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vovikov84 [41]
3 years ago
14

A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo

cused on the man. At what rate is the searchlight rotating when the man is 8 ft from the point on the path closest to the searchlight
Physics
1 answer:
BARSIC [14]3 years ago
4 0

We are given that,

\frac{dx}{dt} = 4ft/s

We need to find \frac{d\theta}{dt} when x=8ft

The equation that relates x and \theta can be written as,

\frac{x}{6} tan\theta

x = 6tan\theta

Differentiating each side with respect to t, we get,

\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}

\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}

\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}

Replacing the value of the velocity

\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2

\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta

The value of cos \theta could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

cos\theta = \frac{6}{10}

\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2

\frac{d\theta}{dt} = \frac{24}{25}

Search light is rotating at a rate of 0.96rad/s

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A 1.2 kg ball drops vertically onto a floor from a height of 32 m, and rebounds with an initial speed of 10 m/s.
iren [92.7K]

Explanation:

Given that,

Mass of the ball, m = 1.2 kg

Initial speed of the ball, u = 10 m/s

Height of the floor from ground, h = 32 m

(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :

\dfrac{1}{2}mv^2=mgh

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 32}

v = -25.04 m/s (negative as it rebounds)

The impulse acting on the ball is equal to the change in momentum. It can be calculated as :

J=m(v-u)

J=1.2\times (-25.04-10)

J = -42.048 kg-m/s

(b) Time of contact, t = 0.02 s

Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :

J=\dfrac{F}{t}

F=J\times t

F=42.048\times 0.02

F = 0.8409 N

Hence, this is the required solution.

5 0
3 years ago
A contestants spin a wheel when it is their turn in a game show. One contestant gives the wheel an initial angular speed of 3.40
guajiro [1.7K]

Answer:

4.62 s

Explanation:

We are given that

Initial angular speed,\omega=3.4 rad/s

\theta=1\frac{1}{4} rev=\frac{5}{4}\times 2\pi=2.5\pi rad

\omega'=0

\omega'^2-\omega^2=2\alpha \theta

Substitute the values

0-(3.4)^2=2\times 2.5\pi \alpha

\alpha=\frac{-(3.4)^2}{2\times 2.5\pi}=-0.736 rad/s^2

\omega'=\omega+\alpha t

0=3.4-0.736 t

-0.736t=-3.4

t=\frac{-3.4}{-0.736}=4.62 s

Hence, the wheel takes 4.62 s to come to rest.

3 0
3 years ago
2. Two toy cars are involved in a race. Car A has mass m while car B has mass 2m. a. The two cars have the same force applied to
ArbitrLikvidat [17]

Answer:

a) The kinetic energy of the two cars is the same

the moment of car 2 is greater than the moment of car 1

b)  the kinetic energy of car 1 is greater than that of car 2

the moment of the two cars is the same

Explanation:

a) to know the kinetic energy of each car, we must find the speed, use Newton's second law to find the acceleration

Car 1

     F = m a

    a = F / m

Let's use kinematics to find the velocity after x = 1 m

       v² = v₀² + 2 a x

The initial speed is zero

       v = √ (2 F/m  x)

For the distance of x = 1 m

        v₁ = √ (2 F / m)

Car 2

      F = 2m a

      a = F / 2m

      v² = 2 a x

      v = √ (F/m  x)

 For x = 1 m

       v₂ = √(F / m)

Let's calculate the kinetic energy of each car

Car 1

      K₁ = ½ m v₁²

      K₁ = ½ m 2F / m

      K₁ = F

Car 2

      K₂ = ½ 2m v₂²

      K₂ = ½ 2m F / m

      K₂ = F

The kinetic energy of the two cars is the same

Let's calculate the moment

Car 1

   P₁ = m v₁

   P₁ = m √ (2F / m)

Car 2

    P₂ = 2m v²

    P₂ = 2m √(F / m)

We see that the moment of car 2 is greater than the moment of car 1

b) in this part the force is applied by t = 10 s

Acceleration is the same, let's find the speed

Car1

          v = v₀ + a t

          v = F / m t

          v₁ = F / m 10

Car 2

           v₂ = F / 2m 10

           v₂ = F / m 5

Let's calculate the kinetic energy of each car

Car 1

           K₁ = ½ m v₁²

           K₁ = ½ m (F / m 10)²

           K₁ = 50 F² / m

Car2

         K₂ = ½ 2m v₂²

         K₂ = m (F / m 5)²

         K₂ = 25 F² / m

In this case we see that the kinetic energy of car 1 is greater than that of car 2

Let's calculate the moment

Car 1

         P₁ = m v₁

         P₁ = m F / m 10

         P₁ = 10 F

 

Car 2

        P₂ = 2m v₂

        P₂ = 2m F / m 5

        P₂ = 10 F

In this case the moment of the two cars is the same

7 0
3 years ago
You throw a ball downward from a window at a speed of 2.0m/s. How far has it fallen in 3 seconds?
bekas [8.4K]
Let's just assume that you throw the ball with an initial speed of 2 m/s instead of dropping it like free falling.
a=9.81 m/s^2
Vi= 2 m/s
t= 3 x

we use the formula
d = (Vi)(t) + (1/2)(a)(t)^2
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Please find attached photograph for your answer. Hope it helps! Please do comment

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