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Andrej [43]
3 years ago
8

Which of the following is true about a family psychologist?

Physics
2 answers:
viva [34]3 years ago
7 0

Answer:

1. A family psychologist would help deal with developmental and behavior problems in children and teenagers

3. Family psychologists would typically counsel a couple on the edge of divorce

Explanation:

ikadub [295]3 years ago
3 0
Are there answer choices?
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Comparativo entre a máquina a vapor de Neu carmen e James Watt!
Bas_tet [7]

Answer:

jame watt

Explanation:

James Watt tiña máis importancia que outro home

3 0
4 years ago
I'll mark brainliest
irga5000 [103]

Answer:

A) 35 ft

B) 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

Explanation:

A) Total distance covered by the dog = 20 + 15

                                  = 35 ft

B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;

total displacement of the dog = 20 - 15

                                  = 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick

But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.

Thus,

Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

7 0
3 years ago
2. A powerful experimental sewing machine is powered by a mass-spring system. This
Alexus [3.1K]

We have that the Number of stitches per sec and he mass of  oscillation motion is mathematically given as

a) Nt=25stitches per sec

b) m=2.033e-5kg

<h3>Number of stitches per sec and he mass of  oscillation motion</h3>

Question Parameters:

This <u>sewing </u>machine is capable of stitching 1,500 stiches in one minute.

If the <em>sewing </em>machine has a spring constant of 0.5 N/m,

Generally the equation for the Number of stitches per sec  is mathematically given as

Nt=N/t

Therefore

Nt=1500/60

Nt=25stitches per sec

b)

Generally the equation for the Time t  is mathematically given as

T=2\pi\sqrt{\frac{m}{k}}

Therefore

0.04=2\pi\sqrt{\frac{m}{0.5}}\\\\m=\frac{0.5*0.04^2}{4\pi^2}

m=2.033e-5kg

For more information on Mass visit

brainly.com/question/15959704

7 0
2 years ago
How do I know if i’m doing number 2 right?
Ksju [112]

We are given an object that is speeding up on a level ground.

Let's remember that the gravitational energy depends on the change in height, therefore, if the object is not changing its height it means that the gravitational energy remains constant.

The kinetic energy depends on the velocity. If the velocity is increasing this means that the kinetic energy is also increasing.

Now, every change in velocity requires acceleration and acceleration requires a force. The force and the distance that the object moves are equivalent to the work that is transferred to the object and therefore, the change in kinetic energy. This means that the total energy of the system increases as work is transferred to the mass.

We have that the total energy of the system increases in the form of kinetic energy and that the gravitational potential energy remains constant. Therefore, the diagrams should look like pie charts that grow but the area of the segment of the potential energy stays the same. It should look similar to the following.

8 0
1 year ago
Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103
son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

PE = -\frac{GMm}{R}

As M=m, then

PE = -\frac{Gm^2}{R}

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

U = -\frac{2Gm^2}{R}

dU = -\frac{2Gm^2}{R}

dKE = -dU

Therefore replacing we have that,

\frac{1}{2}mv^2 =\frac{Gm^2}{2R}

Re-arrange to find v,

v= \sqrt{\frac{Gm}{R}}

v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}

v = 816.7m/s

Therefore the  velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

2r = 2*10^3m

Therefore

dU = Gm^2(\frac{1}{R}-\frac{1}{2r})

v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}

v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}

v = 1.83*10^7m/s

Therefore the velocity when they are about to collide is 1.83*10^7m/s

7 0
3 years ago
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