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Oduvanchick [21]
2 years 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]2 years 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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Sketch the circuit labeling the meter and bulb as two separate resistors connected in parallel to the voltage source. Then show
Ksenya-84 [330]

Answer:

Show attached picture

Explanation:

Let's call V the voltage provided by the battery in the circuit. M is the multimeter (let's call R_M its internal resistance) and R indicates the resistance of the light bulb.

We know that the meter's internal resistance is 1000 times higher than the bulb's resistance:

R_M = 1000 R (1)

Both  the meter and the bulb are connected in parallel to the battery, so they both have same potential difference at their terminals:

V_M = V_R

Using Ohm's law, V=RI, we can rewrite the previous equation as:

R_M I_M = R I_R

where

I_M is the current in the meter

I_R is the current in the bulb

Using (1), this equation becomes

(1000 R) I_M = R I_R \rightarrow I_M = \frac{I_R}{1000}

so, the current in the meter is 1000 times less than through the bulb.

5 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
Sholpan [36]

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

5 0
2 years ago
Two disks with the same rotational inertia i are spinning about the same frictionless shaft, with the same angular speed ω, but
valentina_108 [34]

Answer:

3. none of these

Explanation:

The rotational kinetic energy of an object is given by:

K=\frac{1}{2}I \omega^2

where

I is the moment of inertia

\omega is the angular speed

In this problem, we have two objects rotating, so the total rotational kinetic energy will be the sum of the rotational energies of each object.

For disk 1:

K_1 = \frac{1}{2}I (\omega)^2 = \frac{1}{2}I\omega^2

For disk 2:

K_2 = \frac{1}{2}I(-\omega)^2 = \frac{1}{2}I\omega^2

so the total energy is

K=K_1 + K_2 = \frac{1}{2}I\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2

So, none of the options is correct.

5 0
2 years ago
When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What
ser-zykov [4K]

Answer:

Electric field, E=4\times 10^5\ N/C

Explanation:

It is given that,

Magnitude of charge, q_o=2\ nC=2\times 10^{-9}\ C

Force experienced, F=8\times 10^{-4}\ N

We need to find the electric field at the origin. It is given by :

F=q_o\times E

E=\dfrac{F}{q_o}

E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}

E=4\times 10^5\ N/C

So, the electric field at the origin is 4\times 10^5\ N/C. Hence, this is the required solution.

3 0
2 years ago
A jogger runs 10.0 blocks do east, 5.0 blocks due South, and another two. Zero blocks do east. Assume all blocks are equal size,
Annette [7]

Jogger moves in three displacements

d1 = 10 blocks East

d2 = 5 blocks South

d3 = 2 blocks East

now we can say

total displacement towards East direction will be

d_x = 10 + 2= 12 blocks

Total displacement towards South

d_y = 5 block

now to find the net displacement we can use vector addition

d = \sqrt{d_x^2 + d_y^2}

d = \sqrt{12^2 + 5^2}

d = 13 blocks

<em>so magnitude of net displacement will be equal to 13 blocks</em>

6 0
2 years ago
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