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ivanzaharov [21]
3 years ago
7

A race car driver loses his leg in an accident. Apply Lamarck's and Darwin's theories of evolution to predict whether this disab

ility will be carried forward by his children.
Physics
1 answer:
ASHA 777 [7]3 years ago
7 0
It will not be carried to children ... because genetic defect can be carried to children, not the defect during his lifetime
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Metal sphere A has a charge of -2 units and an identical sphere B has a charge of -4 units. If the spheres are brought into cont
shusha [124]

Answer:

 q1 = q₂=  -3

therefore each sphere has the same charge of -3 untis

Explanation:

The metallic spheres have mobile charge, so when the two spheres come into contact the total charge

           Q_total = q₁ + q₂

           Q_total = -2 -4

   

          Q_total = -6 units

it is distributed in between the two spheres evenly since the charges of the same sign repel each other.

When the spheres separate each one has

            q₁ = -6/2

            q1 = q₂=  -3

therefore each sphere has the same charge of -3 untis

5 0
3 years ago
A proposed space station includes living quarters in a circular ring 50.0 m in diameter. At what angular speed should the ring r
mel-nik [20]
Given that the space station is free of gravitational force, it is required that it spins an certain speed to acquire centripetal acceleration.

In this case, you want that the centripetal acceleration, Ac, equals g (gravitational acceleration on the earth), becasue this will cause a centripetal force equal to the weight on earth.

The formula for centripetal acceleration is Ac = [angular velocity]^2*R

where R = [1/2]50.0m = 25.0 m

Ac = 9.81 m/s^2

=> [angular velocity]^2 = Ac/R = 9.81m/s^2v/ 25.0m = 0.3924 (rad/s)^2

[angular velocity] = √(0.3924) rad/s = 0.63 rad/s

Answer: 0.63 rad/s

 

5 0
3 years ago
when a temparature of a coin is 75°C, the coin's diameter increases. if the original diameter of a coin is 1.8*10^-2 m and its c
andrey2020 [161]

Answer:

ΔD = 2.29 10⁻⁵ m

Explanation:

This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation

          ΔA = 2α A ΔT

the area is

         A = π r² = π D² / 4

we substitute

         ΔA = 2α π D² ΔT/4

as they do not indicate the initial temperature, we assume that ΔT = 75ºC

    α = 1.7 10⁻⁵ ºC⁻¹

we calculate

          ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4

          ΔA = 6.49 10⁻⁷ m²

by definition

           ΔA = A_f- A₀

           A_f = ΔA + A₀

           A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4

           A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴

           A_f = 2,551 10⁻⁴ m²

the area is

           A_f = π D_f² / 4

           A_f = \sqrt{4  A_f /\pi }

           D_f = \sqrt{4 \ 2.551 10^{-4} /\pi }

           D_f = 1.80229 10⁻² m

the change in diameter is

           ΔD = D_f - D₀

           ΔD = (1.80229 - 1.8) 10⁻² m

           ΔD = 0.00229 10⁻² m

           ΔD = 2.29 10⁻⁵ m

5 0
3 years ago
A 15.0 g bullet is moving to the right with speed 270 m/s when it hits a target and travels an additional 25.0 cm into the targe
Fantom [35]

Answer:

F=2187\ N

Explanation:

Given:

  • mass of bullet,m=15\ g=0.015\ kg
  • initial velocity of bullet, u=270\ m.s^{-1}
  • displacement of the bullet in the target, s=25\ cm=0.25\ m

Here as given in the question the bullet penetrates the target by the given displacement of the bullet into it. During this process it faces deceleration and hence it comes to rest.

  • so, final velocity of the bullet, v=0\ m.s^{-1}

Now using the equation of motion:

v^2=u^2+2a.s

where:

a= acceleration of the bullet

0^2=270^2+2a\times 0.25

a=145800\ m.s^{-2}

<u>Now the force of resistance offered by the target in stopping it:</u>

F=m.a

F=0.015\times145800

F=2187\ N

5 0
3 years ago
IN nuclear in oder to slow down fast neutron the target material should contain
stealth61 [152]

Answer:

it depends if it is alpha ,beta or gamma

Explanation:

if alpha use paper

if beta use some clothing or aluminium

if it is gamma use lead

8 0
2 years ago
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