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Illusion [34]
4 years ago
10

How do Newton's laws of motion apply to the motion of an animal, such as a cat that is running?

Physics
2 answers:
Bad White [126]4 years ago
8 0
An object in motion will still be in motion and an object at rest will stay at rest unless being acted on by an unbalanced force
Zolol [24]4 years ago
3 0
According to newton's first law of motion a stationary or moving object will be in that state forever until an external force is exerted
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How does the principle of moments affect the couple in physics
ivanzaharov [21]
When you push a door closed it doesn't travel in a straight line it turns around the hinges this is an example of a line
3 0
3 years ago
Explain how heat is related to temperature and thermal energy...
daser333 [38]

Answer:

Thermal energy is directly related to heat because thermal energy is how heat is transfered.

5 0
3 years ago
The small ball of mass m = 0.5 kg is attached to point A via string and is moving at constant speed in a horizontal circle of ra
babymother [125]

Answer:

d = 2.45 meters

Explanation:

Mass of the ball, m = 0.5 kg

Radius of the circle, r = 0.16 m

The angular speed of the ball around the circle is, \omega=2\ rad/s

The attached figure shows the whole scenario. Let F_t is the force acting on the ball in tangential direction. The forces will balanced each other at equilibrium.

In horizontal direction,

T\ sin\theta=F_t=mr\omega^2................(1)

In vertical direction,

T\ cos\theta=mg...............(2)

From equation (1) and (2) :

tan\theta=\dfrac{r\omega^2}{g}

Also,

tan\theta=\dfrac{r}{d}

d=\dfrac{g}{\omega^2}d=\dfrac{9.8}{2^2}

d = 2.45 meters

So, the value of d is 2.45 meters. Hence, this is the required solution.  

3 0
4 years ago
A ball of 0.5kg slows down from 5m/s to 3m/s. Calculate the work done inthe process.
dlinn [17]

Answer:

9.8 Joules (rounded to 2 significant figures)

Explanation:

Work done (J)= Force(N) x distance changed (m)

  • Force= 9.80665 x 0.5kg
  • Force= 4.90332 Newtons

  • Distance changed= 5-3
  • distance changed= 2m/s

--> work done= 4.90332 x 2

work done= 9.8 Joules

5 0
3 years ago
A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo
Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

W = \Delta U = U_2 - U_1

where the potential energy is defined as follows

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Let's first calculate the distance 'r' for both positions.

r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m

Now, we can calculate the potential energies for both positions.

U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J

Finally, the total work done on the moving particle can be calculated.

W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J

4 0
3 years ago
Read 2 more answers
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