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vagabundo [1.1K]
3 years ago
15

The diagram shows two charged objects, X and Y.

Physics
2 answers:
masya89 [10]3 years ago
4 0

The missing diagram is in the attachments.

Answer: X: positive Y: positive

Explanation: Electric field is a vector quantity, which means it can be represented by a vector arrow: the arrow points in the direction of electric field and its length represents the magnitude at a given location. There are another representation of the electric field called electric field lines, <u>in which the line points away from a positively charged source and towards a negatively charged source</u>. This occurs because it follows a pattern, where the lines points in the direction that a positive test charge would have if it is accelerating on the line.

Analyzing the diagram, it can be observed that the lines are pointing away from both of the charged objects. Therefore, both X and Y are <u>positively charged</u>.

posledela3 years ago
3 0

Answer:

D on edge

Explanation:

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A distance of 0.002 m separates two objects of equal mass. If the gravitational force between them is
Alika [10]

Answer: 24.97 kg

Explanation:

The gravitational force between two objects of masses M1, and M2 respectively, and separated by a distance R, is:

F = G*(M1*M2)/R^2

Where G is the gravitational constant:

G = 6.67*10^-11 m^3/(kg*s^2)

In this case, we know that

R = 0.002m

F = 0.0104 N

and that M1 = M2 = M

And we want to find the value of M, then we can replace those values in the equation to get

0.0104 N = (6.67*10^-11 m^3/(kg*s^2))*(M*M)/(0.002m)^2

(0.0104 N)*(0.002m)^2/(6.67*10^-11 m^3/(kg*s^2)) = M^2

623.69 kg^2 = M^2

√(623.69 kg^2) = M = 24.97 kg

This means that the mass of each object is 24.97 kg

6 0
3 years ago
A 75-m-long train begins uniform acceleration from rest. the the train has a speed of 23 m/s when it passes a railway worker tan
Lapatulllka [165]

Answer:

acceleration a = 1.04 m/s2

Explanation:

Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:

s = 75 + 180 = 255 m

We can use the following equation of motion to find out the distance traveled by the car:

v^2 - v_0^2 = 2aswhere v = 23 m/s is the velocity of the car when it passes the worker, v_0 = 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.

23^2 - 0^2 = 2*a*255

510a = 529

a = 529 / 510 = 1.04 m/s^2

8 0
3 years ago
Which of these statements best describes how energy is transferred when an object experiences friction?
Slav-nsk [51]
<h2>Answer:</h2><h2> b hopefully this helps you with work </h2>
3 0
3 years ago
A car of mass 1167 kg accelerates on a flat highway from 10 m/s to 28.0 m/s. How much work does the car's engine do on the car?
denis23 [38]

Answer:

Workdone = 465766038 Joules.

Explanation:

<u>Given the following data;</u>

Mass = 1167

Initial velocity = 10m/s

Final velocity =28m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

W = ½MVf² - ½MVi²

Substituting into the equation

W = ½(1167)*28² - ½(1167)*10²

W = ½ * 1361889* 784 - ½ * 1361889 * 100

W = 533860488 - 68094450

Workdone = 465766038 Joules.

7 0
3 years ago
You hang a heavy ball with a mass of 30 kg from a tungsten rod 2.8 m long by 1.5 mm by 2.6 mm. You measure the stretch of the ro
guajiro [1.7K]

Answer:

Young's modulus (Y) = 3.56×10^11 N/m^2

The speed of sound in tungsten = 6166.4 m/s

Explanation:

Young's modulus (Y) = stress/strain

Stress = force/area

Force = mg = 30×9.8 = 294 N

Area = 1.5 × 2.6 = 3.9 mm^2 = 3.9/10^6 = 3.9×10^-6 m^2

Stress = 294/3.9×10^-6 = 7.54×10^7 N/m^2

Strain = extension/length

Extension = 0.000594 m

Length = 2.8 m

Strain = 0.000594/2.8 = 2.12×10^-4

Y = 7.54×10^7/2.12×10^-4 = 3.56×10^11 N/m^2

Y = h × rho × g

rho = 18.7 g/cm^3 = 18.7 g/cm^3 × 1 kg/1000 g × (100 cm/1 m)^3 = 18,700 kg/m^3

h = 3.56×10^11/(18,700×9.8) = 1.94×10^6 m

From the equations of motion

v^2 = u^2 + 2gh =

Initial speed (u) = 0 m/s

v = sqrt (2×9.8×1.94×10^6)

v = 6166.4 m/s

7 0
4 years ago
Read 2 more answers
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