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Rzqust [24]
3 years ago
5

To get off a frozen lake, a 70 kg person removes his shoe of mass 0.175 kg and throws it horizontally away from the shore at a v

elocity of 3.2 m/s. If the person is 5.15 m from the shore, how long do they take to reach the shore?
Physics
1 answer:
MakcuM [25]3 years ago
3 0

Answer:

the person will be in the shore at 10.73 minutes after launch the shoe.

Explanation:

For this we will use the law of the lineal momentum.

L_i = L_f

Also,

L = MV

where M is de mass and V the velocity.

replacing,

M_i V_i = M_{fp}V_{fp} + M_{fz}V_{fz}

wher Mi y Vi are the initial mass and velocity, Mfp y Vfp are the final mass and velocity of the person and Mfz y Vfz are the final mass and velocity of the shoe.

so, we will take the direction where be launched the shoe as negative. then:

(70)(0) = (70-0.175)(V_fp) + (0.175)(-3.2m/s)

solving for V_fp,

V_fp = \frac{(3.2)(0.175)}{69.825}

V_fp = 0.008m/s

for know when the person will be in the shore we will use the rule of three as:

1 second -------------- 0.008m

t seconds-------------- 5.15m

solving for t,

t = 5.15m/0.008m

t = 643.75 seconds = 10.73 minutes

 

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An electric motor rotating a workshop grinding wheel at 1.06 102 rev/min is switched off. Assume the wheel has a constant negati
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Answer:

t = 106π / 30*2.1

Explanation:

w_{i} = 1.06*10^{2}    => 106

    => 106 x 2π/60

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∴ (w_{f} - w_{i}) / ∝ = t

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A toy car starts from the rest and accelerates at 1.50m/s2 [E] for 2.25s. What is the final velocity, of the car
V125BC [204]

3.375m/s is the final velocity of the car.

<h3>How do you find final velocity?</h3>

The final velocity depends on how large the acceleration is and the distance over which it acts.

Initial velocity of an object, you can multiply the acceleration due to a force by the time the force is applied and add it to the initial velocity to get the final velocity.

According to the question,

A toy car starts from the rest and accelerates

So the acceleration = 1.50m/s²

Time =  2.25s

x=x_{0}  + vt

x = 0 + ( 1.50m/s^2*2.25s)

x = 3.375m/s

The final velocity, of the car is 3.375 m/s.

Learn more about velocity here:brainly.com/question/18084516

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3 years ago
Read 2 more answers
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and
Leya [2.2K]

Answer:

A) 0.9844 s

B) x2 = 0.4587 m

C) v = 6.657 m/s

Explanation:

We are given;

Height of take off point above pool; x1 = 1.8 m

Initial take off velocity; u = 3 m/s

Final velocity at highest point before free fall; v = 0 m/s

B) To find the highest point above the board her feet reaches means the distance from take off to the top of the motion just before free fall.

Thus, we will be using equation of motion and we have;

v² = u² + 2gs

Now, let s = x2 which will be the distance between take off and the top before free fall.

So;

v² = u² + 2g(x2)

Now,since the motion is against gravity, g will be negative.

Thus;

v² = u² + 2(-9.81)(x2)

Plugging in the relevant values to give;

0² = 3² - (19.62x2)

19.62(x2) = 9

x2 = 9/19.62

x2 = 0.4587 m

A) We want to find how long her feet is in air.. It means we want to find out the time to get to a distance of x1 and also the time to achieve the distance (x1 + x2) on free-fall.

Thus, using equation of motion;

v = u + gt

Again, g = -9.81

Thus;

0 = 3 - 9.81t1

9.81t1 = 3

t1 = 3/9.81

t1 = 0.3058 s

Now, for the time taken to achieve the distance (x1 + x2) on free-fall, we will use the formula;

s = ut + ½gt²

Where s = (x1 + x2) = 1.8 + 0.4587 = 2.2587 m

And now, u = 0 m/s because the start of the free fall is from maximum height with velocity of 0 m/s. Again, g = - 9.81 m/s²

Thus;

2.2587 = 0 - ½(-9.81)(t2)²

2.2587 = 4.905(t2)²

(t2)² = 2.2587/4.905

(t2)² = 0.4605

t2 = √0.4605

t2 = 0.6786 s

Thus, total time of feet in air = t1 + t2 = 0.3058 + 0.6786 = 0.9844 s

C) Velocity when feet hit the water would be given by;

v = u + gt

Where u = 0 m/s and t = t2 = 0.6786

Since it's in direction of gravity, g = 9.81 m/s

v = 0 + (0.6786 × 9.81)

v = 6.657 m/s

4 0
3 years ago
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