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KIM [24]
2 years ago
11

I need heelp can anyone heelp me plz

Physics
1 answer:
GuDViN [60]2 years ago
7 0

Answer:

e) 1.04

Explanation:

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Write Newton's 3rd law. How would this law relate to a rocket ship taking off from the earth? Would this law affect the rocket s
irina1246 [14]

Answer:

Part A

Newton's 3rd law states that action and reaction are equal and opposite, mathematically, we have;

F_A = -F_B

Where;

F_A = The action force

F_B = The reaction force

Part B

The law indicates that the force with which a rocket ship uses in taking off from the Earth, F_A is equal in magnitude, and opposite in direction to the reaction force of the Earth to the motion of the rocket, (-)F_B

Part C

The law is a universal law, and it will also affect the rocket ship in space, as the force of the jet from the exhaust is directed towards Earth while in space, the rocket is propelled deeper into space

Explanation:

6 0
3 years ago
Three identical train cars, coupled together, are rolling east at 1.8 m/s . A fourth car traveling east at 4.5 m/s catches up wi
Vedmedyk [2.9K]

Answer:

v = 1.98\ m/s

Explanation:

consider the mass of each train car be m

m₁ = m₂ = m₃ = m

speed of the three identical train

u₁ = u₂ = u₃ = 1.8 m/s

m₄ = m             u₄ = 4.5 m/s

m₅ = m              u₅ = 0 (initial velocity )

final velocity

v₁ = v₂ = v₃ = v₄ = v₅ = v

using conservation of momentum

m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅

m (1.8 + 1.8 + 1.8 +4.5) = 5 m v

v = \dfrac{9.9}{5}

v = 1.98\ m/s

5 0
3 years ago
I am trying to find the magnitude of a resultant vector. Do i take inconsideration the negatives when i find the x & y compo
attashe74 [19]
Absolutely !  If you have two vectors with equal magnitudes and opposite
directions, then one of them is the negative of the other.  Their correct
vector sum is zero, and that's exactly the magnitude of the resultant vector.

(Think of fifty football players pulling on each end of the rope in a tug-of-war. 
Their forces are equal in magnitude but opposite in sign, and the flag that
hangs from the middle of the rope goes nowhere, because the resultant
force on it is zero.)

This gross, messy explanation is completely applicable when you're totaling up
the x-components or the y-components.
4 0
3 years ago
A 60kg bicyclist (including the bicycle) is pedaling to the
Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight (W=mg): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force (F_A): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag (R): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

F_{net}=ma

where

F_{net} is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

a=3.1 m/s^2 is its acceleration

Substituting, we find the net force on the bicyclist:

F_{net}=(60)(3.1)=186 N

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

F_{net}=F_a-R

where:

F_{net} is the net force

F_a is the applied force (forward)

R is the air drag (backward)

In this problem we have:

F_{net}=186 N is the net force (found in part b)

R=60 N is the magnitude of the air drag

Solving for F_a, we find the force produced by the bicyclist while pedaling:

F_a=F_{net}+R=186+60=246 N

3 0
3 years ago
Explanation for question i & ii. Thank you.
suter [353]
Do u know Chinese ?if yes I can explain to u easily
6 0
3 years ago
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