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lions [1.4K]
3 years ago
7

The is the sac-like structure that holds the testes.

Physics
2 answers:
Brilliant_brown [7]3 years ago
5 0

Answer:

Scrotum

Explanation:

Nuetrik [128]3 years ago
4 0

Answer: Scrotum

Explanation: edge

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Hi please help with this question! Need the workings.
andreyandreev [35.5K]
<span>Mass of the copper penny m = 2.6 g Atomic mass of copper = 63.55, Atomic number = 29, So the number of neutrons = Atomic mass - Atomic number = 63 - 29 = 34 a. Neutron mass = 34 x (2.6 / 63.55) = 1.4 grams Copper atoms per mole = 6.040 x 10^23 atoms/mol moles of copper = 2.6 / 63.06 = 0.04123 mol Total atoms in the copper = 6.040 x 10^23 atoms/mol x 0.04123 mol = 0.25 x 10^23 atoms Number of electrons in the copper = 29 per atom Mass of the electron = 9.085 x 10^-28 g b. Electron mass = 0.25 x 10^23 x 29 x 9.085 x 10^-28 = 65.86 x 10^-5 g</span>
6 0
3 years ago
Question 14 of 25
natali 33 [55]

the answer is sueist

Explanation:

7 0
3 years ago
Read 2 more answers
A car travels from point A to point B, moving in the same direction but with a non-constant speed. The first half of the distanc
Dmitrij [34]

Answer:

Explanation:

From A to B

distance traveled with velocity v_1  in time t_1

\frac{d}{2}=v_1t_1----1

from B to C

distance traveled is 0.5 d with v_2  and v_3  velocity for half-half time

\frac{d}{2}=\frac{v_2t_2}{2}+\frac{v_3t_3}{2}----2

divide 1 and 2 we get

\frac{1}{1}=\frac{2v_1t_1}{v_2t_2+v_3t_3}

\frac{t_1}{t_2}=\frac{v_2+v_3}{2v_1}

Now average velocity is given by

v_{avg}=\frac{d}{t_1+t_2}

taking t_1  common

v_{avg}=\frac{2v_1t_1}{t_1(1+\frac{t_2}{t_1})}

v_{avg}=\frac{2v_1}{1+\frac{2v_1}{v_2+v_3}}

v_{avg}=\frac{2v_1(v_2+v_3)}{2v_1+v_2+v_3}  

6 0
3 years ago
When the speed of the bottle is 2 m/s, the KE is kg m2/s2. When the speed of the bottle is 3 m/s, the KE is kg m2/s2. When the s
d1i1m1o1n [39]

mass of the bottle in each case is M = 0.250 kg

now as per given speeds we can use the formula of kinetic energy to find it

1) when speed is 2 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(2)^2 = 0.5 J

2) when speed is 3 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(3)^2 = 1.125 J

3) when speed is 4 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(4)^2 = 2 J

4) when speed is 5 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(5)^2 = 3.125 J

5) when speed is 6 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(6)^2 = 4.5 J

3 0
3 years ago
Read 2 more answers
A body is projected from the ground at an angle of 30° with the horizontal at an initial speed of 128 ft/s. Ignoring air frictio
Elanso [62]

Answer

given,

v = 128 ft/s

angle made with horizontal = 30°

now,

horizontal component of velocity

vx = v cos θ = 128 x cos 30° = 110.85 ft/s

vertical component of velocity

vy = v sin θ = 128 x sin 30° = 64 m/s

time taken to strike the ground

using equation of motion

v = u + at

0 =-64 -32 x t

t = 2 s

total time of flight is equal to

T = 2 t = 2 x 2 = 4 s

b) maximum height

using equation of motion

 v² = u² + 2 a h

 0 = 64² - 2 x 32 x h

 64 h = 64²

  h = 64 ft

c) range

R = v_x × time of flight

R = 110.85 × 4

R = 443.4 ft

4 0
3 years ago
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