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3 years ago

True or false Scientific endeavor is driven by both simple curiosity as well as societal demands

Physics

1 answer:

ICE Princess25 [194]3 years ago

6 0

Answer:

its true that Scientific endeavor is driven by both simple curiosity as well as societal demands.

Explanation:

When a scientist has a curiosity about something he carried out a research. and when their is a demand of something in society that time scientific research is carried out. Therefore its true that a scientific endeavor is driven by simple curiosity or societal demand.

For example

in society, there is demand of a medicine which can completely kill the cancer and a scientist has curiosity to know how to kill cancer cell. In this way a scientific endeavor for cancer medicine can be carried out by both simple curiosity as well as societal demands.

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A 40 g ball rolls around a 30 cm -diameter L-shaped track, shown in the figure, (Figure 1)at 60 rpm . What is the magnitude of t

levacccp [35]

Answer:

0.47 N

Explanation:

Here we have a ball in motion along a circular track.

For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.

This force is called centripetal force, and its magnitude is given by:

$F=m\omega^2 r$

where

m is the mass of the object

$\omega$ is the angular velocity

r is the radius of the circle

For the ball in this problem we have:

m = 40 g = 0.04 kg is the mass of the ball

$\omega =60 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s$ is the angular velocity

r = 30 cm = 0.30 m is the radius of the circle

Substituting, we find the force:

$F=(0.040)(6.28)^2(0.30)=0.47 N$

3 0

2 years ago

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago

The kinetic energy of a book on a shelf is equal to the work done to lift the book to the shelf. t/f

Svet_ta [14]

The kinetic energy of a book on a shelf is equal to the work done to lift the book to the shelf is false. The kinetic energy on the shelf is zero because it is not in action.

7 0

3 years ago

Read 2 more answers

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

2 years ago

A B C D Plz help me.

Montano1993 [528]

Answer:

The correct option is;

${}$ Man doing most work ${}$ Box gaining the most energy

D ${}$ Y Q

Explanation:

The given parameters of the question are;

The distance the box P is pushed by the Man X = 0

The force the Man X applies to the box P = x N

The distance the box Q is lifted by the Man Y = h > 0 meters

The minimum force the Man Y applies to the box Q = W, the weight of the box

Work done = Force × Distance

Energy gained = Potential energy + Kinetic Energy = (Mass × Gravity × Height = Weight × Height = W × h) + 1/2 × Mass × Velocity²

The final velocity of either box = 0 m/s (The boxes are at rest on the ground or the shelf)

Therefore, Kinetic energy = 0 J

The work done by Man X = 0 × x = 0 J

The energy gained by the box P = W × 0 = 0 J

The work done by Man Y = W × h = W·h J

The energy gained by the box P = W × h = W·h J

We have, the work done by the man Y = W·h J is more than the work done by the man X = 0 J

The energy gained by the box P = W·h J is more than the energy gained by the box Q = 0 J

Therefore, the correct option is D, Man doing the most work is Y, box gaining the most energy is Q.

8 0

2 years ago