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Setler79 [48]
3 years ago
8

Please help! This is due tomorrow and I absolutely need help.

Physics
1 answer:
zysi [14]3 years ago
4 0

Answer:

Correct answer:  11. Total distance d = 200m ; 12. Vav = 3.63m/s ;

13. Total displacement Dt = 0m ; 14. V₂(10s-15s) = 0 m/s ;

15. V₃(15s-40s) = 4 m/s ; 16. V₁(0s-10s) = 6 m/s > V₄(40s-55s) = 2.67 m/s

Explanation:

The whole movement can be divided into four stages.

In the first stage the subject moves 60m in a positive direction for 10s,

in the other it is stationary for 5s, in the third it moves 100m in the opposite (negative) direction for 25s and in the fourth in the positive 40m for 15s.

11. Total distance = 60 + 0 + 100 + 40 = 200m

12. The formula for calculating the average speed (velocity) is

Vav = (S₁ + S₂ + S₃ + S₄) / (t₁ + t₂ + t₃ + t₄)

Vav = (60 + 0 + 100 + 40)/ (10 + 5 + 25 + 15) = 200/55 = 3.63 m/s

13. The movement started from the origin and ended at the origin

Total displacement is zero meters.

14. The speed between 10s and 15s is zero, because he did not move.

15. V₃ = S₃/t₃ = 100/25 = 4 m/s

16. V₁ = S₁/t₁ = 60/10 = 6 m/s   and V₄ = S₄/t₄ = 40/15 = 2.67 m/s

V₁ > V₄

God is with you!!!

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Explanation:

It is given that,

Mass of the passenger, m = 75 kg

Acceleration of the rocket, a=49\ m/s^2

(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :

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F = 3675 N

Ratio, R=\dfrac{F}{W}

R=\dfrac{3675}{75\times 9.8}=5

So, the ratio between the horizontal force and the weight is 5 : 1.

(b) The magnitude of total force the seat exerts against his body is F' i.e.

F'=\sqrt{F^2+W^2}

F'=\sqrt{(3675)^2+(75\times 9.8)^2}

F' = 3747.7 N

The direction of force is calculated as :

\theta=tan^{-1}(\dfrac{W}{F})

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\theta=11.3^{\circ}

Hence, this is the required solution.

8 0
4 years ago
At one point in space, the electric potential energy of a 15 nC charge is 42 μJ . Part A) What is the electric potential at this
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Answer:

Part A:

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Part B:

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Explanation:

<u> Part A:</u>

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The potential energy at a point due to a charge is defined as

\rm U=qV.

<em>where</em>,

V = electric potential at that point.

Therefore,

\rm V=\dfrac{U}{q}=\dfrac{42\times 10^{-6}}{15\times 10^{-9}}=2.8\times 10^3\ Volts.

<u>Part B:</u>

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\rm U=q_1V = 20\times 10^{-9}\times 2.8\times 10^3=5.6\times 10^{-5}\ J.

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3 years ago
A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of 2.00 m/s . The coefficient o
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2.04 meters distance is traveled by the sled before stopping.

Mass of the sled = m

The initial speed of the sled = 2 m/s

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Gravity = 9.8 m/ s²

Let the initial kinetic energy sled be

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The work done by the frictional force is,

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W _{f} = μ_{k}mgd

Work done by frictional force= Initial kinetic energy of the sled

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Therefore, the distance traveled by the sled before stopping is 2.04 meters.

To know more about work done, refer to the below link:

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Answer:

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Rate of CO_2 outflow

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Therefore rate of change of CO_2 is \frac{1}{100}-\frac{p}{50}

% rate of change is100\times \ Rate of \ Change=\frac{1}{10}-2p

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Integrate both:

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