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3 years ago

What is primary sex organs called

2 answers:

6 0

The primary sex organs for men are called testes and ovaries for the women. They produce sex cells called gametes and they secrete sex hormones.

atroni [7]3 years ago

5 0

They're called the <span>reproductive organs.

</span>For a male >> testicles or testes

For a female >> Ovaries

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A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r

sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana $m_{b}$ = 8.97 kg

at distance r = 4/5 { radius of platform }

mass of monkey $m_{m}$ = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + $m_{b}$ ( $\frac{4}{5}$ R)² + $m_{m}$ R² ]w

m/2 × ω₀ = [ m/2 + $m_{b}$ ( $\frac{4}{5}$ )² + $m_{m}$ ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

8 0

2 years ago

Which person will most likely hear the loudest sound?

Nataliya [291]

Person standing on A will hear the loudest sound

Explanation:

The intensity of a sound wave (which is proportional to the loudness of the sound) follows an inverse square law, which is:

$I\propto \frac{1}{r^2}$

where

I is the intensity of the wave

r is the distance from the source of the sound

This equation means that the intensity of the sound wave (and therefore, its loudness) is inversely proportional to the square of the distance from the source: therefore,

As we get closer to the source of sound, the loudness increases
As we move away from the source of sound, the loudness decreases

Therefore, the person that will hear the loudest sound is the one standing closer to the source, and therefore person A.

Learn more about sound waves:

brainly.com/question/4899681

#LearnwithBrainly

6 0

3 years ago

Read 2 more answers

Susan found a box of old rocks in her garage. They had large crystals and were labeled

ASHA 777 [7]

Calcium rocks I do believe. Jk

3 0

3 years ago

Read 2 more answers

The Burj Khalifa is the tallest building in the world at 828 m. How much work would a man with a weight of 700 N do if he climbe

Hoochie [10]

Answer:

579600J

Explanation:

Given parameters:

Height of the building = 828m

Weight of the man = 700N

Unknown:

Work done by the man = ?

Solution:

The work done by the man is the same as the potential energy expended.

Work done:

Work done = Weight x height = 700 x 828

Work done = 579600J

6 0

2 years ago

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$