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Artyom0805 [142]
2 years ago
6

Which of the following is true of children with chronic illness? a.) They are all eligible to recievie special education service

s. b.) Very few such children are enrolled in public schools c.) Their eligibility for soeical education services depends on whether their conditions adversely affect their educational functioning. d.) They represent a large proportion of the children eligible for speical education services.
Physics
1 answer:
ss7ja [257]2 years ago
5 0

Answer:

c.) Their eligibility for social education services depends on whether their conditions adversely affect their educational functioning.

Explanation:

Chronic Illness is a human health condition in which a particular (or number of) illness is persistent in the body and the effects on the body are long-lasting and are often resistant to treatment. The word chronic is usually used when the disease/illness/sickness and its effects stay in the body for more than three months.

The likeliest answer from the options given is option C because before social education services are given, it has to be decided if their health condition adversely affects their education.

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A concave mirror has a focal length of 30.0 CM. an object is placed 15.0 CM from the mirror. what is the radius of curvature of
Nana76 [90]

Answer:

60 cm

Explanation:

We are given;

  • Focal length of a concave mirror as  30.0 cm
  • Object distance is 15.0 cm

We are required to determine the radius of curvature.

We need to know that the radius of a curvature is the radius of a circle from which the curved mirror is part.

We also need to know that the radius of curvature is twice the focal length of a curved mirror.

Therefore;

Radius of curvature = 2 × Focal length

Therefore;

Radius of curvature = 2 × 30 cm

                                 = 60 cm                                  

7 0
3 years ago
A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa
Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

Q=P\tan\phi...(I)

We need to calculate the \phi

Using formula of \phi

\phi=\cos^{-1}(Power\ factor)

Put the value into the formula

\phi=\cos^{-1}(0.6)

\phi=53.13^{\circ}

Put the value of Φ in equation (I)

Q=600\tan(53.13)

Q=799.99\ kVAR

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

Q'=600\tan(0.95)

Q'=9.94\ KVAR

We need to calculate the difference between Q and Q'

Q''=Q-Q'

Put the value into the formula

Q''=799.99-9.94

Q''=790.05\ KVAR

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0
3 years ago
The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when the diameter is 60 mm?
jok3333 [9.3K]
V = [4/3]π r^3 => [dV / dr ] = 4π r^2

[dV/dt] = [dV/dr] * [dr/dt]

[dV/dt] = [4π r^2] * [ dr/ dt]

r = 60 mm, [dr / dt] = 4 mm/s

[dV / dt ] = [4π(60mm)^2] * 4mm/s = 180,955.7 mm/s 


5 0
3 years ago
Consider the data collected in science class. Different masses were thrown with varied amounts of
lesya [120]

Answer:

A: In all cases, the acceleration was the same.

Explanation:

I know this because its a clear obvious answer not only that it was one of my USA TESTPREP questions and it was right.

All you mainly have to do is the math - F=ma , In each case , the acceleration is 5 m/s squared

4 0
3 years ago
Read 2 more answers
A​ right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 ​lb/ft3
tresset_1 [31]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ ft^{3} to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = \pi r^{2} =  \pi 4^{2} =16\pi ft^{2}

Now,

Work done required = \int_{0}^{8} 52\times 16\pi (21 - y)dy

                                  = 832\pi \int_{0}^{8} (21 - y)dy

                                  = 832\pi([ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\)

                                  = 113152\pi = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

7 0
3 years ago
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