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Debora [2.8K]
3 years ago
5

A speeding motorist traveling down a straight highway at 100 km/h passes a parked police car. It takes the police constable 1.0

s to take a radar reading and to start up his car. The police vehicle accelerates from rest at 2 m/s2 and finally catches up with the speeder. a) How much time has elapsed when the two cars meet?
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
6 0

Answer:

t = 7.5 s

Explanation:

The distance traveled by the car at the time of meeting of the two cars must be the same. First, we calculate the distance traveled by the police car. For that we use 2nd equation of motion. Here, we take the time when police car starts to be reference. So,

s₁ = Vi t + (0.5)gt²

where,

s₁ = distance traveled by police car

Vi = Initial Velocity = 0 m/s

t = time taken

Therefore,

s₁ = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

s₁ = 4.9 t²

Now, we calculate the distance traveled by the car. For constant speed and time to be 1 second more than the police car time, due to car starting time, we get:

s₂ = Vt = V(t + 1)

where,

s₂ = distance traveled by car

V = Velocity of car = (100 km/h)(1000 m/1 km)(1 h/ 3600 s) = 27.78 m/s

Therefore,

s₂ = 27.78 t + 27.78

Now, we know that at the time of meeting:

s₁ = s₂

4.9 t² = 27.78 t + 27.78

4.9 t² - 270.78 t - 27.78 = 0

solving the equation and choosing the positive root:

t = 6.5 s

since, we want to know the time from the moment car crossed police car. Therefore, we add 1 second of starting time in this.

t = 6.5 s + 1 s

<u>t = 7.5 s</u>

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Fudgin [204]

Answer:

\mu_k = 0.15

Explanation:

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v^{2} - u^{2} = 2aS

Where

u is initial velocity  = 0 m/s

a = acceleration

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final velocity = 1.0 m/s

a = \frac {v^{2}}{2S}

a = \frac {1{2}}{2*8.6}

a = 0.058 m/s^2

from newton second law

Net force = ma

f_{net} = ma

F - f = ma

25 - \mu_kN = ma

25 - \mu_kmg = ma

\frac {25 - ma}{mg} =\mu_k

\frac {25 - 16*0.058}{16*9.81} = 0.15

\mu_k = 0.15

4 0
3 years ago
A force Ě = F, î + Fy h acts on a particle
AVprozaik [17]

Answer:

Pt 1: W=39J

Pt 2: θ = 19.7

Explanation:

Part 1:

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W=(10i-1j)(4i+3j)

W=(10*4)-(1*1)

W=40-1

W=39J

Part 2:

|F|=\sqrt{10^2+(-1)^2}

|F|=\sqrt{101}

|S|=\sqrt{4^2+1^2}

|S|=\sqrt{17}

W=|F||S|cosθ

39=(\sqrt{101})(\sqrt{17} )cosθ

θ = 19.7

Hope it helps

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A girl sits on a tire swing and is pushed in a circular motion by her father. Her tangential speed is 2.8 m/s and the girl trave
Jet001 [13]
Formula: Ca=Vt^2/r

centripetal acceleration(Ca)= ?
tangential speed(Vt)= 2.8m/s
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Substitute: Ca=2.8^2/2
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3 0
3 years ago
Read 2 more answers
The eagle drops the trout a height of 6.1 m the fish travels 7.9 m horizontaly before hitting the water what is the velocity of
Semenov [28]

The velocity of the eagle is 7.0 m/s

Explanation:

The motion of the fish is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction, where the horizontal velocity of the fish is equal to the horizontal velocity of the eagle

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

We start by analyzing the vertical motion, using the following suvat equation:

s=ut+\frac{1}{2}at^2

where:

s = 6.1 m is the vertical displacement of the fish

u = 0 is the initial vertical velocity of the fish

a=g=9.8 m/s^2 is the acceleration of gravity

t is the time

Solving for t, we find the time of flight of the fish:

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(6.1)}{9.8}}=1.12 s

Now we know that during this time, the fish travels a horizontal distance of

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Therefore, the horizontal velocity of the fish is

v_x = \frac{x}{t}=\frac{7.9}{1.12}=7.0 m/s

And therefore, this is the initial velocity of the eagle.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
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A strong lightning bolt transfers an electric charge of about 16 C to Earth (or vice versa). How many electrons are transferred?
zzz [600]

Answer:

Number of electrons, n=9.98\times 10^{19}

Explanation:

A strong lightning bolt transfers an electric charge of about 16 C to Earth, q = 16 C

We need to find the number of electrons that transferred. Let there are n electrons transferred. It is given by using quantization of electric charge as :

q = ne

n=\dfrac{q}{e}

e is elemental charge

n=\dfrac{16}{1.602\times 10^{-19}}

n=9.98\times 10^{19}

So, there are 9.98\times 10^{19} electrons that gets transferred. Hence, this is the required solution.

3 0
3 years ago
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