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zubka84 [21]
3 years ago
8

Newton's first law of motion states that an object at rest will stay at rest unless acted

Physics
1 answer:
Ludmilka [50]3 years ago
7 0

Answer:

All are examples

Explanation:

Football acted upon by the force of the foot

Bicycle acted on by imbalanced force of gravity

Friction of dirt on player slows his run.

stick and water acted on by imbalanced force of gravity

You might be interested in
What is the voltage drop across the 10.0 2 resistor?
amid [387]

Answer:

the answer is equal to 2.00v

4 0
2 years ago
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
A 10 kg block is attached to a light cord that is wrapped around the pulley of an electric motor, as shown above. If the motor r
Anika [276]

Answer:

156.8 Watts

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 10 kg

Height (h) = 8 m

Time (t) = 5 s

Power (P) =?

Next, we shall determine the energy used by the motor to raise the block. This can be obtained as follow:

Mass (m) = 10 kg

Height (h) = 8 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

E = mgh

E = 10 × 9. 8 × 8

E = 784 J

Finally, we shall determine the power output of the motor. This can be obtained as illustrated below:

Time (t) = 5 s

Energy (E) = 784 J

Power (P) =?

P = E/t

P = 784 / 5

P = 156.8 Watts

Therefore, the power output of the motor is 156.8 Watts

7 0
3 years ago
An aquifer pump test was conducted in a confined aquifer where the initial piezometric surface was at elevation 12.45 m, and wel
murzikaleks [220]

Answer:

T = 0.0088 m²/s

Explanation:

given,

initial piezometric elevation = 12.5 m

thickness of aquifer = 14 m

discharge = 28.24 L/s = 0.02824 m³/s

we know                              

k = \dfrac{qln(\dfrac{R_2}{R_1})}{2\pi D(H_2-H_1)}

k = \dfrac{0.02824 \times ln(\dfrac{75}{25})}{2\pi \times 14 (11.45-10.89)}                                                                        

k = 0.629 mm/sec

Transmissibilty

T = k × H                          

T = 0.629 × 14 × 10⁻³

T = 0.0088 m²/s

4 0
3 years ago
When light passes into a more dense material, it bends away from the
asambeis [7]

Answer: true

Explanation:

7 0
3 years ago
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