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3 years ago

2 Points

1 answer:

5 0

The advantage is that we do not run out of resources and a disadvantage is that is dangerous when a “human” gets too close and gets sick by the radiation.

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a car initially at 39.40 m/s accelerates at -3.00 m/s^2 how far has a car travel when it comes to a stop 9.35 seconds later

Kipish [7]

Answer:

499.523. meter

I hope it's helps you

6 0

3 years ago

The Force of friction between an object and the surface upon which it is sliding is 12N. The weight of the object is 20N. What i

shtirl [24]

Answer: The coefficient of kinetic friction is μ = 0.6

Explanation:

For an object of mass M, the weight is:

W = M*g

where g is the gravitational acceleration: g = 9.8m/s^2

And the friction force between this object and the surface can be written as:

F = W*μ

where μ is the coefficient of friction (kinetic if the object is moving, and static if the object is not moving, usually the static coefficient is larger)

In this case, the weight is:

W = 20N

And the friction force is:

F = 12N

Replacing these values in the equation for the friction force we get:

12N = 20N*μ

(12N/20N) = μ = 0.6

The coefficient of kinetic friction is μ = 0.6

7 0

3 years ago

Read 2 more answers

The diagram shows a person holding a bow and arrow.

denis-greek [22]

Releasing the string

8 0

4 years ago

Read 2 more answers

An electron enters a region between two large parallel plates made of aluminum separated by a distance of 2.0 cm and kept at a p

kotykmax [81]

Answer:

a) $v=1.77\times 10^6\ m/s$

b) $v=3.872\times 10^6\ m/s$

c) $v=5.5\times 10^6\ m/s$

d) $v=6.7\times 10^6\ m/s$

e) $v=7.7\times 10^6\ m/s$

Explanation:

Given that

d = 2 cm

V = 200 V

$u=4\times 10^5\ m/s$

We know that

F = E q

F = m a

E = V/d

m a = q .V/d b

$a=\dfrac{q.V}{m.d}$ ---------1

The mass of electron

$m=9.1\times 10^{-31}\ kg$

The charge on electron

$q=1.6\times 10^{-19}\ C$

Now by putting the all values in equation 1

$a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2$

$a=1.5\times 10^{15}\ m/s^2$

We know that

$v^2=u^2+2as$

s = 0.1 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}$

$v=\sqrt{3.16\times 10^{12}}\ m/s$

$v=1.77\times 10^6\ m/s$

s = 0.5 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}$

$v=\sqrt{1.5\times 10^{13}}\ m/s$

$v=3.872\times 10^6\ m/s$

s = 1 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1\times 10^{-2}$

$v=\sqrt{3.06\times 10^{13}}\ m/s$

$v=5.5\times 10^6\ m/s$

s = 1.5 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1.5\times 10^{-2}$

$v=\sqrt{4.5\times 10^{13}}\ m/s$

$v=6.7\times 10^6\ m/s$

s = 2 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 2\times 10^{-2}$

$v=\sqrt{6.06\times 10^{13}}\ m/s$

$v=7.7\times 10^6\ m/s$

5 0

4 years ago

Read 2 more answers

You are given four resistors, 2 ohms, 3 ohms, 5 ohms, and 10 ohms. Your friend say you can connect them so you obtain an equival

AlexFokin [52]

If we will connect the resistors 2ohms, 3ohms, 5ohms in series and the 10ohms resistance parallel then we get equivalent resistance of 5 ohms.

The equivalent circuit is,

R equivalent for the series connection is,

$\begin{gathered} Req(S)=2+3+5=10ohms \\ Now,\text{ } \\ Req\text{ }for\text{ }10\text{ }ohms\text{ }and\text{ }10\text{ }ohms\text{ }is, \\ \frac{1}{Req}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5} \\ So, \\ Req=5\text{ ohms} \end{gathered}$

The equivalent resistance is 5 ohms.

So your friend is saying true.

6 0

1 year ago