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2 years ago

Ann is driving down a street at 64 km/h.

Physics

2 answers:

aksik [14]2 years ago

8 0

Answer:

13.5 m

Explanation:

First, convert km/h to m/s:

64 km/h × (1000 m / km) × (1 h / 3600 s) = 17.8 m/s

Distance = rate × time

d = 17.8 m/s × 0.761 s

d = 13.5 m

Kobotan [32]2 years ago

6 0

Answer:

13.5 m

Explanation:

64 km * 1000 m/km

-----------------------

1 hr * 3600 s/hr

= 17.8 m/s

d = vt

= (17.8 m/s)(0.761 s)

= 13.5 m

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A child rides a wagon down a hill. Eventually, the wagon comes to a stop. Which is most responsible for causing the wagon to sto

Dvinal [7]

Answer:

well, the hill isn't constantly going down hill, there's an ending point or goes back up hill making a v/u shape or there's nothing helping the wagon being pushed or pull currently

Explanation:

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3 years ago

A child and sled with a combined mass of 50.0 kg slide down a frictionless slope. if the sled starts from rest and has a speed o

Furkat [3]

<span> Using conservation of energy

Potential Energy (Before) = Kinetic Energy (After)

mgh = 0.5mv^2

divide both sides by m

gh = 0.5v^2

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h = (0.5*2.2^2)/9.81

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4 0

2 years ago

Read 2 more answers

Please, pretty please help me!

Andrews [41]

Answer:

umm

Explanation:

1223565 578633 =334675

8 0

2 years ago

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What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL

lozanna [386]

Answer:

$\displaystyle \Delta x=1.74\ m$

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

$\displaystyle \Delta x=\swrt{\frac{2PE}{k}}$

Substituting:

$\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}$

Calculating:

$\displaystyle \Delta x=\sqrt{3.0175}$

$\boxed{\displaystyle \Delta x=1.74\ m}$

6 0

3 years ago

A rabbit is moving in the positive x-direction at 1.10 m/s when it spots a predator and accelerates to a velocity of 10.9 m/s al

anzhelika [568]

Answer:

aₓ = 0 , ay = -6.8125 m / s²

Explanation:

This is an exercise that we can solve with kinematics equations.

Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.

x axis

vₓ = v₀ₓ = 1.10 m / s

aₓ = 0

y axis

initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration

$v_{y}$ = v_{oy} -ay t

ay = (v_{oy} -v_{y}) / t

ay = (0 -10.9) / 1.6

ay = -6.8125 m / s²

the sign indicates that the acceleration goes in the negative direction of the y axis

8 0

3 years ago