Ask question

Ask question

All categories

3 years ago

10

Consider the expression below. Assume m is an integer. 6m(2m + 18) Enter an expression in the box that uses the variable m and m

akes the equation true. (Simplify your answer completely. If no expression exists, enter DNE.)

1 answer:

Anika [276]3 years ago

8 0

6x2=12m
6x18=108
12m+108
Simplified: m+9 bc 12/12 and 108/12

You might be interested in

Which of these scientists had the greatest contribution to early microscopy?

Alisiya [41]

B. I think is the correct answer

5 0

2 years ago

A 0.150 kg baseball has 118 j of KE. how fast is the ball moving?(unit=m/s)

MrRissso [65]

Answer:

Explanation 118 = (1/2) * 0.15 * v² 118 = 0.075 * v² v² = 1573.33 m/s ... since KE = m/2*V^2 , then : V = √2KE/m = √20*118/1.5 = 39.67 m//sec ( 142.8 km/h ; 88.75 mph).:

4 0

3 years ago

One hundred jumping beans are placed along the center line of a gymnasium floor at six-inch intervals. Twelve hours later, the d

stira [4]

Answer:

a) Diffusion coefficient, D = 1.5 in/hr

b) Mean jump frequency, f = 0.0833 Hz

Explanation:

a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:

$^{2} = 2Dt$ ..........(1)

Where <r> = mean displacement

D = Diffusion coefficient

t = time = 12 hrs

sum of the squares of the distance divided by 100 is 36 in2.

<r>²= 36 in²

Substituting these values into equation (1) above

$36 = 2 * D *12\\36 = 24 D\\D = 36/24\\D = 1.5 in/hr$

b) Mean jumping distance, <r> = 0.1 inches

Applying equation (1) again

Where D = 1.5 in/hr

$^{2} = 2Dt$

$0.1^{2} = 2 * 1.5t\\0.01 = 3t\\t = 0.01/3\\t = 0.0033 hrs\\t = 0.0033 * 3600\\t = 12 seconds$

The mean jump frequency, f = 1/t

f = 1/12

f = 0.0833 Hz

8 0

3 years ago

A particle has a charge of +1.5 µC and moves from point A to point B, a distance of 0.15 m. The particle experiences a constant

kirill [66]

Answer:

Part a)

$F = 6 \times 10^{-3} N$

Direction of force is along the motion of charge

Part b)

$E = 4000 N/C$

direction of electric field is along the direction of motion

Explanation:

Part a)

As we know that the change in electric potential energy is equal to the work done by electric field

$W = EPE_A - EPE_B$

$W = 9.0 \times 10^{-4} J$

now from the equation of work done we know that

$W = F.d$

$(9.0 \times 10^{-4}) = F(0.15)$

$F = 6 \times 10^{-3} N$

Direction of force is along the motion of charge

Part b)

As we know the relation between electrostatic force and electric field given as

$F = qE$

$(6 \times 10^{-3}) = 1.5 \times 10^{-6} E$

$E = 4000 N/C$

direction of electric field is along the direction of motion

8 0

3 years ago

How should the magnetic field lines be drawn for the magnets shown below?

sergeinik [125]

Option B is the correct answer that show how magnetic field lines should be drawn for the magnets shown in the figure.

<h3>What is Magnetic Line of Force ?</h3>

The Magnetic Line of Force of a magnet is defined as the line along which a free N - pole would tend to move if placed in the field of a line such that the tangent to it at any point gives the direction of the field at that point.

When the two unlike poles are placed to each other, there will be attraction. And when the two like poles are placed to each other, there will be repulsion. The reason is that the line of force tend to move from the north pole to the south pole.

From the given diagram, the two magnets are of the same south pole. They are of like pole and there will be repulsion between the two magnets.

Therefore, Option B is the correct answer that show how magnetic field lines should be drawn for the magnets shown in the figure.

Learn more about Magnetic Field Lines here: brainly.com/question/17011493

#SPJ1

5 0

2 years ago