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4 years ago

1ST ANSWER WILL BE BRAINLIEST

Physics

2 answers:

Maslowich4 years ago

5 0

All of the above <em>I think but if you have on choice I think it is d </em>

zloy xaker [14]4 years ago

4 0

All of the above. They all right

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Why could the beam of particles not have a neutral charge?

tino4ka555 [31]

Answer:

The positive and negative charges balance each other. Overall, the atom is uncharged (neutral). However, if something happens to make an atom lose or gain an electron then the atom will no longer be neutral. An atom that gains or loses an electron becomes an ion.

Explanation:

I think this is the answer : D

4 0

3 years ago

LOOK AT THE PICTURES 2!

Nadusha1986 [10]

E = 0.5mv^2 = 0.5*44*10^2 = 2200J

5 0

4 years ago

Read 2 more answers

An object moves with constant acceleration 3.10 m/s2 and over a time interval reaches a final velocity of 12.4 m/s. If its initi

Harrizon [31]

Answer:

Explanation:

Given:

a = 3.10 m/s^2

vf = 12.4 m/s

vi = -6.2 m/s

t = (vf - vi)/a

= (12.4 + 6.2)/3.1

= 6 s

displacement = (vf - vi)*t

= (12.4 + 6.2) * 6

= 111.6 m.

3 0

3 years ago

Read 2 more answers

A sample contains radioactive atoms of two types, A and B. Initially there are five times as many A atoms as there are B atoms.

victus00 [196]

Answer:

Explanation:

Initially no of atoms of A = N₀(A)

Initially no of atoms of B = N₀(B)

5 X N₀(A) = N₀(B)

N = N₀ $e^{-\lambda t}$

N is no of atoms after time t , λ is decay constant and t is time .

For A

N(A) = N(A)₀ $e^{-\lambda_1 t}$

For B

N(B) = N(B)₀ $e^{-\lambda_2 t}$

N(A) = N(B) , for t = 2 h

N(A)₀ $e^{-\lambda_1 t}$ = N(B)₀ $e^{-\lambda_2 t}$

N(A)₀ $e^{-\lambda_1 t}$ = 5 x N₀(A) $e^{-\lambda_2 t}$

$e^{-\lambda_1 t}$ = 5 $e^{-\lambda_2 t}$

$e^{\lambda_2 t}$ = 5 $e^{\lambda_1 t}$

half life = .693 / λ

For A

.77 = .693 / λ₁

λ₁ = .9 h⁻¹

$e^{\lambda_2 t}$ = 5 $e^{\lambda_1 t}$

Putting t = 2 h , λ₁ = .9 h⁻¹

$e^{\lambda_2\times 2}$ = 5 $e^{.9\times 2}$

$e^{\lambda_2\times 2}$ = 30.25

2 x λ₂ = 3.41

λ₂ = 1.7047

Half life of B = .693 / 1.7047

= .4065 hours .

= .41 hours .

6 0

3 years ago

A block of mass m1 = 290 g is at rest on a plane that makes an angle θ = 30° above the horizontal. The coefficient of kinetic fr

GrogVix [38]

Answer:

The speed of the block when is fallen 30cm

v=0.726 $\frac{m}{s}$

Explanation:

∑F= (m2)g - ƒ - (m1)g*sin(θ) = (m1)a

g = 9.81 m/s²

ƒ = μN = μ(m1)g

$(m2)*g - u*(m1)*g - (m1)*g*sin(\alpha ) = (m1)*a$

(0.200)(9.81) - (0.1)(0.290)(9.81) - (0.290)(9.81)sin(30°) = (0.290)a

$(0.200kg)(9.81\frac{m}{s^{2}}) - (0.1)(0.290kg)(9.81\frac{m}{s^{2}}) - (0.290kg)(9.81\frac{m}{s^{2}})sin(30°) = (0.290kg)*a$

$0.511101=0.29*a\\a=0.879\frac{m}{s^{2} }$

$v_{f}^{2}=v_{o}^{2}+2*a(x_{f}^{2}-v_{o}^{2})\\v_{o}=0\\v_{o}=0\\v_{f}^{2}=2*a(x_{f})\\v_{f}=\sqrt{2*a(x_{f})}\\v_{f}=\sqrt{2*0.879\frac{m}{s^{2}}*0.30m} \\v_{f}=0.726 \frac{m}{s}$

5 0

4 years ago