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aev [14]
2 years ago
13

2) A skier stands at rest and begins to ski downhill with an acceleration of 3.0 m/s² {downhill). What is

Physics
1 answer:
Finger [1]2 years ago
5 0

Answer:

337.5m

Explanation:

<u>Kinematics</u>

Under constant acceleration, the kinematic equation holds:

s=\frac{1}{2}at^2+v_ot+s_o, where "s" is the position at time "t", "a" is the constant acceleration, "v_o" is the initial velocity, and s_o is the initial position.

<u>Defining Displacement</u>

Displacement is the difference in positions: s-s_o or \Delta s
s=\frac{1}{2}at^2+v_ot+s_o

s-s_o=\frac{1}{2}at^2+v_ot

\Delta s=\frac{1}{2}at^2+v_ot

<u>Using known information</u>

Given that the initial velocity is zero ("skier stands at rest"), and zero times anything is zero, and zero plus anything remains unchanged, the equation simplifies further to the following:

\Delta s=\frac{1}{2}at^2+v_ot

\Delta s=\frac{1}{2}at^2+(0)*t

\Delta s=\frac{1}{2}at^2+0

\Delta s=\frac{1}{2}at^2

So, to find the displacement after 15 seconds, with a constant acceleration of 3.0 m/s², substitute the known values, and simplify:

\Delta s=\frac{1}{2}at^2

\Delta s=\frac{1}{2}(3.0[\frac{m}{s^2}])(15.0[s])^2

\Delta s=337.5[m]

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Explanation:

6 0
3 years ago
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A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

6 0
4 years ago
.2, A car starting from rest has an acceleration of
riadik2000 [5.3K]

Answer: 2.5 m/s and 6.25 m

Explanation:

u = 0

a = 0.5 m/s²

t = 5 s

v = u + at

=  0 + 0.5 × 5

= <u>2.5 m/s</u>

s = ut + 1/2 at²

= 1/2 × 2.5 × 5

=<u> 6.25 m</u>

7 0
4 years ago
A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be
Artyom0805 [142]

Explanation:

It is known that relation between force and acceleration is as follows.

                      F = m \times a

I is given that, mass is 1090 kg and acceleration is 21 m/s. Therefore, we will calculate force as follows.

              F = m \times a      

                 = 1090 \times (\frac{21}{16})

                 = 1430.625 N

Also, it is known that

      sin(\theta) = \frac{\text{Force car can exert}}{\text{Force gravity pulls car}}

      sin(\theta) = \frac{1430.625 N}{(1090 \times 9.8) N}

        \theta = 7.70 degrees

Thus, we can conclude that the maximum steepness for the car to still be able to accelerate is 7.70 degrees.

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3 years ago
Suppose the carts collided on a surface that had a slight incline to it. Would you expect the momentum to be conserved
lesantik [10]

Momentum is conserved when carts are collided on a slanting plane.

To find the answer, we need to know about the conversation of momentum.

<h3>What's the conversation of momentum?</h3>
  • Conservation of linear momentum says the total momentum before the collision and after the collision remains the same.
  • Mathematically, m1u1+m2u2 = m1v1+m2v2
<h3>How is the momentum conserved when collision occurs on a slanting plane?</h3>
  • On a slanting plane, the velocity has two components,
  1. horizontal component
  2. horizontal component Vertical component
  • So, its momentum has also similar two components.
  • The momentum is conserved along horizontal direction and vertical direction separately.

Thus, we can conclude that the momentum is conserved when carts are collided on a slanting plane.

Learn more about the conversation of momentum here:

brainly.com/question/7538238

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7 0
2 years ago
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