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3 years ago

A mother has the genotype Bb for brother green eyes, the months did heterozygous hair, what is the genotype of the father ?

Physics

1 answer:

natali 33 [55]3 years ago

7 0

Assuming brown hair (B) color is completely dominant over blond hair (b), then the father's genotype must be bb.

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The minimum stopping distance of a car moving at 20.5 mi/h is 11.6 m. Under the same conditions (so that the maximum braking for

pshichka [43]

Answer:

d = 69 .57 meter

Explanation:

First case

Speed of car ( v ) = 20.5 mi/h = 9.164 M/S

distance ( d ) = 11.6 meter ( m = mass of the car )

Work done = 0.5 m v² = 0.5 * 9.164² * m J = 41.99 m J

Force = ( workdone /distance ) = ( 41.99 m / 11.6 ) = 3.619 m N

Second case

v = 50.2 mi/h = 22.44135 m/s

d = ?

Work done = 0.5 * 22.44² * m J = 251.7768 * m J

Since the braking force remains the same .

3.619 m = ( 251.7768 m / d )

d = 69 .57 meter

7 0

3 years ago

Why are peer reviews important?

riadik2000 [5.3K]

Peer review is important because it is used by scientists to decided which results should be published in a scientific journal

8 0

3 years ago

Read 2 more answers

An object has a momentum of 4,000 kg-m/s and a mass of 115 kg. It crashes into another object that has a mass of 100 kg, and the

ANTONII [103]

Answer:

18.60 m/s

Explanation:

Original momentum = mv = 4000 with m = 115

after collision m = 115 + 100 = 215 kg

but the total momentum is still the same (conserved)

4000 = 215 v shows v = 18.60 m/s

4 0

2 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

2 years ago

Read 2 more answers

A horizontally oriented pipe has a diameter of 5.8 cm and is filled with water. The pipe draws water from a reservoir that is in

Gnoma [55]

Answer:

v₂ = 97.4 m / s

Explanation:

Let's write the Bernoulli equation

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Index 1 is for tank and index 2 for exit

We can calculate the pressure in the tank with the equation

P = F / A

Where the area of a circle is

A = π r²

E radius is half the diameter

r = d / 2

A = π d² / 4

We replace

P = F 4 / π d²2

P₁ = 397 4 /π 0.058²

P₁ = 1.50 10⁵ Pa

The water velocity in the tank is zero because it is at rest (v1 = 0)

The outlet pressure, being open to the atmosphere is P1 = 1.13 105 Pa

Since the pipe is horizontal y₁ = y₂

We replace on the first occasion

P₁ = P₂ + ½ ρ v₂²

v₂ = √ (P1-P2) 2 / ρ

v₂ = √ [(1.50-1.013) 10⁵ 2/1000]

v₂ = 97.4 m / s

6 0

3 years ago