All categories

2 years ago

Why is the picture above not an accurate representation of charges and their field lines?

Physics

2 answers:

pochemuha2 years ago

7 0

Because field lines don’t all go in to the pole. The bottom half go outwards not inwards.

Liula [17]2 years ago

3 0

Answer:

C, There should be more field lines around +2Q than the -Q

Explanation:

just did the test its correct x

You might be interested in

A rock falls off a cliff. How fast will it be going after falling for 4.33 seconds?

bixtya [17]

Answer:42.43m/s

Explanation:According to vf=vi+at, we can calculate it since v0 equals to 0. vf=0+9.8m/s^2*4.33s= 42.434m/s

4 0

1 year ago

An arrow movirg 48.3 m/s has 5.22<br> kg•m/s of momentum. What is its<br> mass?

spin [16.1K]

Answer:

0.11 kg

Explanation:

Ft = MV

Ft = momentum 5.22kg m/s

M = mass

V = velocity 48.3m/s

Therefore

5.22 = M x 48.3

Divide both sides by 48.3

5.22/48.3 = M x 48.3/48.3

0.11 = M

M = 0.11kg

6 0

3 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

3 0

4 years ago

MY LAST ONE<,Brainliest for best answer!

grigory [225]

_Award brainliest if helped!
Mechanical Advantage = Force by Hammer / Force by Nail = 160/40 = 4

8 0

3 years ago

Read 2 more answers

To which category does galaxy #1 belong? Why does it belong in this category?

vodka [1.7K]

Answer:

This galaxy belongs to the elliptical galaxy category. This is because it does not have spiral arms.

5 0

3 years ago