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butalik [34]
3 years ago
10

Leta Stetter Hollingworth conducted pioneering work on __________. A. identity development in ethnic minorities B. cognitive pro

cesses in animals and humans C. racism and its effects on child development D. adolescent development and gifted children  
Physics
2 answers:
dalvyx [7]3 years ago
7 0

This is the only answer in the choices.

D. adolescent development and gifted children  

rusak2 [61]3 years ago
5 0
I believe the correct answer from the choices listed above is option D. Leta Stetter Hollingworth conducted pioneering work on adolescent development and gifted children. She <span>was an American psychologist who conducted pioneering work in the early 20th century. </span>
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A block of weight mg sits on an inclined plane as shown in (Figure 1) . A force of magnitude F1 is applied to pull the block up
svlad2 [7]

Answer:

W = (F1 - mg sin θ) L,   W = -μ  mg cos θ L

Explanation:

Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane

Y Axis  

       N - W_{y} =

       N = W_{y}

X axis

       F1 - fr - Wₓ = 0

       fr = F1 - Wₓ

Let's use trigonometry to find the components of the weight

     sin θ = Wₓ / W

     cos θ = W_{y} / W

      Wₓ = W sin θ

      W_{y} = W cos θ

We substitute

      fr = F1 - W sin θ

Work is defined by

        W = F .dx

        W = F dx cos θ

The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1

         

        W = -fr x

        W = (F1 - mg sin θ) L

Another way to calculate is

         fr = μ N

         fr = μ W cos θ

the work is

         W = -μ  mg cos θ L

4 0
3 years ago
A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'
Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = \frac{1}{2} I ω^2 + \frac{1}{2} mv^{2}

where

I = \frac{1}{2} mr^{2} and ω = \frac{v}{r}

thus,

KE = \frac{1}{2} \frac{1}{2} mr^{2} (\frac{v}{r})^2 + \frac{1}{2} mv^{2}

     = \frac{1}{2} \frac{1}{2} mr^{2} \frac{v^{2} }{r^{2}} + \frac{1}{2} mv^{2}

     = \frac{1}{4} mv^{2} + \frac{1}{2} mv^{2}

     = \frac{3}{4} mv^{2}

     = \frac{3}{4} (0.46) (1.1)^{2}

     = 0.42 J

8 0
3 years ago
Umar has two copper pans, each containing 500cm3 of water. Pan A has a mass of 750g and pan B has a mass of 1.5kg. Which pan wil
Olin [163]

Answer:

heat required in pan B is more than pan A

Explanation:

Heat required to raise the temperature of the substance is given by the formula

Q = ms\Delta T

now we know that both pan contains same volume of water while the mass of pan is different

So here heat required to raise the temperature of water in Pan A is given as

Q_1 = (m_w s_w + m_ps_p)\delta T

Q_1 = (0.5(4186) + 0.750(s))\Delta T

Now similarly for other pan we have

Q_2 = (m_w s_w + m_ps_p)\delta T

Q_2 = (0.5(4186) + 1.50(s))\Delta T

So here by comparing the two equations we can say that heat required in pan B is more than pan A

3 0
3 years ago
Phương trình trạng thái tổng quát của khí lí tưởng diễn tả là​
tatyana61 [14]

Expla

pV/T=const goodluke

6 0
3 years ago
Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on
leva [86]

Answer:48 V

Explanation:

Given

Three charged particle with charge

q_1=50\ nC at y=6\ m

q_2=-80\ nC at x=-4\ m

q_3=70\ nC at y=-6\ m

Electric Potential is given by

V=\frac{kQ}{r}

Distance of q_1 from x=8\ m

d_1=\sqrt{6^2+8^2}

d_1=\sqrt{36+64}

d_1=10\ m

similarly d_2=8-(-4)

d_2=12\ m

d_3=\sqrt{(-6)^2+8^2}

d_3=\sqrt{36+64}

d_3=10\ m

Potential at x=8\ m is

V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}

V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]

V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}

V_{net}=9\times 5.33

V_{net}=47.97\approx 48\ V

5 0
3 years ago
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