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Katena32 [7]
3 years ago
11

What was the purpose of the little Albert experiment?

Physics
1 answer:
Murljashka [212]3 years ago
3 0

Answer:

to condition a phobia in an emotionally stable child

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A block with mass M = 3 kg is moving on a flat surface with constant speed v1 =
Alchen [17]

Answer:

this makes no since so i cant help you here sorry

5 0
1 year ago
You have a bowling ball with a mass of 4kg. You throw it with an acceleration of 10 m/s/s. With how much force will it hit the p
Eddi Din [679]

Answer:

<h3>The answer is 40 N</h3>

Explanation:

The force acting on an object can be found by using the formula

<h3>force = mass × acceleration</h3>

From the question

mass = 4 kg

acceleration = 10 m/s²

So we have

force = 4 × 10

We have the final answer as

<h3>40 N</h3>

Hope this helps you

6 0
3 years ago
A +2e charge is at the point (-1,0) mm in the x,y plane. A –e charge is at the point (0,1) mm. What is the electric field at the
Gennadij [26K]

Answer:

Let I and j be the unit vector along x and y axis respectively.

Electric field at origin is given by

E= kq1/r1^2 i + kq2/r2^2j

= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)

= (2.88i + 1.44j)*10^-3 N/C

Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3

F= (1.382 i + 0.691 j) *10^-21

 

 Goodluck

Explanation:

4 0
1 year ago
You are standing on the SCALE<br> The scale is on the floor.<br> What are the forces on the SCALE?
ehidna [41]

Answer:

The Normal and Gravitational Force

Explanation:

The normal force pushes up and is between the ground and the scale. The gravitational force is the force exerted on the ground.

3 0
3 years ago
Read 2 more answers
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
2 years ago
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