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allochka39001 [22]
3 years ago
13

If a person is standing erect and flexes the trunk on the hip, the center of mass will move __________ and the line of gravity m

oves __________ within the base of support.
Physics
1 answer:
marta [7]3 years ago
8 0

Answer:

anterior

anterior

Explanation:

In the given question is asked that

If a person is standing erect and flexes the trunk on the hip, the center of mass will move ___________ and the line of gravity moves___________ within the base of support.

The current answer to the blanks will be

anterior

anterior

hope this helps any further query can be asked in comment section.

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For a home sound system, two small speakers are located so that one is 52 cm closer to the listener than the other. What is the
galina1969 [7]

Answer:

The first frequency of audible sound in the speed sound is

f = 662 Hz

Explanation:

vs = 344 m/s

x =  52 cm * 1 / 100m = 0.52m

The wave length is the distance between the peak and peak so

d = 2x

d = 2*0.52 m

d = 1.04 m

So the frequency in the speed velocity is

f = 1 / T

f = vs / x = 344 m/s / 0.52m

f ≅ 662 Hz

7 0
4 years ago
Which of the following is a true statement for a child's toy spinning in a circle at constant speed?
ASHA 777 [7]
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7 0
3 years ago
A crate falls from an airplane flying horizontally at an altitude of 2,000 m.
Hatshy [7]

Answer:

20.2 seconds

Explanation:

The airplane (and therefore the crate) initially has no vertical velocity, so v₀ = 0 m/s.

The crate is in free fall, so a = -9.8 m/s².

The crate falls downward, so Δx = -2000 m.

Find: t, the time it takes for the crate to land.

Δx = v₀ t + ½ at²

-2000 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 20.2 s

It takes 20.2 seconds for the crate to land.

7 0
3 years ago
A certain kind of glass has an index of refraction of 1.640 for blue light of wavelength 440 nm and an index of 1.595 for red li
seraphim [82]

Answer: 18.27°

Explanation:

Given

Index of refraction of blue light, n(b) = 1.64

Wavelength of blue light, λ(b) = 440 nm

Index of refraction of red light, n(r) = 1.595

Wavelength of red light, λ(r) = 670 nm

Angle of incident, θ = 30°

Angle of refraction of red light is

θ(r) = sin^-1 [(n(a)* sin θ) / n(r)], where n(a) = index of refraction of air = 1

So that,

θ(r) = sin^-1 [(1 * sin 30) / 1.595]

θ(r) = sin^-1 (0.5 / 1.595)

θ(r) = sin^-1 0.3135

θ(r) = 18.27°

4 0
3 years ago
A uniform, solid sphere of radius 2.50 cm and mass 4.75 kg starts with a purely translational speed of 3.00 m/s at the top of an
allsm [11]

Answer:

The final translational seed at the bottom of the ramp is approximately 4.84 m/s

Explanation:

The given parameters are;

The radius of the sphere, R = 2.50 cm

The mass of the sphere, m = 4.75 kg

The translational speed at the top of the inclined plane, v = 3.00 m/s

The length of the inclined plane, l = 2.75 m

The angle at which the plane is tilted, θ = 22.0°

We have;

K_i + U_i = K_f + U_f

K = (1/2)×m×v²×(1 + I/(m·r²))

I = (2/5)·m·r²

K =  (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²

U = m·g·h

h = l×sin(θ)

h = 2.75×sin(22.0°)

∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75×v_f² + 0

7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93

∴ 77.93 ≈ 7/10 × 4.75×v_f²

v_f² = 77.93/(7/10 × 4.75)

v_f ≈ √(77.93/(7/10 × 4.75)) ≈ 4.84

The final translational seed at the bottom of the ramp, v_f ≈ 4.84 m/s.

3 0
3 years ago
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