\v_f^2 \ - v_i^2 = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ - v_i^2 = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\$

$v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\$

Thus, the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

To obtain the change in the final speed of object B, apply the following equations.

$K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B = \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\$

$\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i$

Thus, the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

3 0

3 years ago

When you drop a object on earth, earth gets pulled slightly toward that object as well. How do I calculate the force if the eart

kiruha [24]

The gravitation force with which the earth is being pulled can be determined by applying Newton's law of universal gravitation.

<h3>What is gravitation force?</h3>

According Newton's law of universal gravitation, the force exerted between two objects in the universe is directly proportional to the product of masses of the two objects and inversely proportional to the square of the distance between the two objects.

Mathematically, the formula for gravitation force is given as;

F = GmM/R²

where;

m is the mass of the object
M is mass of earth
R is the distance of the object from earth
G is universal gravitation constant

If the mass of the object is know and the distance between earth and the object is also known, the force with which the earth is being pulled can be calculated by applying Newton's law of universal gravitation as shown in the above equation.

Thus, the force with which the earth is being pulled can calculated as well.

Learn more about gravitation force here: brainly.com/question/27943482

#SPJ1

7 0

1 year ago

The quantity of charge passing through a surface of area 1.82 cm2 varies with time as q = q1 t 3 + q2 t + q3 , where q1 = 5.2 C/

Brrunno [24]

Answer:

Current through the surface at t = 1.1 s is 21.37 A.

Explanation:

The charge is passing through a surface of area varies with time as :

$q=q_1t^3+q_2t+q_3$

Here,

$q_1=5.2\ C/s^3\\\\q_2=2.5\ C/s\\\\q_3=6.5\ C$

t is in seconds

$q=5.2t^3+2.5t+6.5$

The rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dq}{dt}\\\\I=\dfrac{d(5.2t^3+2.5t+6.5)}{dt}\\\\I=15.6t^2+2.5$

At t = 1.1 s, Current,

$I=15.6(1.1)^2+2.5\\\\I=21.37\ A$

So, the instantaneous current through the surface at t = 1.1 s is 21.37 A.

3 0

3 years ago

A man weighing 750 n and a woman weighing 500 n have the same momentum. what is the ratio of the man's kinetic energy km to that

miss Akunina [59]

Because weight W = M g, the ratio of weights equals the ratio of masses.

(M_m g)/ (M_w g) = [ (p^2 Man )/ (2 K_man)] / [ (p^2 Woman )/ (2 K_woman)

but p's are equal, so

K_m/K_m = (M_w g)/(M_m g) = W_woman / W_man = 450/680 = 0.662

4 0

3 years ago