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2 years ago

While on an ice-fishing trip, you find yourself holding a 10 kg fish but stuck in the middle of

Physics

1 answer:

Anna007 [38]2 years ago

3 0

The forward force you exert on the fish and your backward action will allow you to reach the shore.

<h3>Newton's third law of motion</h3>

Newton's third law of motion states that for every action, there is an equal and opposite reaction.

Fa = -Fb

Let's assume the fish is held in the hook, this will give you the opportunity to throw the fish forward while still holding it.

When the the fish is thrown forward, you will move backwards with an equal force based on Newton's third law. Your backward momentum towards the shore will help to maintain equal linear momentum between you and the fish.

Thus, this forward force of the fish and your backward action will allow you to reach the shore.

Learn more Newton's third law of motion here: brainly.com/question/25998091

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6. An elevator takes 33k) of energy to lift a group of people up 6 m. What is the total mass of

Xelga [282]

Answer:

350 Kg

Explanation:

We use the Work Formula:

W = Force . Distance

33.000 Joules = Force . 6

Force = 5500 Newtons

Force = Mass . Acceleration

5500 = Mass . 10

Mass = 550

550 - 200 = 350 Kg

5 0

3 years ago

A) How much does an 85g apple weight on the surface of the Earth? (gEarth = 9.8 m/s2)

liq [111]

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3 years ago

A thin spherical shell with radius R1 = 4.00 cm is concentric with a larger thin spherical shell with radius R2 = 6.00 cm . Both

kakasveta [241]

Answer:

The right response will be "450 volts".

Explanation:

The given values are:

R1 = 4.00 cm

R2 = 6.00 cm

q1 = +6.00 nC

q2 = −9.00 nC

As we know,

The potential difference between the two shell's difference will be:

⇒ $\Delta V=K[(\frac{q1}{R1}+\frac{q2}{R2})-(\frac{q1}{R1} +(\frac{q2}{R2}))]$

$=K[\frac{q1}{R2}-\frac{q1}{R1} ]$

On substituting the values, we get

$=(9\times 10^9)[\frac{6\times 10^{-9}}{0.04}-\frac{6\times 10^{-9}}{0.06}]$

Δ $=450 \ volts$

3 0

3 years ago

In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 3.40 mm . what is the differenc

Strike441 [17]

The solution would be like this for this specific problem:

Given:

diffraction grating slits = 900 slits per centimeter

interference pattern that is observed on a screen from the grating = 2.38m

maxima for two different wavelengths = 3.40mm

slit separation .. d = 1/900cm = 1.11^-3cm = 1.111^-5 m 

Whenas n = 1, maxima (grating equation) sinθ = λ/d
Grant distance of each maxima from centre = y ..
As sinθ ≈ y/D y/D = λ/d λ = yd / D 

∆λ = (λ2 - λ1) = y2.d/D - y1.d/D
∆λ = (d/D) [y2 -y1]

∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m

6 0

3 years ago

express the following in metres (1) 52fm (2) 26 Mm (3)12am (4) 69 pm (5) 85 mm

AlekseyPX

Answer:........u are so sweet

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3 years ago