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Oduvanchick [21]
1 year ago
5

A truck traveling down the highway collides with a slower moving mosquito traveling in the same direction. Which of the followin

g statements is true during the collision? A. the truck exerts a larger force on the mosquito exerts on the truck. B. The truck and mosquito exert the same size force on each other. C. The mosquito exerts a larger force on the truck than the truck exerts on the mosquito. What law of physics did you use in determining your answer?
Physics
1 answer:
Ipatiy [6.2K]1 year ago
6 0

Answer:

B. The truck and mosquito exert the same size force on each other.

Explanation:

Newton's third law (law of action-reaction) states that

"When an object A exerts a force (action) on an object B, then object B exerts an equal and opposite force (reaction) on object A"

In this case, we can call

object A = the truck

object B = the mosquito

Thereforce according to Newton's third law, the force exerted by the truck on the mosquito is equal in magnitude to the force exerted by the mosquito on the truck (and in opposite direction).

The reason for which the mosquito will experience much more damage is the fact that the mosquito's mass is much smaller than the truck's mass, and since the acceleration is inversely proportional to the mass:

a=\frac{F}{m}

the mosquito will experience a much larger deceleration than the truck, therefore much more damage.

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Answer:

Position xf is farther away from the sensor than x0, and ax is negative

Explanation:

                     Area of trapezoidal are=

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                            =\frac{1}{2} *(2.25-1.75*0.5)

                            =0.6875 m

As the area is positive therefore displacement from xo is positive

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3 0
2 years ago
: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop
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Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

  1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
  2. On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒Fa*sin(50)-Fb*sin(20)=0

⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950  

⇒Fa*cos(20)+Fb*cos(50)=950

⇒2.24*Fb*cos(20)+Fb(50)=950

⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

⇒Fb=\frac{950}{2.24*0.94+0.64}

⇒ Fb=\frac{950}{2.75}=345.5

⇒Fa=2.24*Fb

      =2.24*345.5

      =773.93

Therefore approximately, Fa=774 N and Fb=346 N

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Answer:  Sean is standing still, and Rhea is running toward Sean while   kicking the ball

Explanation: Your welcome :)

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Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m      

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(15)^2}{22.5}

a=10\ m/s^2

We know that, g=9.8\ m/s^2

a=\dfrac{10\times g}{9.8}

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Hence, this is the required solution.                                              

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