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Anarel [89]
3 years ago
11

Where in the cell can you find the location of inherited traits?

Physics
2 answers:
riadik2000 [5.3K]3 years ago
8 0
D.n.a is the answer
sashaice [31]3 years ago
5 0

Answer:

D.N.A

Explanation:

The d.n.a is where all the heriditary information is stored.

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Plz help me I’m a desperate little boy
Neporo4naja [7]

It is the first one. But if you have this question you are not a little boy

5 0
3 years ago
The speed of a wave on a violin A string is 288 m/s and on the G string is 128 m/s. The force exerted on the ends of the string
Katyanochek1 [597]

Answer:

\dfrac{\mu_A}{\mu_G}=0.197

Explanation:

given,

Speed of a wave on violin A = 288 m/s

Speed on the G string = 128 m/s

Force at the end of string G  = 110 N

Force at the end of string A = 350 N

the ratio of mass per unit length of the strings (A/G). = ?

speed for string A

 v_A = \sqrt{\dfrac{F_A}{\mu_A}}.......(1)

speed for string G

 v_G = \sqrt{\dfrac{F_G}{\mu_G}}........(2)

Assuming force is same in both the string

now,

dividing equation (2)/(1)

\dfrac{v_G}{v_A}=\dfrac{\sqrt{\dfrac{F_G}{\mu_G}}}{\sqrt{\dfrac{F_A}{\mu_A}}}

\dfrac{v_G}{v_A}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}

\dfrac{128}{288}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}

\dfrac{\mu_A}{\mu_G}=0.197

5 0
3 years ago
a man is walking inside a bus which is travelling at 56.2 km/h.if the speed of the man relative to the ground is 55.8 km/h,is he
nlexa [21]

Answer:

rawr

Explanation:

6 0
2 years ago
A penguin slides at a constant velocity of 3.57 m/s down an icy incline. The incline slopes above the horizontal at an angle of
igomit [66]

Answer:t=0.81 s

Explanation:

Given

Penguin slides down with constant velocity of 3.57 m/s

as the Penguin Slides with constant velocity therefore F_{net} is zero on Penguin

F_{net}=mg\sin \theta -f_r

f_r=friction Force

f_r=\mu mg

\mu =coefficient of Kinetic friction

mg\sin \theta =\mu mg

\tan \theta =\mu

\mu =\tan 9.85=0.45

after reaching on floor final velocity of penguin will be zero after time t

thus

v=u+at

here a=\mu g

a=0.45\times 9.8=4.41  (deceleration)

0=3.57-4.41\times t

t=\frac{3.57}{4.41}=0.809

t=0.81 s

8 0
3 years ago
A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r
Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

Explanation:

Given that,

Angular acceleration \alpha=\pi\ rad/s^2

Angular displacement \theta=\pi\ rad

Angular velocity \omega =2\pi\ rad/s

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}

\omega_{0}^2=\omega^2-2\alpha\theta

Where, \alpha = angular acceleration

\omega = angular velocity

Put the value into the formula

\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi

\omega=\sqrt{2\pi^2}

\omega_{0}=\pi\sqrt{2}\ rad/s

Hence, The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

3 0
3 years ago
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