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2 years ago

Charles Darwin,

Physics

2 answers:

Roman55 [17]2 years ago

8 0

The answer is B. Artificial Selection :)

grandymaker [24]2 years ago

3 0

The answer is B artificial selection

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At its widest point, the diameter of a bottlenose dolphin is 0.50 m. Bottlenose dolphins are particularly sleek, having a drag c

fiasKO [112]

Answer:

497.00977 N

3742514.97005

Explanation:

$\rho$ = Density of water = 1000 kg/m³

C = Drag coefficient = 0.09

v = Velocity of dolphin = 7.5 m/s

r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m

A = Area

Drag force

$F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N$

The drag force on the dolphin's nose is 497.00977 N

at 20°C

$\mu$ = Dynamic viscosity = $1.002\times 10^{-3}\ Pas$

Reynold's Number

$Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005$

The Reynolds number is 3742514.97005

8 0

2 years ago

I need help with this physics question

insens350 [35]

It's impossible to describe WHERE a place is without mentioning ANOTHER place.

... Across the street from -- the bank.
... Next door to -- my house.
... 30 miles west of -- Chicago.
... Up above -- the tree.
... Two days ride out of -- Tulsa.
... Halfway home from -- school.
... Twice as far from Earth as -- the moon is.
... The first seat in -- the second row.
... Behind -- the dog's left ear.
... At the bottom of -- the pool.
... On the tip of -- my tongue.
... In the front seat of -- the car.
... I saw you in -- my dream.
... You're always on -- my mind.

The question is trying to get you to realize that to get from a reference point to a certain position, you have to know

How far
and
In what direction.

4 0

3 years ago

Current what can do work in an electric device

Serggg [28]

It could help transport important information

5 0

2 years ago

Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m

otez555 [7]

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

$T = 2 \pi \sqrt\frac{m}{k}$

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

$T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s$

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

5 0

2 years ago

After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg

kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s$

v₂ = 4.2 m/s

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)$

v₂ = 5 m/s

5 0

2 years ago