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nignag [31]
2 years ago
14

Which of the following is true about mutations in somatic cells?

Physics
1 answer:
S_A_V [24]2 years ago
6 0

Answer:

B.) They affect the reproductive cells.

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Timed! I would really appreciate some help! thank you!
GenaCL600 [577]

Answer:

x = 5[km]

Explanation:

We must convert the time from minutes to hours.

t=30[min]*\frac{1h}{60min}= 0.5[h]\\

We know that speed is defined as the relationship between space and time.

v=x/t

where:

x = space [m]

t = time = 0.5 [h]

v = velocity [m/s]

Now replacing:

x = 10[\frac{km}{h} ]*0.5[h]\\x=5[km]

4 0
2 years ago
Priyadarshini means..........​
Blizzard [7]

Answer:

A submission from India says the name Priyadharshini means "Beloved and pleasing to look at" and is of Sanskrit origin. According to a user from India, the name Priyadharshini is of Indian (Sanskrit) origin and means "God gift".

6 0
2 years ago
Read 2 more answers
A skydiver jumps out of a plane flying at an altitude of 5,500 meters. How fast is the skydiver going when they get to the groun
andrey2020 [161]

Answer:

2,500 feet (760 meters)

Explannation: <em>At about 2,500 feet (760 meters), the skydiver throws out a pilot chute, and it deploys the parachute. Its used to control the fall rate.</em>

4 0
3 years ago
The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5
ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E =  K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) =  7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) =  4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.89e5)^{2}  + (7.89e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) net magnitude =  1.115e6\frac{N}{C} @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

 E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.88e5)^{2}  + (7.88e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) and (b) the net magnitude =  1.242e6\frac{N}{C} @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0
3 years ago
In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen i
pantera1 [17]

Answer:

Δx = 4.68 x 10⁻³ m = 4.68 mm

Explanation:

The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:

Δx = λD/d

So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:

Δx = 4λD/d

where,

Δx = distance between eighth order maximum and fourth order maximum=?

λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m

d = slit separation = 0.2 mm = 2 x 10⁻⁴ m

D = Distance between slits and screen = 48 cm = 0.48 m

Therefore,

Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)

<u>Δx = 4.68 x 10⁻³ m = 4.68 mm</u>

5 0
3 years ago
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