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3 years ago

Your environment can play a role in whether or not a specific gene is expressed

Physics

2 answers:

andrey2020 [161]3 years ago

5 0

True is the answer
hope it helped!

Hatshy [7]3 years ago

4 0

I don't think so. How are you supposed to write the answer?

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A rotating wheel requires a time Δt = [01]_____________________ to rotate 37.0 revolutions. Its angular speed at the end of the

Sidana [21]

Answer:

Explanation:

Initial angular velocity ω₁ = 0 , final angular velocity ω₂ = 75.9 rad /s

angle rotated = θ

= 37 x 2π

= 74 π

The formula for angular velocity

ω₂² = ω₁² + 2αθ , α is angular acceleration

75.9² = 0 + 2 α x 74 π

α = 75.9² / 2 x 74 π

= 12.396 rad / s²

8 0

3 years ago

A lightweight string is wrapped several times around the rim of a small hoop. If the free end of the string is held in place and

MariettaO [177]

Answer:

Explanation:

Let T be the tension

For linear motion of hoop downwards

mg -T = ma , m is mass of the hoop . a is linear acceleration of CG of hoop .

For rotational motion of hoop

Torque by tension

T x R , R is radius of hoop.

Angular acceleration be α,

Linear acceleration a = α R

So TR = I α

= I a / R

a = TR² / I

Putting this value in earlier relation

mg -T = m TR² / I

mg = T ( 1 + m R² / I )

T = mg / ( 1 + m R² / I )

mg / ( 1 + R² / k² )

Tension is less than mg or weight because denominator of the expression is more than 1.

5 0

3 years ago

SI UNIT OF GRAVITY is the

pickupchik [31]

Answer:

Explanation:

4 0

3 years ago

If you push any floating object down from equilibrium and release it, it bobs up and down. That looks like an oscillation, so le

GarryVolchara [31]

Answer:

$F_{y}$ = ( ρ_fluid g A) y

Explanation:

This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force

for the first part, let's write Newton's equilibrium equation

B₀ - W = 0

B₀ = W

ρ_fluid g V_fluid = W

the volume of the fluid is the area of the cube times the height it is submerged

V_fluid = A y

For the second part, the body introduces a quantity and below this equilibrium point, the equation is

B - W = m a

ρ_fluid g A (y₀ + y) - W = m a

ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a

ρ_fluid g A y + (B₀-W) = ma

the part in parentheses is zero since it is the force when it is in equilibrium

ρ_fluid g A y = m a

this equation the net force is

$F_{y}$ = ( ρ_fluid g A) y

we can see that this force varies linearly the distance and measured from the equilibrium position

8 0

3 years ago

A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is

malfutka [58]

Answer: A) $6.38(10)^{6} m$

Explanation:

The equation for the moment of inertia $I$ of a sphere is:

$I=\frac{2}{5}mr^{2}$ (1)

Where:

$I=9.74(10)^{37}kg m^{2}$ is the moment of inertia of the planet (assumed with the shape of a sphere)

$m=5.98(10)^{24}kg$ is the mass of the planet

$r$ is the radius of the planet

Isolating $r$ from (1):

$r=\sqrt{\frac{5I}{2m}}$ (2)

Solving:

$r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}}$ (3)

Finally:

$r=6381149.077m \approx 6.38(10)^{6} m$

Therefore, the correct option is A.

4 0

3 years ago