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katovenus [111]
3 years ago
14

True or false. Lateral to the inguinal region is the coxal region, while superior to the scapular region is the dorsal region.

Physics
1 answer:
mars1129 [50]3 years ago
4 0

Answer:

false

Explanation:

false

false

The right answer is false

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A liquid is used to make a mercury-type barometer. The barometer is intended for space-faring astronauts. At the surface of the
Anarel [89]

Answer:

Density of liquid = 4730 kg/m³

Atmospheric pressure on planet X = 8401.7 N/m²

Explanation:

Pressure, P = ρgh where ρ = density of liquid, g =9.8 m/s² and h = height of column at earth's surface = 2185 mm. Since P = atmospheric pressure, for mercury, P = ρ₁gh₁ where ρ₁ = 13.6 g/cm³ and h₁ = 760 mm

So, ρgh = ρ₁gh₁

ρ = ρ₁h₁/h = 13.6 g/cm³ × 760/2185 = 4.73 g/cm³ = 4730 kg/m³

The atmospheric pressure on planet X

P = ρg₁h₃     g₁ = g/4 and h₃ = 725 mm = 0.725 m

on planet X

P = ρg₁h₃ = (4730 kg/m³ × 9.8 m/s² × 0.725 m)/4 = 8401.7 N/m²

6 0
3 years ago
Help!!! Asap!!!
KiRa [710]

Answer:

Probs paper cuz the rest are metals

4 0
1 year ago
A truck is using a hook to tow a car whose mass is one quarter that of the truck. If the force exerted by the truck on the car i
LenKa [72]

Answer:

Force exerted by car on truck will be 6000 N in opposite direction

Explanation:

It is given that mass of the car is one quarter of the mass of the truck

Force exerted by the truck on the car is 6000 N

We have to find the force exerted by car on the truck

According to newtons third law for any action there is equal and opposite reaction

So force exerted by car on the truck will be equal to 6000 N in opposite direction

7 0
2 years ago
Are Styrofoam and wood the same temperature, if they are not what are each others temperature
Olenka [21]
They are almost the same temperature!!
3 0
3 years ago
A spring has a force constant of 310.0 N/m. (a) Determine the potential energy stored in the spring when the spring is stretched
skad [1K]

Answer:

(a) 0.2618 J

(b)  0.1558 J

(c) 0 J

Explanation:

from Hook's Law,

The energy stored in a stretched spring = 1/2ke²

Ep = 1/2ke² ......................... Equation 1

Where k = spring constant, e = extension, E p = potential energy stored in the spring.

(a) When The spring is stretched to 4.11 cm,

Given: k = 310 N/m, e = 4.11 cm = 0.0411 m

Substituting these values into equation 1

Ep = 1/2(310)(0.0411)²

Ep = 155(0.0016892)

Ep =155×0.0016892

Ep = 0.2618 J.

(b) When the spring is stretched 3.17 cm

e = 3.17 cm = 0.0317 m.

Ep = 1/2(310)(0.0317)²

Ep = 155(0.0317)²

Ep = 155(0.0010049)

Ep = 0.155758 J

Ep ≈ 0.1558 J.

(c) When the spring is unstretched,

e = 0 m, k = 310 N/m

Ep = 1/2(310)(0)²

Ep = 0 J.

6 0
3 years ago
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