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2 years ago

briefly discuss the phenomenon of childhood maltreatment in the context of five different psychological perspectives

Physics

1 answer:

Butoxors [25]2 years ago

6 0

Answer:

Child maltreatment or child abuse is a phenomenon of sexual, physical, and/or psychological abuse of children.

Five different psychological perspectives of childhood maltreatment are as follows:

Poverty is one of the biggest cause of childhood maltreatment that impact once psychological health.
Child do not get proper care from their parents.
Childrens are emotionally weak and anything negative thought can depress them.
Lack of communication sometimes make children underconfident.
Social descrimination on the basis of race and gender can impact a child psychological balance.

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A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.

fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force $N= mg cos 22$

$\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N$

8 0

2 years ago

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A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the fol

qaws [65]

Answer:

Part a)

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

$d = \frac{1}{2}gt^2$

Part c)

$v_f = v_o - gt$

Part d)

$d = \frac{1}{2}gt^2$

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall

So we have

$v_y = -gt$

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

Distance from helicopter is same as the distance of free fall

so we will have

$d = \frac{1}{2}gt^2$

Part c)

If helicopter is rising upwards with uniform speed

then final speed of the package after time t is given as

$v_f = v_i + at$

$v_f = v_o - gt$

Part d)

distance from helicopter

$d = \frac{1}{2}gt^2$

8 0

2 years ago

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Kenny and Candy decided to sit on a see-saw while visiting a local play park. Candy, of mass

pochemuha

Answer:

(i) 208 cm from the pivot

(ii) Move further from the pivot

Explanation:

(i) Sum of the moments about the pivot of the seesaw is zero.

∑τ = Iα

(50 kg) (10 N/kg) (2.5 m) + (60 kg) (10 N/kg) x = 0

1250 Nm + 600 N x = 0

x = -2.08 m

Kenny should sit 208 cm on the other side of the pivot.

(ii) To increase the torque, Kenny should move away from the pivot.

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2 years ago

Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci

Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

$\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W$

$\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})$

$h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}$

$2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]$

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

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2 years ago

How do you do projectile motion problems

PtichkaEL [24]

The key projectile motion is that gravity allows downward only

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2 years ago

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