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inn [45]
3 years ago
11

A 1.2 x10 3 kilogram automobile in motion strikes a 1.0 x 10 -4 kilogram insect as a result the insect is accelerated at a rate

of 1.0 x 10 2 meters per second 2 what is the magnitude of the force the insect exerts on the car
Physics
1 answer:
Mariana [72]3 years ago
6 0
According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
F=ma
where F is the magnitude of the force, m is the mass of the object and a its acceleration.

In this problem, the object is the insect, with mass m=1.0 \cdot 10^{-4} kg. The acceleration of the insect is a=1.0 \cdot 10^2 m/s^2, therefore we can calculate the force exerted by the car on the insect:
F=ma=(1.0 \cdot 10^{-4} kg)(1.0 \cdot 10^2 m/s^2)=0.01 N

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.
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You are the pilot of a spaceship (9.1835 × 104 kg) and you enter into elliptical orbit around an uncharted planet, Rosseforp. At
Kaylis [27]

Answer:

Check the explanation

Explanation:

The orbital period of a satellite that is given as (T) and the mean distance from the central body (R) are connected by the following equation: representing T as the satellite period, R will be represented as the average radius of orbit for the satellite (which is the distance from center of central planet)

Kindly check the attached image below to see the step by step explanation to the above question.

7 0
3 years ago
A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is
gregori [183]

Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is d = 0.313 \ m

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is \theta = 25 ^o

         sin \theta  = \frac{h}{d}

Where h = h_b - h_a

     So      d sin \theta  = h_b - h_a

From the question we are told that

      The mass of the block is  m = 20 \ kg

       The mass of the pendulum is  m_p = 2 \ kg

       The velocity of the pendulum at the bottom of swing is v_p = 15 m/s

        The coefficient of restitution is  e =0.7

         The coefficient of kinetic friction is  \mu _k = 0.5

The velocity of the block after the impact is mathematically represented as

            v_2 f = \frac{m_b - em_p}{m_b + m_p}  * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p

Where  v_2 i is the velocity of the block  before collision which is  0

                  = \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20}   * 15

Substituting value

                   v_2 f = 2.310\  m/s

According to conservation of energy principle

      The energy at point a  =  energy at point b

So    PE_A + KE _A = PE_B + KE_B  +  E_F

Where  

         PE_A is the potential energy at A which is mathematically represented as

          PE_A = m_b gh_a = 0 at the bottom

      KE _A is the kinetic energy at A  which is mathematically represented as

               K_A = \frac{1}{2} m_b * v_2f^2                  

         PE_B is the potential energy at B which is mathematically represented as  

            PE_B = m_b gh

From the diagram h = h_b -h_a

       PE_B = m_b g(h_b - h_a)

KE _B is the kinetic energy at B  which is 0 (at the top )

Where is E_F is the workdone against velocity  which from the diagram is

      \mu_k m_b g cos 25 *d

So

   \frac{1}{2} m_b v_2 f^2  = m_b g h_b + \mu_k m_b g cos \25 * d

Substituting values

   \frac{1}{2}  * 20 * 2.310^2 = 20 * 9.8 * d sin(25)  + 0.5* 20 * 9.8 * cos 25 * d    

So

       d = 0.313 \ m

       

   

6 0
3 years ago
Jayden was given a marshmallow and a syringe in class to experiment with. She placed the marshmallow in the syringe and sealed t
Tju [1.3M]
<h2>Answer:</h2>

The correct option is A.

A) The increased pressure, pushed the molecules closer together, and caused the marshmallow to shrink.

<h2>Explanation:</h2>

Jayden experimented, she placed the marshmallow in the syringe and sealed the end. When she depressed the plunger of the syringe, the pressure increased and pushed the molecules closer together and causes the marshmallow to shrink.

<h2 />
3 0
3 years ago
A 120-kg roller coaster cart is being tested on a new track, and a crash-test dummy is loaded into it. The roller coaster starts
avanturin [10]

Answer:

vb = 22.13 m/s

So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.

Explanation:

In order to find the speed of roller coaster at Point B, we will use the law of conservation of Energy. In this situation, the law of conservation of energy states that:

K.E at A + P.E at A = K.E at B + P.E at B

(1/2)mvₐ² + mghₐ = (1/2)m(vb)² + mg(hb)

(1/2)vₙ² + ghₐ = (1/2)(vb)² + g(hb)

where,

vₙ = velocity of roller coaster at point a = 0 m/s

hₙ = height of roller coaster at point a = 25 m

g = 9.8 m/s²

vb = velocity of roller coaster at point B = ?

hb = Height of Point B = 0 m (since, point is the reference point)

Therefore,

(1/2)(0 m/s)² + (9.8 m/s²)(25 m) = (1/2)(vb)² + (9.8 m/s²)(0 m)

245 m²/s² * 2 = vb²

vb = √(490 m²/s²)

<u>vb = 22.13 m/s</u>

<u>So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.</u>

5 0
2 years ago
Which energy transfer occurs when a cube of ice is placed in a glass of<br> water?
Daniel [21]
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5 0
2 years ago
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