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olga2289 [7]
2 years ago
13

You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 1515 1515 lb and was

traveling eastward. Car B weighs 1125 1125 lb and was traveling westward at 44.0 44.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 20.5 20.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750 0.750 . How fast (in miles per hour) was car A traveling just before the collision
Physics
2 answers:
krok68 [10]2 years ago
7 0

Answer:

72.8mph

Explanation:

east is positive

1515 E -1125(45) = 2640 V

F = 2640*.75 = -1980 pounds

so

-1980 pounds = m a

but m = 2640/g where g = 32.2 ft/s^2

so

a = -(1980/2640)g = -.75 g = -24.15 ft/s^2

= dv/dt

0 = V-24.15 t so V =24.15 t

22.5 = V t -12.07 t^2

22.5 = 24.15 t^2 - 12.07 t^2 = 12.07 t^2

t = 1.37 seconds

so

V = 24.15 * 1.37 seconds = 33.1 ft/s

convert that to mph

33.1 ft/s = 22.6 mph

now back to momentum

1515 E -1125(45) = 2640 V

1515 E = 50625+59664

E = 72.8 mph

expeople1 [14]2 years ago
3 0

Answer:

Speed = 72.76mph

Explanation:

east direction signifies +

Therefore,

1515E -1125(45) = -2640

F = -2640×0.75 = -1980lbs

so,

-1980lbs = Ma

but m = 2640×g where g = 32.2 ft/s²

so,

a = -(1980/2640g) = -0.75×g = -24.15ft/s² = dv/dt

0 = V-24.15t

V =24.15t

22.5 = Vt -12.07t²

22.5 = 24.15t² - 12.07t² = 12.07t²

t = 1.37 seconds

Therefore,

V = 24.15 × 1.37 seconds = 33.1 ft/s

converting to mph gives

33.1 ft/s = 22.6 mph

now back to momentum

1515 E -1125(45) = 2640 V

1515 E = 50625+59664

Speed = 72.76 mph

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A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta
BartSMP [9]

Answer:

35%

Explanation:

Assuming no external torques present during the collision between the record and the turntable, total angular momentum must be conserved.

For a rotating body with some angular velocity and moment of inertia, the angular momentum can be expressed as follows;

L = I* ω

So, as initial angular momentum and final angular momentum must be the same, we have:

Li = Lf

⇒ I₁ * ω₁ = I₂ * ω₂ (1)

where I₁ is the rotational inertia of the turntable, and I₂, is the combined rotational inertia of the turntable and the record:

I₂ = I₁ + 0.54 I₁ = 1.54 I₁

We can solve (1) for the new common angular speed, as follows:

ω₂ = ω₁ / 1.54 (2)

The initial rotational kinetic energy is given by definition for the following equation:

Kroti = 1/2 * I₁ * ω₁² (3)

The final rotational kinetic  energy takes into account the new rotational inertia and the common final angular speed:

Krotf = 1/2* I₂ * ω₂² = 1/2* 1.54 I₁* (ω₁/1.54)² (4)

Dividing both sides in (3) and (4), we get:

Krotf/Kroti = 1/1.54 = 0.65

This means that the final rotational kinetic energy, has reduced to 0.65 of the initial value, or that has lost 35% of the initial kinetic energy.

8 0
2 years ago
You need to design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a
Svetllana [295]

Answer:

Hello your question has some missing parts and the required diagram attached below is the missing part and the diagram

Digital circuits require actions to take place at precise times, so they are controlled by a clock that generates a steady sequence of rectangular voltage pulses. One of the most widely

used integrated circuits for creating clock pulses is called a 555 timer.  shows how the timer’s output pulses, oscillating between 0 V and 5 V, are controlled with two resistors and a capacitor. The circuit manufacturer tells users that TH, the time the clock output spends in the high (5V) state, is TH =(R1 + R2)*C*ln(2). Similarly, the time spent in the low (0 V) state is TL = R2*C*ln(2). Design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2?

ANSWER : R1 = 144.3Ω,   R2 =  72.2 Ω

Explanation:

Frequency = 10 MHz

Time period = 1 / F =  0.1 <em>u </em>s

Duty cycle = 75% = 0.75

Duty cycle can be represented as :   Ton / T

Also: Ton = Th = 0.75 * 0.1 <em>u </em>s  = 75 <em>n</em> s

TL = T - Th = 100 <em>n</em>s - 75 <em>n</em> s = 25 <em>n</em> s

To find the value of R2 we use the equation for  time spent in the low (0 V) state

TL = R2*C*ln(2)

hence R2 = TL / ( C * In 2 )

c = 500 pF

Hence R2 = 25 / ( 500 pF * 0.693 )  = 72.2 Ω

To find the value of R1 we use the equation for the time the clock output spends in the high (5V) state,

Th = (R1 + R2)*C*ln(2)

  from the equation make R1 the subject of the formula

R1 =  (Th - ( R2 * C * In2 )) / (C * In 2)

R1 = ( 75 ns - ( 72.2 * 500 pF * 0.693)) / ( 500 pF * 0.693 )

R1 = ( 75 ns  - ( 25 ns ) / 500 pf * 0.693

     = 144.3Ω

8 0
2 years ago
A BMX bicycle rider takes off from a ramp at a point 2.4 m above the ground. The ramp is angled at 40 degrees from the horizonta
adoni [48]

Answer:

The BMX lands 5.4 m from the end of the ramp.

Explanation:

Hi there!

The position of the BMX is given by the position vector "r":

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t

x0 = initial horizontal position

v0 = initial velocity

α = jumping angle

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

Please, see the attached graphic for a better understanding of the situation. At final time, when the bicycle reaches the ground, the vector position will be "r final" (see figure). The y-component of the vector "r final" is - 2.4 m (placing the origin of the frame of reference at the jumping point). With that information, we can use the equation of the y-component of the vector "r" (see above) to calculate the time of flight. With that time, we can then obtain the x-component (rx in the figure) of the vector "r final". Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

-2.4 m = 0 m + 5.9 m/s · t · sin 40° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 5.9 m/s · t · sin 40° + 2.4 m

Solving the quadratic equation:

t = 1.2 s

Now, we can calculate the x-component of the vector "r final" that is the horizontal distance traveled by the bicycle:

x = x0 + v0 · t · cos α

x = 0 m + 5.9 m/s · 1.2 s · cos 40°

x = 5.4 m

The BMX lands 5.4 m from the end of the ramp.

Have a nice day!

8 0
2 years ago
A wire along the z axis carries a current of 4.9 A in the z direction Find the magnitude and direction of the force exerted on a
AURORKA [14]

Answer:

<h2>0.069 N, in the X direction</h2>

Explanation:

According to Flemming's left hand rule, it sates that if the first three fingers of the left hand are held mutually at right angles to one another, the fore finger will point in the direction of magnetic field, the middle finger will point in direction of current, while the thumb will point to the direction of force.

Mathematically the law is stated as

F= BIL

given data

Magnetic field B=  0.43T

Current I= 4.9 A

length of conductor L= 3.3cm to meter , 3.3/100=  0.033 m

Applying the formula the force is calculated as

F= 0.43*4.9* 0.033= 0.069 N

According to Flemming's rule the direction of all parameters are mutually perpendicular to one another, then the Force is in the X direction

5 0
2 years ago
An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part
guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

E=\dfrac{kq}{r^2}

k is electrostatic constant

if r = 2 m, electric field is given by :

E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)

If r = 1 m, electric field is given by :

E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)

Dividing equation (1) and (2) we get :

\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0
2 years ago
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