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____ [38]
3 years ago
5

Which one of the following parenting factors is associated with improved IQ scores in children?

Physics
1 answer:
lapo4ka [179]3 years ago
5 0
I would say B. A stable home and varied activities 
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What is the value of work done on an object when a 0.1x102-newton force moves it 30 meters and the angle between the force and t
disa [49]

Answer:306joules

Explanation:

3 0
3 years ago
A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mas
konstantin123 [22]

Answer:

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

Explanation:

The expression for conservation of the angular momentum (L) is

L_{i} = L_{f}  I_{i}\times\omega_{i} = I_{f}\times\omega_{f}

Where

I_{i}\ and \ \omega_{i} initial moment of inertia and angular velocity

I_{f}\ and \ \omega_{f} is the final moment of inertia and angular velocity

The expression of moment of inertia of the satellite (a solid sphere) is

I_{i} = \frac{2}{5}m_{s}r^{2}

Where m_{s} is the satellite mass

r is the  radus of the sphere

Substititute 1900kg for m and 4.6m for r

I_{i} = \frac{2}{5}m_{s}r^{2}\\\\ = \frac{2}{5}\times1900 kg\times (4.6 m)^{2} \\\\= 1.61 \cdot 10^{4} kgm^{2}

The final moment of inertia of the satellite about the centre of mass

I_{f} = I_{i} + 2\timesI_{x} \\\\= 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}m_{x}l^{2}

Where m_{x} is the antenna's mass and

I is the length of the antenna

I_{f} = 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}150.0 kg\times(6.6 m)^{2} \\\\= 2.05 \cdot 10^{4} kgm^{2}

So, the Final rotation rate of the satellite is:

I_{i}\times\omega_{i} = I_{f}\times\omega_{f} \\\\\omega_{f} = \frac{I_{i}\times\omega_{i}}{I_{f}} \\\\= \frac{1.61 \cdot 10^{4} kgm^{2}\times8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kgm^{2}} \\\\= 6.3 rev/s

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

5 0
3 years ago
Can someone help me with these physics questions?
wariber [46]

hi! Glad to help =)

1. would be c

2 would be d

3 would be A

6 0
3 years ago
A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?
Triss [41]

Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

H_m = \frac{u^2sin^2 \theta}{2g}

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta  = sin^{-1}(0.2123)\\\\\theta  = 12.26^0

Therefore, the angle of projection is 12.26⁰.

6 0
2 years ago
Two speakers are spaced 15 m apart and are both producing an identical sound wave. You are standing at a spot as pictured. What
IgorLugansk [536]

Answer:

Frequency is 213.04\ s^{-1}.

Explanation:

Distance between source 1 from the receiver , S_1 =\sqrt{10^2+22^2}=24.17\ m.

Distance between source 2 from the receiver , S_2=\sqrt{5^2+22^2}=22.56\ m.

Now ,

Path difference , r = S_1-S_2=24.17-22.56=1.61\ m.

We know, for constructive interference path difference should be integral multiple of wavelength .  

Therefore, r=n\times \lambda

It is given that n = 1,

Therefore, \lambda=1.61\ m.

Frequency can be found by , \nu=\dfrac{v}{\lambda}= \dfrac{343}{1.61}=   213.04\ s^{-1} .

Hence, this is the required solution.

5 0
3 years ago
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