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2 years ago

A 6 kg brick is pulled across the flat sidewalk by a horizontal force of 35 N. The brick is experiencing a frictional

Physics

1 answer:

SIZIF [17.4K]2 years ago

6 0

<h3>Answer: Approximately 4.67 m/s^2</h3>

==============================================

Explanation:

Let's say you want to push the brick to the right. The free body diagram will have an arrow pointing right on the rectangle (the brick) and the arrow is labeled with 35 N.

Friction always counteracts whatever force you apply. The friction force arrow will point left and be labeled with 7 N.

The net horizontal force is therefore 35-7 = 28 N and the direction is to the right. The positive net force means you've overcome the force of friction and the brick is moving.

F = 28 is the net force

m = 6 is the mass

a = unknown acceleration

F = m*a .... newton's second law

28 = 6a

6a = 28

a = 28/6

a = 4.67

The acceleration of the brick is approximately 4.67 m/s^2

This means that for every second, the brick's velocity is increasing by about 4.67 m/s.

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Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25

mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp = 1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0

2 years ago

Which one of the following types of electromagnetic wave travels through space the fastest?

Setler [38]

Answer:

As a result, light travels fastest in empty space, and travels slowest in solids. In glass, for example, light travels about 197,000 km/s.

Explanation:

4 0

1 year ago

Blake and Sandra are having a rummage sale. Blake drags 3 boxes a distance of 10 meters each. He exerts a force of 20 newtons on

vampirchik [111]

For Blake:

3 boxes at a distance of 10 meters each, each box weighs 20 N

Work done by Blake = 3 * 10m * 20N
= 600 J
Power = 600 J/ 2 min
= 300 J/min

For Sandra:

4 boxes, 15 N each at a distance of 12 meters each.

Work done by Sandra = 4 * 15 N *12m
= 720 J
Power = 720 J/ 4 min
= 180 J/min

Blake does less work than Sandra.
Blake's power is more than Sandra's.

5 0

3 years ago

Read 2 more answers

______ 4. Justin was creating a chart comparing the potential energy between charged particles and the potential energy of magne

fredd [130]

Answer:

The correct option is;

A. The potential energy between both like charges and like poles increases as they move closer together

Explanation:

Here we have that when we move the like poles of two bar magnets close to each other, there is an increased resistance in the continuing motion, therefore for each extra gap closer achieved, there is an increase in potential energy

Similarly, when two like charges are brought closer together, the potential energy, or the energy available to push the two like charges apart increases charge as the as the charges are brought closer together

Therefore, the correct option is the potential energy between both like charges and like poles increases as they move closer together.

4 0

3 years ago

A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. the coefficient of kinetic friction between the box

forsale [732]

We need to see what forces act on the box:

In the x direction:

Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.

In the y direction:

N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.

From N-Gcosα=0 we get:

N=Gcosα, we will need this for the force of friction.

Now to solve for Fh:

Fh=ma + Ff + Gsinα,

Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²

Fh=ma + μmgcosα+mgsinα

Now we plug in the numbers and get:

Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N

The horizontal force for pulling the body up the ramp needs to be Fh=71 N.

8 0

2 years ago