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malfutka [58]
3 years ago
12

What career would need to know Greatest Possible Error and why?

Physics
1 answer:
horsena [70]3 years ago
5 0
<span>uclear Physicist specializing inoffensive weapons systems - pretty obvious why this would be. Highway Engineer - again, pretty obvious reasons.</span>
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Some radar systems detect the size and shape of objects such as aircraft and geological terrain. What is the frequency of such a
vova2212 [387]

Answer:

f=1.57\times 10^9\ Hz

Explanation:

Given that,

A system can detect objects as small as 19.1 cm i.e. 0.191 m. It is the wavelength.

We know that,

Frequency, f=\dfrac{c}{\lambda}

So,

f=\dfrac{3\times 10^8}{0.191}\\\\=1.57\times 10^9\ Hz

So, the frequency of such a system is equal to1.57\times 10^9\ Hz.

4 0
3 years ago
I don’t get this, i’m on a timer so please help
Svetlanka [38]

right angle Answer:

Explanation:

5 0
3 years ago
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A spherically symmetric charge distribution has a charge density given by ρ = a/r , where a is constant. Find the electric field
Airida [17]
Charge dQ on a shell thickness dr is given by 

dQ = (charge density) × (surface area) × dr 

dQ = ρ(r)4πr²dr 

∫ dQ = ∫ (a/r)4πr²dr 

∫ dQ = 4πa ∫ rdr 

Q(r) = 2πar² - 2πa0² 

Q = 2πar² (= total charge bound by a spherical surface of radius r) 

Gauss's Law states: 

(Flux out of surface) = (charge bound by surface)/ε۪ 

(Surface area of sphere) × E = Q/ε۪ 

4πr²E = 2πar²/ε۪ 

<span>E = a/2ε۪


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

</span>
3 0
3 years ago
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An 0.80-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the ba
Dimas [21]

Answer:

a) B=2.9891\times 10^{-5}\,T

b)  west end of the bar is positive.

Explanation:

Given:

  • emf induced in the bar, \epsilon=5.5\times 10^{-4}\,V
  • length of the bar, l=0.8\,m
  • velocity of the bar at the given instant, v=23\,m.s^{-1}

(a)

The magnitude of the horizontal component of the Earth's magnetic field(B):

We know:

\epsilon=B.l.v

5.5\times 10^{-4}=B\times 0.8\times 23

B=2.9891\times 10^{-5}\,T

(b)

Using Fleming's left hand rule we determine that the current is flowing towards east end of the bar i.e. west end of the bar is positive.

4 0
3 years ago
Puck 1 is moving 10 m/s to the left and puck 2 is moving 8 m/s to the right. They have the same mass, m.
Julli [10]

Answer:

(a) the total momentum of the system before the collision = -2m kg.m/s.

(b) the total momentum of the system after the collision = -2m kg.m/s.

(c) puck 1's velocity after the collision in component form = (5.44 i, 2.54 j)

Explanation:

Given;

mass of Puck 1 , = m

mass of Puck 2, = m (since they have the same mass m)

initial velocity of Puck 1, u₁ = 10 m/s to the left

initial velocity of Puck 2, u₂  = 8 m/s to the right

Let the rightward direction be positive direction

Let the leftward direction be negative direction

(a) the total momentum of the system before the collision;

P₁ = (initial momentum of Pluck 1) + (initial momentum of Pluck 2)

P₁ = (-mu₁) + mu₂

P₁ = mu₂ - mu₁

P₁ = m(u₂ - u₁)

P₁  = m(8 - 10)

P₁  = -2m kg.m/s

(b) the total momentum of the system after the collision;

Based on the principle of conservation of linear momentum, the total momentum before collision is equal to the total momentum after collision.

Thus, the total momentum of the system after the collision is -2m kg.m/s.

(c) puck 1's velocity after the collision in component form

v = (v_x, v_y)\\\\v = (vcos \theta , vsin \theta)\\\\v = (6cos 25^0 , 6sin25^0)\\\\v = (5.44i, 2.54j)m/s

8 0
3 years ago
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