a 15kg child is riding a merry go round at 3rpm. what centripetal force must she exert if she is 8m from the center

Answers

Answer 1

Answer:
This is another application of Newton's second law of motion

   Force = (mass) x (acceleration).

The quantities are a little more complicated in circular motion than
they are in plain old straight-line motion. In circular motion, the 'force'
is the centripetal force on the object, always pointing toward the center
of the circle. And the 'acceleration' is the centripetal acceleration, also
pointing toward the center of the circle, and equal to

      (speed)² / (radius).

The only thing we really need to find is the centripetal acceleration, and
then we'll have everything needed to plug into the formula and calculate
the centripetal force.

   Acceleration =    (speed)² / (radius).

   Speed = (distance) / (time) =

   (3 circumferences) / (minute) =

   (3 x 2pi x radius) / minute =

(6 pi x 8 meters) / (60 sec) = 0.8 pi m/s .

   Acceleration = (speed)² / (radius) =

   0.64 pi² m²/s² / (8 meters) = 0.08 pi² m/s² =

   0.79 m/s² (rounded) .

   Force = (mass) x (acceleration) =

   (15 kg) x (0.79 m/s²) = 11.84 newtons
      (about 2.66 pounds) .

I would not say that she "exerts this centripetal force". The fact of
the matter is much simpler: This is the force that something must
exert on her, pointing toward the center, in order to keep her revolving
around the center at that speed. It could be the friction between her
shoes and the platform, if she's standing on the merry-go-round.
It could be a rope tied between her ankle and something at the center
of the rotating platform. It could be the safety belt on the horse that
she's riding. Whatever it is, something has to constantly pull her toward
the center of the platform, with a force of 11.84 newtons, otherwise her
15kg body will not travel in that circular path.

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Where are magnetic fields strongest near a bar magnet

Answers

The answer is letter A or 'top'

At the poles of the magnet.

A long solenoid has 1400 turns per meter of length, and it carries a current of 4.9 A. A small circular coil of wire is placed inside the solenoid with the normal to the coil oriented at an angle of 90.0˚ with respect to the axis of the solenoid. The coil consists of 42 turns, has an area of 1.2 × 10-3 m2, and carries a current of 0.45 A. Find the torque exerted on the coil.

Answers

Answer:

The torque on the coil is $1.955* 10^(- 4)\ N-m$

Solution:

No. of turns per meter length, n = 1400 turns\m

Current, I = 4.9 A

Angle, $\theta = 90.0^(\circ)$

No. of turns of coil, N = 42 turns

Area, A = $1.2* 10^(- 3)m^(2)$

Current in the coil, I' = 0.45 A

Now,

To calculate the exerted torque on the coil:

The magnetic field, B produced inside the coil is given by:

$B = n\mu_(o)I$

$B = 1400* 4\pi times 10^(- 7)* 4.9 = 8.62* 10^(- 3)\ T$

Now, the torque exerted is given by:

$\tau = I'NAB$

$\tau = 0.45* 42* 1.2* 10^(- 3)* 8.62* 10^(- 3) = 1.955* 10^(- 4)\ N-m$

Answer:

$T\approx 1.95* 10^(-4) N.m$

Explanation:

Given:

A long solenoid having

no. of turns per meter, n =1400

current, I = 4.9 A

A small coil of wire placed inside the solenoid

angle of orientation with respect to the axis of the solenoid, $\theta=90\degree$ °

no. of turns in the coil, N = 42

area of the coil, $a= 1.2* 10^(-3) m^2$

current in the coil, $i =0.45 A$

We have for torque:

$T=n.i.a.B. sin\theta$ .......................(1)

∵ $B=\mu_(0) .n.I$ ................................(2)

where:

B= magnetic field

$\mu_0=$ The permeability of free space = $4\pi*10^(-7) T.m.A^(-1)$

Substitute B from eq. (2) into eq. (1) we have:

$T=n.i.a.(\mu_0.N.I ).sin\theta$

putting the respective values in above eq.

$T=42* 0.45* 1.2* 10^(-3)* 4\pi*10^(-7) * 1400* 4.9* sin 90^(\circ)$

$T\approx 1.95* 10^(-4) N.m$

A mass that weighs 8 lb stretches a spring 24 in. The system is acted on by an external force of 4 sin 4t lb. If the mass is pulled down 6 in. and then released, determine the position of the mass at any time. Determine the first four times at which the velocity of the mass is zero

Answers

Answer:

$t = (\pi)/(8), (\pi)/(4), (3\pi)/(8), (3 \pi)/(4)$

Explanation:

The equation of force is

F = 4 sin 4 t

Compare with the standard equation

f = A sin wt

where, w is the angular frequency and A is the amplitude.

Now

w = 4 rad/s

Let the time period is T.

the relation for the time period is

$T = (2\pi)/(w)\n\nT = (2 \pi)/(4)\n\nT = (\pi)/(2)$

the time period is defined as the time taken by the body to complete one oscillation.

So, the velocity is zero at the extreme points where the object is at time, T/4 and its odd T/2, 3T/4, 3T/2, etc.

So, the velocity is zero at time

$t = (\pi)/(8), (\pi)/(4), (3\pi)/(8), (3 \pi)/(4)$

Final answer:

To determine the position of the mass and the times when the velocity is zero in a mass-spring system.

Explanation:

To determine the position of the mass at any time, we need to solve the equation of motion for the mass-spring system. The equation is:

mx'' + kx = Fext

where m is the mass, x is the displacement of the mass from its equilibrium position, k is the spring constant, and Fext is the external force. We can solve this differential equation to find the position of the mass as a function of time.

To determine the first four times at which the velocity of the mass is zero, we need to solve for the velocity of the mass, which is given by:

v = x'

We can find the times at which the velocity is zero by finding the values of t for which x' = 0.

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What is the potential of a 35kg child sitting at the top of a slide that is 3.5m above the ground? What is her kinetic energy if she moves down the slide at a speed of 5.0m/s??

Answers

Ep = mgh
= (35)(9.8)(3.5)
= 1200.5 J

Ek = (1/2)(mv^2)
= (0.5)(35)(5^2)
= 437.5 J

an apple in a tree has a gravitational potential energy of 175J and a mass of 0.36g . how high from the ground is the apple

Answers

Final answer:

The height of the apple from the ground can be calculated using the formula for gravitational potential energy. Given the apple's mass and its potential energy, we find that the apple is approximately 5000 meters from the ground.

Explanation:

The gravitational potential energy of an object can be calculated using the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s² on Earth), and h is the height of the object from the ground. In this case, the apple's gravitational potential energy is given as 175J, its mass is 0.36g (or 0.00036 kg when converted.)

To find the apple's height from the ground, you need to rearrange the formula to solve for h: h = PE / (m * g). Substituting the given values in the formula, h = 175 J / (0.00036 kg * 9.8 m/s²), you get approximately h = 5000 meters.

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23°C = ______ K
23
296
-250
123

Answers

TºC + 273 :

23 + 273 = 296 K

Answer B

hope this helps!

Answer:

the answer is 4 years

Explanation:

4 years passed do u still need this

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