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2 years ago

Newton’s third law explains what happens when two objects ______.

1 answer:

8 0

React? His third law basically states that for every reaction there is an opposite and equal reaction. An example of this is if you and a friend are on an ice skating rink and facing each other. If you give them a push, you will also slide backwards, since they exert a force on you equal to the force that you exert on them.

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elaborate on the difference between a chemical change and a physical change. a) a physical change is irreversible where as a che

lozanna [386]

I think the correct answer from the choices listed above is option D. The difference between a chemical change and a physical change is lies in the fact that a physical change does not break any bonds while a chemical change does form new substances. Hope this answers the question.

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3 years ago

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During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive

olganol [36]

Answer:

The force is $F_c = 789.03 \ N$

Explanation:

From the question we are told that

The tangential resistive force is $F_t = 115 \ N$

The mass of the wheel is m = 1.80 kg

The diameter of the wheel is $d = 50.0 cm = 0.5 \ m$

The diameter of the sprocket is $d_c = 8.50 \ cm =0.085 \ m$

The angular acceleration considered is $\alpha = 4.30\ rad/s^2$

Generally the radius of the wheel is

$r = \frac{d}{2}$

=> $r = \frac{0.5}{2}$

=> $r = 0.25 \ m$

Generally the radius of the sprocket is

$r_c = \frac{d_c}{2}$

=> $r_c = \frac{0.085}{2}$

=> $r_c = 0.0425 \ m$

Generally the moment of inertia of the wheel is mathematically represented as

$I = m * r^2$

=> $I = 1.80 * 0.25^2$

=> $I = 1.1125 \ kg \cdot m^2$

Generally the torque experienced by the wheel due to the forces acting on it is mathematically represented as

$\tau = F_c * r_c - F_t * r$

Here $F_c$ is the force acting on the sprocket

$\tau = F_c * 0.0425 - 115 * 0.25$

$\tau = 0.0425F_c - 28.75$

Generally the torques that will cause the wheel to move with $\alpha = 4.30\ rad/s^2$ is mathematically represented as

$\tau = I * \alpha$

$0.0425F_c - 28.75 = I * \alpha$

$0.0425F_c - 28.75 = 1.1125 *4.30$

$F_c = 789.03 \ N$

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3 years ago

PLZ HELP, BRAINLIEST !!

Tems11 [23]

Density = mass divided by volume so if you use that formula Liquid A turns out to be a higher density so it flows slower.

3 0

3 years ago

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A basketball player can jump 1.6 m off the hardwood floor. With what upward velocity did he leave the floor?

irinina [24]

First do 1.6 m (how far he jumps) 9.8 m/s (what gravity is measured at) then times 2

= 31.36

Sq root = 5.6

3 0

3 years ago

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Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r

cricket20 [7]

Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

Find:

Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

E_net = E_2 + E_4

E_2 = E_4

E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.

Hence,

E_net = 2*E_part(a)*cos(Q)

- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:

Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

- Now, compute the net electric field E_net:

E_net = 2*(99.518)*cos(27.12)

E_net = 177.151 N / C

5 0

3 years ago